Title: Vector Span

Q: Created by: Dr. Pang Chung Yin Amy, Dr. Chiang Shun Ling, and Naw Sai.
Department of Mathematics,
Hong Kong Baptist University (HKBU) © 2025.
A: These decks may be freely shared and distributed among students and educators. We hope they help in your linear algebra journey!

Q: Is $\begin{bmatrix} 7 \\ 6 \end{bmatrix}$ a linear combination of $\begin{bmatrix} 3 \\ 3 \end{bmatrix}$ and $\begin{bmatrix} -1 \\ 0 \end{bmatrix}$?
A: $\begin{bmatrix} 7 \\ 6 \end{bmatrix}$ is a linear combination of $\begin{bmatrix} 3 \\ 3 \end{bmatrix}$ and $\begin{bmatrix} -1 \\ 0 \end{bmatrix}$ if $\begin{bmatrix} 7 \\ 6 \end{bmatrix} = c_{1}\begin{bmatrix} 3 \\ 3 \end{bmatrix} + c_{2}\begin{bmatrix} -1 \\ 0 \end{bmatrix}$ for some scalars $c_1, c_2$.
This is the linear system $\left[\begin{array}{cc|c} 3 & -1 & 7 \\ 3 & 0 & 6 \end{array}\right]$, which row reduces to $\left[\begin{array}{cc|c} 1 & 0 & 2 \\ 0 & 1 & -1 \end{array}\right]$.
$c_{1} = 2$, $c_{2} = -1$ is a solution.
So, $\begin{bmatrix} 7 \\ 6 \end{bmatrix} = 2\begin{bmatrix} 3 \\ 3 \end{bmatrix} -\begin{bmatrix} -1 \\ 0 \end{bmatrix}$.

Q: Is $\begin{bmatrix} -3 \\ 2 \end{bmatrix}$ a linear combination of $\begin{bmatrix} -3 \\ -2 \end{bmatrix}$ and $\begin{bmatrix} 0 \\ -2 \end{bmatrix}$?
A: $\begin{bmatrix} -3 \\ 2 \end{bmatrix}$ is a linear combination of $\begin{bmatrix} -3 \\ -2 \end{bmatrix}$ and $\begin{bmatrix} 0 \\ -2 \end{bmatrix}$ if $\begin{bmatrix} -3 \\ 2 \end{bmatrix} = c_{1}\begin{bmatrix} -3 \\ -2 \end{bmatrix} + c_{2}\begin{bmatrix} 0 \\ -2 \end{bmatrix}$ for some scalars $c_1, c_2$.
This is the linear system $\left[\begin{array}{cc|c} -3 & 0 & -3 \\ -2 & -2 & 2 \end{array}\right]$, which row reduces to $\left[\begin{array}{cc|c} 1 & 0 & 1 \\ 0 & 1 & -2 \end{array}\right]$.
$c_{1} = 1$, $c_{2} = -2$ is a solution.
So, $\begin{bmatrix} -3 \\ 2 \end{bmatrix} = \begin{bmatrix} -3 \\ -2 \end{bmatrix} -2\begin{bmatrix} 0 \\ -2 \end{bmatrix}$.

Q: Is $\begin{bmatrix} 1 \\ -4 \end{bmatrix}$ a linear combination of $\begin{bmatrix} 2 \\ 3 \end{bmatrix}$ and $\begin{bmatrix} 3 \\ -1 \end{bmatrix}$?
A: $\begin{bmatrix} 1 \\ -4 \end{bmatrix}$ is a linear combination of $\begin{bmatrix} 2 \\ 3 \end{bmatrix}$ and $\begin{bmatrix} 3 \\ -1 \end{bmatrix}$ if $\begin{bmatrix} 1 \\ -4 \end{bmatrix} = c_{1}\begin{bmatrix} 2 \\ 3 \end{bmatrix} + c_{2}\begin{bmatrix} 3 \\ -1 \end{bmatrix}$ for some scalars $c_1, c_2$.
This is the linear system $\left[\begin{array}{cc|c} 2 & 3 & 1 \\ 3 & -1 & -4 \end{array}\right]$, which row reduces to $\left[\begin{array}{cc|c} 1 & 0 & -1 \\ 0 & 1 & 1 \end{array}\right]$.
$c_{1} = -1$, $c_{2} = 1$ is a solution.
So, $\begin{bmatrix} 1 \\ -4 \end{bmatrix} = -\begin{bmatrix} 2 \\ 3 \end{bmatrix} \begin{bmatrix} 3 \\ -1 \end{bmatrix}$.

Q: Is $\begin{bmatrix} 9 \\ 5 \end{bmatrix}$ a linear combination of $\begin{bmatrix} 3 \\ 2 \end{bmatrix}$ and $\begin{bmatrix} 3 \\ 1 \end{bmatrix}$?
A: $\begin{bmatrix} 9 \\ 5 \end{bmatrix}$ is a linear combination of $\begin{bmatrix} 3 \\ 2 \end{bmatrix}$ and $\begin{bmatrix} 3 \\ 1 \end{bmatrix}$ if $\begin{bmatrix} 9 \\ 5 \end{bmatrix} = c_{1}\begin{bmatrix} 3 \\ 2 \end{bmatrix} + c_{2}\begin{bmatrix} 3 \\ 1 \end{bmatrix}$ for some scalars $c_1, c_2$.
This is the linear system $\left[\begin{array}{cc|c} 3 & 3 & 9 \\ 2 & 1 & 5 \end{array}\right]$, which row reduces to $\left[\begin{array}{cc|c} 1 & 0 & 2 \\ 0 & 1 & 1 \end{array}\right]$.
$c_{1} = 2$, $c_{2} = 1$ is a solution.
So, $\begin{bmatrix} 9 \\ 5 \end{bmatrix} = 2\begin{bmatrix} 3 \\ 2 \end{bmatrix} \begin{bmatrix} 3 \\ 1 \end{bmatrix}$.

Q: Is $\begin{bmatrix} -1 \\ -5 \end{bmatrix}$ a linear combination of $\begin{bmatrix} 1 \\ 2 \end{bmatrix}$ and $\begin{bmatrix} 0 \\ 3 \end{bmatrix}$?
A: $\begin{bmatrix} -1 \\ -5 \end{bmatrix}$ is a linear combination of $\begin{bmatrix} 1 \\ 2 \end{bmatrix}$ and $\begin{bmatrix} 0 \\ 3 \end{bmatrix}$ if $\begin{bmatrix} -1 \\ -5 \end{bmatrix} = c_{1}\begin{bmatrix} 1 \\ 2 \end{bmatrix} + c_{2}\begin{bmatrix} 0 \\ 3 \end{bmatrix}$ for some scalars $c_1, c_2$.
This is the linear system $\left[\begin{array}{cc|c} 1 & 0 & -1 \\ 2 & 3 & -5 \end{array}\right]$, which row reduces to $\left[\begin{array}{cc|c} 1 & 0 & -1 \\ 0 & 1 & -1 \end{array}\right]$.
$c_{1} = -1$, $c_{2} = -1$ is a solution.
So, $\begin{bmatrix} -1 \\ -5 \end{bmatrix} = -\begin{bmatrix} 1 \\ 2 \end{bmatrix} -\begin{bmatrix} 0 \\ 3 \end{bmatrix}$.

Q: Is $\begin{bmatrix} 0 \\ 2 \end{bmatrix}$ a linear combination of $\begin{bmatrix} 0 \\ 2 \end{bmatrix}$ and $\begin{bmatrix} 0 \\ 1 \end{bmatrix}$?
A: $\begin{bmatrix} 0 \\ 2 \end{bmatrix}$ is a linear combination of $\begin{bmatrix} 0 \\ 2 \end{bmatrix}$ and $\begin{bmatrix} 0 \\ 1 \end{bmatrix}$ if $\begin{bmatrix} 0 \\ 2 \end{bmatrix} = c_{1}\begin{bmatrix} 0 \\ 2 \end{bmatrix} + c_{2}\begin{bmatrix} 0 \\ 1 \end{bmatrix}$ for some scalars $c_1, c_2$.
This is the linear system $\left[\begin{array}{cc|c} 0 & 0 & 0 \\ 2 & 1 & 2 \end{array}\right]$, which row reduces to $\left[\begin{array}{cc|c} 1 & 0.50 & 1 \\ 0 & 0 & 0 \end{array}\right]$.
$c_{1} = 1$, $c_{2} = 0$ is a solution.
So, $\begin{bmatrix} 0 \\ 2 \end{bmatrix} = \begin{bmatrix} 0 \\ 2 \end{bmatrix}$.

Q: Is the vector $\begin{bmatrix} -2 \\ -1 \end{bmatrix}$ in the span of $\left\{ \begin{bmatrix} 1 \\ -3 \end{bmatrix}, \begin{bmatrix} -3 \\ 2 \end{bmatrix} \right\}$?
A: $\begin{bmatrix} -2 \\ -1 \end{bmatrix}$ is in the span of $\begin{bmatrix} 1 \\ -3 \end{bmatrix}$ and $\begin{bmatrix} -3 \\ 2 \end{bmatrix}$ if $\begin{bmatrix} -2 \\ -1 \end{bmatrix} = c_{1}\begin{bmatrix} 1 \\ -3 \end{bmatrix} + c_{2}\begin{bmatrix} -3 \\ 2 \end{bmatrix}$ for some scalars $c_1, c_2$.
This is the linear system $\left[\begin{array}{cc|c} 1 & -3 & -2 \\ -3 & 2 & -1 \end{array}\right]$, which row reduces to $\left[\begin{array}{cc|c} 1 & 0 & 1 \\ 0 & 1 & 1 \end{array}\right]$.
$c_{1} = 1$, $c_{2} = 1$ is a solution.
So, $\begin{bmatrix} -2 \\ -1 \end{bmatrix} = \begin{bmatrix} 1 \\ -3 \end{bmatrix} \begin{bmatrix} -3 \\ 2 \end{bmatrix}$.

Q: Is $\begin{bmatrix} 2 \\ -2 \end{bmatrix}$ a linear combination of $\begin{bmatrix} 1 \\ -3 \end{bmatrix}$ and $\begin{bmatrix} 0 \\ 2 \end{bmatrix}$?
A: $\begin{bmatrix} 2 \\ -2 \end{bmatrix}$ is a linear combination of $\begin{bmatrix} 1 \\ -3 \end{bmatrix}$ and $\begin{bmatrix} 0 \\ 2 \end{bmatrix}$ if $\begin{bmatrix} 2 \\ -2 \end{bmatrix} = c_{1}\begin{bmatrix} 1 \\ -3 \end{bmatrix} + c_{2}\begin{bmatrix} 0 \\ 2 \end{bmatrix}$ for some scalars $c_1, c_2$.
This is the linear system $\left[\begin{array}{cc|c} 1 & 0 & 2 \\ -3 & 2 & -2 \end{array}\right]$, which row reduces to $\left[\begin{array}{cc|c} 1 & 0 & 2 \\ 0 & 1 & 2 \end{array}\right]$.
$c_{1} = 2$, $c_{2} = 2$ is a solution.
So, $\begin{bmatrix} 2 \\ -2 \end{bmatrix} = 2\begin{bmatrix} 1 \\ -3 \end{bmatrix} + 2\begin{bmatrix} 0 \\ 2 \end{bmatrix}$.

Q: Is $\begin{bmatrix} 6 \\ 2 \end{bmatrix}$ a linear combination of $\begin{bmatrix} 2 \\ 2 \end{bmatrix}$ and $\begin{bmatrix} -2 \\ 0 \end{bmatrix}$?
A: $\begin{bmatrix} 6 \\ 2 \end{bmatrix}$ is a linear combination of $\begin{bmatrix} 2 \\ 2 \end{bmatrix}$ and $\begin{bmatrix} -2 \\ 0 \end{bmatrix}$ if $\begin{bmatrix} 6 \\ 2 \end{bmatrix} = c_{1}\begin{bmatrix} 2 \\ 2 \end{bmatrix} + c_{2}\begin{bmatrix} -2 \\ 0 \end{bmatrix}$ for some scalars $c_1, c_2$.
This is the linear system $\left[\begin{array}{cc|c} 2 & -2 & 6 \\ 2 & 0 & 2 \end{array}\right]$, which row reduces to $\left[\begin{array}{cc|c} 1 & 0 & 1 \\ 0 & 1 & -2 \end{array}\right]$.
$c_{1} = 1$, $c_{2} = -2$ is a solution.
So, $\begin{bmatrix} 6 \\ 2 \end{bmatrix} = \begin{bmatrix} 2 \\ 2 \end{bmatrix} -2\begin{bmatrix} -2 \\ 0 \end{bmatrix}$.

Q: Is $\begin{bmatrix} -4 \\ -3 \end{bmatrix}$ a linear combination of $\begin{bmatrix} 3 \\ 3 \end{bmatrix}$ and $\begin{bmatrix} -1 \\ 0 \end{bmatrix}$?
A: $\begin{bmatrix} -4 \\ -3 \end{bmatrix}$ is a linear combination of $\begin{bmatrix} 3 \\ 3 \end{bmatrix}$ and $\begin{bmatrix} -1 \\ 0 \end{bmatrix}$ if $\begin{bmatrix} -4 \\ -3 \end{bmatrix} = c_{1}\begin{bmatrix} 3 \\ 3 \end{bmatrix} + c_{2}\begin{bmatrix} -1 \\ 0 \end{bmatrix}$ for some scalars $c_1, c_2$.
This is the linear system $\left[\begin{array}{cc|c} 3 & -1 & -4 \\ 3 & 0 & -3 \end{array}\right]$, which row reduces to $\left[\begin{array}{cc|c} 1 & 0 & -1 \\ 0 & 1 & 1 \end{array}\right]$.
$c_{1} = -1$, $c_{2} = 1$ is a solution.
So, $\begin{bmatrix} -4 \\ -3 \end{bmatrix} = -\begin{bmatrix} 3 \\ 3 \end{bmatrix} \begin{bmatrix} -1 \\ 0 \end{bmatrix}$.

Q: Is the vector $\begin{bmatrix} 5 \\ 2 \end{bmatrix}$ in the span of $\left\{ \begin{bmatrix} -2 \\ -2 \end{bmatrix}, \begin{bmatrix} -1 \\ 2 \end{bmatrix} \right\}$?
A: $\begin{bmatrix} 5 \\ 2 \end{bmatrix}$ is in the span of $\begin{bmatrix} -2 \\ -2 \end{bmatrix}$ and $\begin{bmatrix} -1 \\ 2 \end{bmatrix}$ if $\begin{bmatrix} 5 \\ 2 \end{bmatrix} = c_{1}\begin{bmatrix} -2 \\ -2 \end{bmatrix} + c_{2}\begin{bmatrix} -1 \\ 2 \end{bmatrix}$ for some scalars $c_1, c_2$.
This is the linear system $\left[\begin{array}{cc|c} -2 & -1 & 5 \\ -2 & 2 & 2 \end{array}\right]$, which row reduces to $\left[\begin{array}{cc|c} 1 & 0 & -2 \\ 0 & 1 & -1 \end{array}\right]$.
$c_{1} = -2$, $c_{2} = -1$ is a solution.
So, $\begin{bmatrix} 5 \\ 2 \end{bmatrix} = -2\begin{bmatrix} -2 \\ -2 \end{bmatrix} -\begin{bmatrix} -1 \\ 2 \end{bmatrix}$.

Q: Is $\begin{bmatrix} 1 \\ -4 \end{bmatrix}$ a linear combination of $\begin{bmatrix} -1 \\ -1 \end{bmatrix}$ and $\begin{bmatrix} 2 \\ -3 \end{bmatrix}$?
A: $\begin{bmatrix} 1 \\ -4 \end{bmatrix}$ is a linear combination of $\begin{bmatrix} -1 \\ -1 \end{bmatrix}$ and $\begin{bmatrix} 2 \\ -3 \end{bmatrix}$ if $\begin{bmatrix} 1 \\ -4 \end{bmatrix} = c_{1}\begin{bmatrix} -1 \\ -1 \end{bmatrix} + c_{2}\begin{bmatrix} 2 \\ -3 \end{bmatrix}$ for some scalars $c_1, c_2$.
This is the linear system $\left[\begin{array}{cc|c} -1 & 2 & 1 \\ -1 & -3 & -4 \end{array}\right]$, which row reduces to $\left[\begin{array}{cc|c} 1 & 0 & 1 \\ 0 & 1 & 1 \end{array}\right]$.
$c_{1} = 1$, $c_{2} = 1$ is a solution.
So, $\begin{bmatrix} 1 \\ -4 \end{bmatrix} = \begin{bmatrix} -1 \\ -1 \end{bmatrix} \begin{bmatrix} 2 \\ -3 \end{bmatrix}$.

Q: Is the vector $\begin{bmatrix} -1 \\ 3 \end{bmatrix}$ in the span of $\left\{ \begin{bmatrix} -1 \\ -3 \end{bmatrix}, \begin{bmatrix} 0 \\ 3 \end{bmatrix} \right\}$?
A: $\begin{bmatrix} -1 \\ 3 \end{bmatrix}$ is in the span of $\begin{bmatrix} -1 \\ -3 \end{bmatrix}$ and $\begin{bmatrix} 0 \\ 3 \end{bmatrix}$ if $\begin{bmatrix} -1 \\ 3 \end{bmatrix} = c_{1}\begin{bmatrix} -1 \\ -3 \end{bmatrix} + c_{2}\begin{bmatrix} 0 \\ 3 \end{bmatrix}$ for some scalars $c_1, c_2$.
This is the linear system $\left[\begin{array}{cc|c} -1 & 0 & -1 \\ -3 & 3 & 3 \end{array}\right]$, which row reduces to $\left[\begin{array}{cc|c} 1 & 0 & 1 \\ 0 & 1 & 2 \end{array}\right]$.
$c_{1} = 1$, $c_{2} = 2$ is a solution.
So, $\begin{bmatrix} -1 \\ 3 \end{bmatrix} = \begin{bmatrix} -1 \\ -3 \end{bmatrix} + 2\begin{bmatrix} 0 \\ 3 \end{bmatrix}$.

Q: Is $\begin{bmatrix} -8 \\ -3 \end{bmatrix}$ a linear combination of $\begin{bmatrix} 2 \\ 3 \end{bmatrix}$ and $\begin{bmatrix} 3 \\ 0 \end{bmatrix}$?
A: $\begin{bmatrix} -8 \\ -3 \end{bmatrix}$ is a linear combination of $\begin{bmatrix} 2 \\ 3 \end{bmatrix}$ and $\begin{bmatrix} 3 \\ 0 \end{bmatrix}$ if $\begin{bmatrix} -8 \\ -3 \end{bmatrix} = c_{1}\begin{bmatrix} 2 \\ 3 \end{bmatrix} + c_{2}\begin{bmatrix} 3 \\ 0 \end{bmatrix}$ for some scalars $c_1, c_2$.
This is the linear system $\left[\begin{array}{cc|c} 2 & 3 & -8 \\ 3 & 0 & -3 \end{array}\right]$, which row reduces to $\left[\begin{array}{cc|c} 1 & 0 & -1 \\ 0 & 1 & -2 \end{array}\right]$.
$c_{1} = -1$, $c_{2} = -2$ is a solution.
So, $\begin{bmatrix} -8 \\ -3 \end{bmatrix} = -\begin{bmatrix} 2 \\ 3 \end{bmatrix} -2\begin{bmatrix} 3 \\ 0 \end{bmatrix}$.

Q: Is $\begin{bmatrix} 4 \\ -5 \end{bmatrix}$ a linear combination of $\begin{bmatrix} 3 \\ -1 \end{bmatrix}$ and $\begin{bmatrix} -2 \\ -3 \end{bmatrix}$?
A: $\begin{bmatrix} 4 \\ -5 \end{bmatrix}$ is a linear combination of $\begin{bmatrix} 3 \\ -1 \end{bmatrix}$ and $\begin{bmatrix} -2 \\ -3 \end{bmatrix}$ if $\begin{bmatrix} 4 \\ -5 \end{bmatrix} = c_{1}\begin{bmatrix} 3 \\ -1 \end{bmatrix} + c_{2}\begin{bmatrix} -2 \\ -3 \end{bmatrix}$ for some scalars $c_1, c_2$.
This is the linear system $\left[\begin{array}{cc|c} 3 & -2 & 4 \\ -1 & -3 & -5 \end{array}\right]$, which row reduces to $\left[\begin{array}{cc|c} 1 & 0 & 2 \\ 0 & 1 & 1 \end{array}\right]$.
$c_{1} = 2$, $c_{2} = 1$ is a solution.
So, $\begin{bmatrix} 4 \\ -5 \end{bmatrix} = 2\begin{bmatrix} 3 \\ -1 \end{bmatrix} \begin{bmatrix} -2 \\ -3 \end{bmatrix}$.

Q: Is $\begin{bmatrix} -8 \\ -1 \end{bmatrix}$ a linear combination of $\begin{bmatrix} 3 \\ 0 \end{bmatrix}$ and $\begin{bmatrix} -2 \\ -1 \end{bmatrix}$?
A: $\begin{bmatrix} -8 \\ -1 \end{bmatrix}$ is a linear combination of $\begin{bmatrix} 3 \\ 0 \end{bmatrix}$ and $\begin{bmatrix} -2 \\ -1 \end{bmatrix}$ if $\begin{bmatrix} -8 \\ -1 \end{bmatrix} = c_{1}\begin{bmatrix} 3 \\ 0 \end{bmatrix} + c_{2}\begin{bmatrix} -2 \\ -1 \end{bmatrix}$ for some scalars $c_1, c_2$.
This is the linear system $\left[\begin{array}{cc|c} 3 & -2 & -8 \\ 0 & -1 & -1 \end{array}\right]$, which row reduces to $\left[\begin{array}{cc|c} 1 & 0 & -2 \\ 0 & 1 & 1 \end{array}\right]$.
$c_{1} = -2$, $c_{2} = 1$ is a solution.
So, $\begin{bmatrix} -8 \\ -1 \end{bmatrix} = -2\begin{bmatrix} 3 \\ 0 \end{bmatrix} \begin{bmatrix} -2 \\ -1 \end{bmatrix}$.

Q: Is the vector $\begin{bmatrix} 3 \\ 5 \end{bmatrix}$ in the span of $\left\{ \begin{bmatrix} 2 \\ 2 \end{bmatrix}, \begin{bmatrix} 1 \\ 3 \end{bmatrix} \right\}$?
A: $\begin{bmatrix} 3 \\ 5 \end{bmatrix}$ is in the span of $\begin{bmatrix} 2 \\ 2 \end{bmatrix}$ and $\begin{bmatrix} 1 \\ 3 \end{bmatrix}$ if $\begin{bmatrix} 3 \\ 5 \end{bmatrix} = c_{1}\begin{bmatrix} 2 \\ 2 \end{bmatrix} + c_{2}\begin{bmatrix} 1 \\ 3 \end{bmatrix}$ for some scalars $c_1, c_2$.
This is the linear system $\left[\begin{array}{cc|c} 2 & 1 & 3 \\ 2 & 3 & 5 \end{array}\right]$, which row reduces to $\left[\begin{array}{cc|c} 1 & 0 & 1 \\ 0 & 1 & 1 \end{array}\right]$.
$c_{1} = 1$, $c_{2} = 1$ is a solution.
So, $\begin{bmatrix} 3 \\ 5 \end{bmatrix} = \begin{bmatrix} 2 \\ 2 \end{bmatrix} \begin{bmatrix} 1 \\ 3 \end{bmatrix}$.

Q: Is the vector $\begin{bmatrix} 5 \\ -4 \end{bmatrix}$ in the span of $\left\{ \begin{bmatrix} -3 \\ 0 \end{bmatrix}, \begin{bmatrix} -1 \\ 2 \end{bmatrix} \right\}$?
A: $\begin{bmatrix} 5 \\ -4 \end{bmatrix}$ is in the span of $\begin{bmatrix} -3 \\ 0 \end{bmatrix}$ and $\begin{bmatrix} -1 \\ 2 \end{bmatrix}$ if $\begin{bmatrix} 5 \\ -4 \end{bmatrix} = c_{1}\begin{bmatrix} -3 \\ 0 \end{bmatrix} + c_{2}\begin{bmatrix} -1 \\ 2 \end{bmatrix}$ for some scalars $c_1, c_2$.
This is the linear system $\left[\begin{array}{cc|c} -3 & -1 & 5 \\ 0 & 2 & -4 \end{array}\right]$, which row reduces to $\left[\begin{array}{cc|c} 1 & 0 & -1 \\ 0 & 1 & -2 \end{array}\right]$.
$c_{1} = -1$, $c_{2} = -2$ is a solution.
So, $\begin{bmatrix} 5 \\ -4 \end{bmatrix} = -\begin{bmatrix} -3 \\ 0 \end{bmatrix} -2\begin{bmatrix} -1 \\ 2 \end{bmatrix}$.

Q: Is $\begin{bmatrix} -5 \\ -4 \end{bmatrix}$ a linear combination of $\begin{bmatrix} 2 \\ 3 \end{bmatrix}$ and $\begin{bmatrix} -1 \\ 2 \end{bmatrix}$?
A: $\begin{bmatrix} -5 \\ -4 \end{bmatrix}$ is a linear combination of $\begin{bmatrix} 2 \\ 3 \end{bmatrix}$ and $\begin{bmatrix} -1 \\ 2 \end{bmatrix}$ if $\begin{bmatrix} -5 \\ -4 \end{bmatrix} = c_{1}\begin{bmatrix} 2 \\ 3 \end{bmatrix} + c_{2}\begin{bmatrix} -1 \\ 2 \end{bmatrix}$ for some scalars $c_1, c_2$.
This is the linear system $\left[\begin{array}{cc|c} 2 & -1 & -5 \\ 3 & 2 & -4 \end{array}\right]$, which row reduces to $\left[\begin{array}{cc|c} 1 & 0 & -2 \\ 0 & 1 & 1 \end{array}\right]$.
$c_{1} = -2$, $c_{2} = 1$ is a solution.
So, $\begin{bmatrix} -5 \\ -4 \end{bmatrix} = -2\begin{bmatrix} 2 \\ 3 \end{bmatrix} \begin{bmatrix} -1 \\ 2 \end{bmatrix}$.

Q: Is $\begin{bmatrix} 3 \\ -3 \end{bmatrix}$ a linear combination of $\begin{bmatrix} -3 \\ 1 \end{bmatrix}$ and $\begin{bmatrix} 0 \\ -1 \end{bmatrix}$?
A: $\begin{bmatrix} 3 \\ -3 \end{bmatrix}$ is a linear combination of $\begin{bmatrix} -3 \\ 1 \end{bmatrix}$ and $\begin{bmatrix} 0 \\ -1 \end{bmatrix}$ if $\begin{bmatrix} 3 \\ -3 \end{bmatrix} = c_{1}\begin{bmatrix} -3 \\ 1 \end{bmatrix} + c_{2}\begin{bmatrix} 0 \\ -1 \end{bmatrix}$ for some scalars $c_1, c_2$.
This is the linear system $\left[\begin{array}{cc|c} -3 & 0 & 3 \\ 1 & -1 & -3 \end{array}\right]$, which row reduces to $\left[\begin{array}{cc|c} 1 & 0 & -1 \\ 0 & 1 & 2 \end{array}\right]$.
$c_{1} = -1$, $c_{2} = 2$ is a solution.
So, $\begin{bmatrix} 3 \\ -3 \end{bmatrix} = -\begin{bmatrix} -3 \\ 1 \end{bmatrix} + 2\begin{bmatrix} 0 \\ -1 \end{bmatrix}$.

Q: Is the vector $\begin{bmatrix} -4 \\ 2 \end{bmatrix}$ in the span of $\left\{ \begin{bmatrix} -1 \\ 3 \end{bmatrix}, \begin{bmatrix} 3 \\ 1 \end{bmatrix}, \begin{bmatrix} -4 \\ -8 \end{bmatrix} \right\}$?
A: $\begin{bmatrix} -4 \\ 2 \end{bmatrix}$ is in the span of $\begin{bmatrix} -1 \\ 3 \end{bmatrix}$, $\begin{bmatrix} 3 \\ 1 \end{bmatrix}$, and $\begin{bmatrix} -4 \\ -8 \end{bmatrix}$ if $\begin{bmatrix} -4 \\ 2 \end{bmatrix} = c_{1}\begin{bmatrix} -1 \\ 3 \end{bmatrix} + c_{2}\begin{bmatrix} 3 \\ 1 \end{bmatrix} + c_{3}\begin{bmatrix} -4 \\ -8 \end{bmatrix}$ for some scalars $c_1, c_2, c_3$.
This is the linear system $\left[\begin{array}{ccc|c} -1 & 3 & -4 & -4 \\ 3 & 1 & -8 & 2 \end{array}\right]$, which row reduces to $\left[\begin{array}{ccc|c} 1 & 0 & -2 & 1 \\ 0 & 1 & -2 & -1 \end{array}\right]$.
$c_{1} = 1$, $c_{2} = -1$, $c_{3} = 0$ is a solution.
So, $\begin{bmatrix} -4 \\ 2 \end{bmatrix} = \begin{bmatrix} -1 \\ 3 \end{bmatrix} -\begin{bmatrix} 3 \\ 1 \end{bmatrix}$.

Q: Is $\begin{bmatrix} -3 \\ -1 \end{bmatrix}$ a linear combination of $\begin{bmatrix} -3 \\ -1 \end{bmatrix}$, $\begin{bmatrix} -3 \\ -1 \end{bmatrix}$, and $\begin{bmatrix} 6 \\ 2 \end{bmatrix}$?
A: $\begin{bmatrix} -3 \\ -1 \end{bmatrix}$ is a linear combination of $\begin{bmatrix} -3 \\ -1 \end{bmatrix}$, $\begin{bmatrix} -3 \\ -1 \end{bmatrix}$, and $\begin{bmatrix} 6 \\ 2 \end{bmatrix}$ if $\begin{bmatrix} -3 \\ -1 \end{bmatrix} = c_{1}\begin{bmatrix} -3 \\ -1 \end{bmatrix} + c_{2}\begin{bmatrix} -3 \\ -1 \end{bmatrix} + c_{3}\begin{bmatrix} 6 \\ 2 \end{bmatrix}$ for some scalars $c_1, c_2, c_3$.
This is the linear system $\left[\begin{array}{ccc|c} -3 & -3 & 6 & -3 \\ -1 & -1 & 2 & -1 \end{array}\right]$, which row reduces to $\left[\begin{array}{ccc|c} 1 & 1 & -2 & 1 \\ 0 & 0 & 0 & 0 \end{array}\right]$.
$c_{1} = 1$, $c_{2} = 0$, $c_{3} = 0$ is a solution.
So, $\begin{bmatrix} -3 \\ -1 \end{bmatrix} = \begin{bmatrix} -3 \\ -1 \end{bmatrix}$.

Q: Is $\begin{bmatrix} -5 \\ -7 \end{bmatrix}$ a linear combination of $\begin{bmatrix} 3 \\ 3 \end{bmatrix}$, $\begin{bmatrix} -1 \\ 1 \end{bmatrix}$, and $\begin{bmatrix} -1 \\ 1 \end{bmatrix}$?
A: $\begin{bmatrix} -5 \\ -7 \end{bmatrix}$ is a linear combination of $\begin{bmatrix} 3 \\ 3 \end{bmatrix}$, $\begin{bmatrix} -1 \\ 1 \end{bmatrix}$, and $\begin{bmatrix} -1 \\ 1 \end{bmatrix}$ if $\begin{bmatrix} -5 \\ -7 \end{bmatrix} = c_{1}\begin{bmatrix} 3 \\ 3 \end{bmatrix} + c_{2}\begin{bmatrix} -1 \\ 1 \end{bmatrix} + c_{3}\begin{bmatrix} -1 \\ 1 \end{bmatrix}$ for some scalars $c_1, c_2, c_3$.
This is the linear system $\left[\begin{array}{ccc|c} 3 & -1 & -1 & -5 \\ 3 & 1 & 1 & -7 \end{array}\right]$, which row reduces to $\left[\begin{array}{ccc|c} 1 & 0 & 0 & -2 \\ 0 & 1 & 1 & -1 \end{array}\right]$.
$c_{1} = -2$, $c_{2} = -1$, $c_{3} = 0$ is a solution.
So, $\begin{bmatrix} -5 \\ -7 \end{bmatrix} = -2\begin{bmatrix} 3 \\ 3 \end{bmatrix} -\begin{bmatrix} -1 \\ 1 \end{bmatrix}$.

Q: Is $\begin{bmatrix} 3 \\ 4 \end{bmatrix}$ a linear combination of $\begin{bmatrix} 3 \\ 2 \end{bmatrix}$, $\begin{bmatrix} 3 \\ 0 \end{bmatrix}$, and $\begin{bmatrix} -3 \\ -4 \end{bmatrix}$?
A: $\begin{bmatrix} 3 \\ 4 \end{bmatrix}$ is a linear combination of $\begin{bmatrix} 3 \\ 2 \end{bmatrix}$, $\begin{bmatrix} 3 \\ 0 \end{bmatrix}$, and $\begin{bmatrix} -3 \\ -4 \end{bmatrix}$ if $\begin{bmatrix} 3 \\ 4 \end{bmatrix} = c_{1}\begin{bmatrix} 3 \\ 2 \end{bmatrix} + c_{2}\begin{bmatrix} 3 \\ 0 \end{bmatrix} + c_{3}\begin{bmatrix} -3 \\ -4 \end{bmatrix}$ for some scalars $c_1, c_2, c_3$.
This is the linear system $\left[\begin{array}{ccc|c} 3 & 3 & -3 & 3 \\ 2 & 0 & -4 & 4 \end{array}\right]$, which row reduces to $\left[\begin{array}{ccc|c} 1 & 0 & -2 & 2 \\ 0 & 1 & 1 & -1 \end{array}\right]$.
$c_{1} = 2$, $c_{2} = -1$, $c_{3} = 0$ is a solution.
So, $\begin{bmatrix} 3 \\ 4 \end{bmatrix} = 2\begin{bmatrix} 3 \\ 2 \end{bmatrix} -\begin{bmatrix} 3 \\ 0 \end{bmatrix}$.

Q: Is $\begin{bmatrix} -2 \\ -6 \end{bmatrix}$ a linear combination of $\begin{bmatrix} 0 \\ 3 \end{bmatrix}$, $\begin{bmatrix} 1 \\ 0 \end{bmatrix}$, and $\begin{bmatrix} -1 \\ -6 \end{bmatrix}$?
A: $\begin{bmatrix} -2 \\ -6 \end{bmatrix}$ is a linear combination of $\begin{bmatrix} 0 \\ 3 \end{bmatrix}$, $\begin{bmatrix} 1 \\ 0 \end{bmatrix}$, and $\begin{bmatrix} -1 \\ -6 \end{bmatrix}$ if $\begin{bmatrix} -2 \\ -6 \end{bmatrix} = c_{1}\begin{bmatrix} 0 \\ 3 \end{bmatrix} + c_{2}\begin{bmatrix} 1 \\ 0 \end{bmatrix} + c_{3}\begin{bmatrix} -1 \\ -6 \end{bmatrix}$ for some scalars $c_1, c_2, c_3$.
This is the linear system $\left[\begin{array}{ccc|c} 0 & 1 & -1 & -2 \\ 3 & 0 & -6 & -6 \end{array}\right]$, which row reduces to $\left[\begin{array}{ccc|c} 1 & 0 & -2 & -2 \\ 0 & 1 & -1 & -2 \end{array}\right]$.
$c_{1} = -2$, $c_{2} = -2$, $c_{3} = 0$ is a solution.
So, $\begin{bmatrix} -2 \\ -6 \end{bmatrix} = -2\begin{bmatrix} 0 \\ 3 \end{bmatrix} -2\begin{bmatrix} 1 \\ 0 \end{bmatrix}$.

Q: Is the vector $\begin{bmatrix} 6 \\ 2 \end{bmatrix}$ in the span of $\left\{ \begin{bmatrix} 1 \\ -1 \end{bmatrix}, \begin{bmatrix} -2 \\ -2 \end{bmatrix}, \begin{bmatrix} 1 \\ 3 \end{bmatrix} \right\}$?
A: $\begin{bmatrix} 6 \\ 2 \end{bmatrix}$ is in the span of $\begin{bmatrix} 1 \\ -1 \end{bmatrix}$, $\begin{bmatrix} -2 \\ -2 \end{bmatrix}$, and $\begin{bmatrix} 1 \\ 3 \end{bmatrix}$ if $\begin{bmatrix} 6 \\ 2 \end{bmatrix} = c_{1}\begin{bmatrix} 1 \\ -1 \end{bmatrix} + c_{2}\begin{bmatrix} -2 \\ -2 \end{bmatrix} + c_{3}\begin{bmatrix} 1 \\ 3 \end{bmatrix}$ for some scalars $c_1, c_2, c_3$.
This is the linear system $\left[\begin{array}{ccc|c} 1 & -2 & 1 & 6 \\ -1 & -2 & 3 & 2 \end{array}\right]$, which row reduces to $\left[\begin{array}{ccc|c} 1 & 0 & -1 & 2 \\ 0 & 1 & -1 & -2 \end{array}\right]$.
$c_{1} = 2$, $c_{2} = -2$, $c_{3} = 0$ is a solution.
So, $\begin{bmatrix} 6 \\ 2 \end{bmatrix} = 2\begin{bmatrix} 1 \\ -1 \end{bmatrix} -2\begin{bmatrix} -2 \\ -2 \end{bmatrix}$.

Q: Is $\begin{bmatrix} 3 \\ 8 \end{bmatrix}$ a linear combination of $\begin{bmatrix} -1 \\ -3 \end{bmatrix}$, $\begin{bmatrix} -1 \\ -2 \end{bmatrix}$, and $\begin{bmatrix} 1 \\ 2 \end{bmatrix}$?
A: $\begin{bmatrix} 3 \\ 8 \end{bmatrix}$ is a linear combination of $\begin{bmatrix} -1 \\ -3 \end{bmatrix}$, $\begin{bmatrix} -1 \\ -2 \end{bmatrix}$, and $\begin{bmatrix} 1 \\ 2 \end{bmatrix}$ if $\begin{bmatrix} 3 \\ 8 \end{bmatrix} = c_{1}\begin{bmatrix} -1 \\ -3 \end{bmatrix} + c_{2}\begin{bmatrix} -1 \\ -2 \end{bmatrix} + c_{3}\begin{bmatrix} 1 \\ 2 \end{bmatrix}$ for some scalars $c_1, c_2, c_3$.
This is the linear system $\left[\begin{array}{ccc|c} -1 & -1 & 1 & 3 \\ -3 & -2 & 2 & 8 \end{array}\right]$, which row reduces to $\left[\begin{array}{ccc|c} 1 & 0 & 0 & -2 \\ 0 & 1 & -1 & -1 \end{array}\right]$.
$c_{1} = -2$, $c_{2} = -1$, $c_{3} = 0$ is a solution.
So, $\begin{bmatrix} 3 \\ 8 \end{bmatrix} = -2\begin{bmatrix} -1 \\ -3 \end{bmatrix} -\begin{bmatrix} -1 \\ -2 \end{bmatrix}$.

Q: Is the vector $\begin{bmatrix} 5 \\ 7 \end{bmatrix}$ in the span of $\left\{ \begin{bmatrix} -1 \\ 1 \end{bmatrix}, \begin{bmatrix} 3 \\ 3 \end{bmatrix}, \begin{bmatrix} 2 \\ 4 \end{bmatrix} \right\}$?
A: $\begin{bmatrix} 5 \\ 7 \end{bmatrix}$ is in the span of $\begin{bmatrix} -1 \\ 1 \end{bmatrix}$, $\begin{bmatrix} 3 \\ 3 \end{bmatrix}$, and $\begin{bmatrix} 2 \\ 4 \end{bmatrix}$ if $\begin{bmatrix} 5 \\ 7 \end{bmatrix} = c_{1}\begin{bmatrix} -1 \\ 1 \end{bmatrix} + c_{2}\begin{bmatrix} 3 \\ 3 \end{bmatrix} + c_{3}\begin{bmatrix} 2 \\ 4 \end{bmatrix}$ for some scalars $c_1, c_2, c_3$.
This is the linear system $\left[\begin{array}{ccc|c} -1 & 3 & 2 & 5 \\ 1 & 3 & 4 & 7 \end{array}\right]$, which row reduces to $\left[\begin{array}{ccc|c} 1 & 0 & 1 & 1 \\ 0 & 1 & 1 & 2 \end{array}\right]$.
$c_{1} = 1$, $c_{2} = 2$, $c_{3} = 0$ is a solution.
So, $\begin{bmatrix} 5 \\ 7 \end{bmatrix} = \begin{bmatrix} -1 \\ 1 \end{bmatrix} + 2\begin{bmatrix} 3 \\ 3 \end{bmatrix}$.

Q: Is $\begin{bmatrix} -4 \\ 2 \end{bmatrix}$ a linear combination of $\begin{bmatrix} 0 \\ 1 \end{bmatrix}$, $\begin{bmatrix} 2 \\ 0 \end{bmatrix}$, and $\begin{bmatrix} -4 \\ 0 \end{bmatrix}$?
A: $\begin{bmatrix} -4 \\ 2 \end{bmatrix}$ is a linear combination of $\begin{bmatrix} 0 \\ 1 \end{bmatrix}$, $\begin{bmatrix} 2 \\ 0 \end{bmatrix}$, and $\begin{bmatrix} -4 \\ 0 \end{bmatrix}$ if $\begin{bmatrix} -4 \\ 2 \end{bmatrix} = c_{1}\begin{bmatrix} 0 \\ 1 \end{bmatrix} + c_{2}\begin{bmatrix} 2 \\ 0 \end{bmatrix} + c_{3}\begin{bmatrix} -4 \\ 0 \end{bmatrix}$ for some scalars $c_1, c_2, c_3$.
This is the linear system $\left[\begin{array}{ccc|c} 0 & 2 & -4 & -4 \\ 1 & 0 & 0 & 2 \end{array}\right]$, which row reduces to $\left[\begin{array}{ccc|c} 1 & 0 & 0 & 2 \\ 0 & 1 & -2 & -2 \end{array}\right]$.
$c_{1} = 2$, $c_{2} = -2$, $c_{3} = 0$ is a solution.
So, $\begin{bmatrix} -4 \\ 2 \end{bmatrix} = 2\begin{bmatrix} 0 \\ 1 \end{bmatrix} -2\begin{bmatrix} 2 \\ 0 \end{bmatrix}$.

Q: Is the vector $\begin{bmatrix} 2 \\ 4 \end{bmatrix}$ in the span of $\left\{ \begin{bmatrix} -1 \\ 1 \end{bmatrix}, \begin{bmatrix} -3 \\ -3 \end{bmatrix}, \begin{bmatrix} 4 \\ 2 \end{bmatrix} \right\}$?
A: $\begin{bmatrix} 2 \\ 4 \end{bmatrix}$ is in the span of $\begin{bmatrix} -1 \\ 1 \end{bmatrix}$, $\begin{bmatrix} -3 \\ -3 \end{bmatrix}$, and $\begin{bmatrix} 4 \\ 2 \end{bmatrix}$ if $\begin{bmatrix} 2 \\ 4 \end{bmatrix} = c_{1}\begin{bmatrix} -1 \\ 1 \end{bmatrix} + c_{2}\begin{bmatrix} -3 \\ -3 \end{bmatrix} + c_{3}\begin{bmatrix} 4 \\ 2 \end{bmatrix}$ for some scalars $c_1, c_2, c_3$.
This is the linear system $\left[\begin{array}{ccc|c} -1 & -3 & 4 & 2 \\ 1 & -3 & 2 & 4 \end{array}\right]$, which row reduces to $\left[\begin{array}{ccc|c} 1 & 0 & -1 & 1 \\ 0 & 1 & -1 & -1 \end{array}\right]$.
$c_{1} = 1$, $c_{2} = -1$, $c_{3} = 0$ is a solution.
So, $\begin{bmatrix} 2 \\ 4 \end{bmatrix} = \begin{bmatrix} -1 \\ 1 \end{bmatrix} -\begin{bmatrix} -3 \\ -3 \end{bmatrix}$.

Q: Is $\begin{bmatrix} 5 \\ 2 \end{bmatrix}$ a linear combination of $\begin{bmatrix} 3 \\ 2 \end{bmatrix}$, $\begin{bmatrix} -2 \\ 0 \end{bmatrix}$, and $\begin{bmatrix} -1 \\ -2 \end{bmatrix}$?
A: $\begin{bmatrix} 5 \\ 2 \end{bmatrix}$ is a linear combination of $\begin{bmatrix} 3 \\ 2 \end{bmatrix}$, $\begin{bmatrix} -2 \\ 0 \end{bmatrix}$, and $\begin{bmatrix} -1 \\ -2 \end{bmatrix}$ if $\begin{bmatrix} 5 \\ 2 \end{bmatrix} = c_{1}\begin{bmatrix} 3 \\ 2 \end{bmatrix} + c_{2}\begin{bmatrix} -2 \\ 0 \end{bmatrix} + c_{3}\begin{bmatrix} -1 \\ -2 \end{bmatrix}$ for some scalars $c_1, c_2, c_3$.
This is the linear system $\left[\begin{array}{ccc|c} 3 & -2 & -1 & 5 \\ 2 & 0 & -2 & 2 \end{array}\right]$, which row reduces to $\left[\begin{array}{ccc|c} 1 & 0 & -1 & 1 \\ 0 & 1 & -1 & -1 \end{array}\right]$.
$c_{1} = 1$, $c_{2} = -1$, $c_{3} = 0$ is a solution.
So, $\begin{bmatrix} 5 \\ 2 \end{bmatrix} = \begin{bmatrix} 3 \\ 2 \end{bmatrix} -\begin{bmatrix} -2 \\ 0 \end{bmatrix}$.

Q: Is $\begin{bmatrix} -4 \\ 10 \end{bmatrix}$ a linear combination of $\begin{bmatrix} 1 \\ 3 \end{bmatrix}$, $\begin{bmatrix} 3 \\ -2 \end{bmatrix}$, and $\begin{bmatrix} -4 \\ -1 \end{bmatrix}$?
A: $\begin{bmatrix} -4 \\ 10 \end{bmatrix}$ is a linear combination of $\begin{bmatrix} 1 \\ 3 \end{bmatrix}$, $\begin{bmatrix} 3 \\ -2 \end{bmatrix}$, and $\begin{bmatrix} -4 \\ -1 \end{bmatrix}$ if $\begin{bmatrix} -4 \\ 10 \end{bmatrix} = c_{1}\begin{bmatrix} 1 \\ 3 \end{bmatrix} + c_{2}\begin{bmatrix} 3 \\ -2 \end{bmatrix} + c_{3}\begin{bmatrix} -4 \\ -1 \end{bmatrix}$ for some scalars $c_1, c_2, c_3$.
This is the linear system $\left[\begin{array}{ccc|c} 1 & 3 & -4 & -4 \\ 3 & -2 & -1 & 10 \end{array}\right]$, which row reduces to $\left[\begin{array}{ccc|c} 1 & 0 & -1 & 2 \\ 0 & 1 & -1 & -2 \end{array}\right]$.
$c_{1} = 2$, $c_{2} = -2$, $c_{3} = 0$ is a solution.
So, $\begin{bmatrix} -4 \\ 10 \end{bmatrix} = 2\begin{bmatrix} 1 \\ 3 \end{bmatrix} -2\begin{bmatrix} 3 \\ -2 \end{bmatrix}$.

Q: Is $\begin{bmatrix} -4 \\ 3 \end{bmatrix}$ a linear combination of $\begin{bmatrix} -3 \\ 3 \end{bmatrix}$, $\begin{bmatrix} -2 \\ 3 \end{bmatrix}$, and $\begin{bmatrix} 4 \\ -3 \end{bmatrix}$?
A: $\begin{bmatrix} -4 \\ 3 \end{bmatrix}$ is a linear combination of $\begin{bmatrix} -3 \\ 3 \end{bmatrix}$, $\begin{bmatrix} -2 \\ 3 \end{bmatrix}$, and $\begin{bmatrix} 4 \\ -3 \end{bmatrix}$ if $\begin{bmatrix} -4 \\ 3 \end{bmatrix} = c_{1}\begin{bmatrix} -3 \\ 3 \end{bmatrix} + c_{2}\begin{bmatrix} -2 \\ 3 \end{bmatrix} + c_{3}\begin{bmatrix} 4 \\ -3 \end{bmatrix}$ for some scalars $c_1, c_2, c_3$.
This is the linear system $\left[\begin{array}{ccc|c} -3 & -2 & 4 & -4 \\ 3 & 3 & -3 & 3 \end{array}\right]$, which row reduces to $\left[\begin{array}{ccc|c} 1 & 0 & -2 & 2 \\ 0 & 1 & 1 & -1 \end{array}\right]$.
$c_{1} = 2$, $c_{2} = -1$, $c_{3} = 0$ is a solution.
So, $\begin{bmatrix} -4 \\ 3 \end{bmatrix} = 2\begin{bmatrix} -3 \\ 3 \end{bmatrix} -\begin{bmatrix} -2 \\ 3 \end{bmatrix}$.

Q: Is the vector $\begin{bmatrix} 4 \\ 10 \end{bmatrix}$ in the span of $\left\{ \begin{bmatrix} 1 \\ -2 \end{bmatrix}, \begin{bmatrix} -3 \\ -3 \end{bmatrix}, \begin{bmatrix} 1 \\ -2 \end{bmatrix} \right\}$?
A: $\begin{bmatrix} 4 \\ 10 \end{bmatrix}$ is in the span of $\begin{bmatrix} 1 \\ -2 \end{bmatrix}$, $\begin{bmatrix} -3 \\ -3 \end{bmatrix}$, and $\begin{bmatrix} 1 \\ -2 \end{bmatrix}$ if $\begin{bmatrix} 4 \\ 10 \end{bmatrix} = c_{1}\begin{bmatrix} 1 \\ -2 \end{bmatrix} + c_{2}\begin{bmatrix} -3 \\ -3 \end{bmatrix} + c_{3}\begin{bmatrix} 1 \\ -2 \end{bmatrix}$ for some scalars $c_1, c_2, c_3$.
This is the linear system $\left[\begin{array}{ccc|c} 1 & -3 & 1 & 4 \\ -2 & -3 & -2 & 10 \end{array}\right]$, which row reduces to $\left[\begin{array}{ccc|c} 1 & 0 & 1 & -2 \\ 0 & 1 & 0 & -2 \end{array}\right]$.
$c_{1} = -2$, $c_{2} = -2$, $c_{3} = 0$ is a solution.
So, $\begin{bmatrix} 4 \\ 10 \end{bmatrix} = -2\begin{bmatrix} 1 \\ -2 \end{bmatrix} -2\begin{bmatrix} -3 \\ -3 \end{bmatrix}$.

Q: Is the vector $\begin{bmatrix} -2 \\ -2 \end{bmatrix}$ in the span of $\left\{ \begin{bmatrix} -1 \\ 2 \end{bmatrix}, \begin{bmatrix} -3 \\ 0 \end{bmatrix}, \begin{bmatrix} -1 \\ 2 \end{bmatrix} \right\}$?
A: $\begin{bmatrix} -2 \\ -2 \end{bmatrix}$ is in the span of $\begin{bmatrix} -1 \\ 2 \end{bmatrix}$, $\begin{bmatrix} -3 \\ 0 \end{bmatrix}$, and $\begin{bmatrix} -1 \\ 2 \end{bmatrix}$ if $\begin{bmatrix} -2 \\ -2 \end{bmatrix} = c_{1}\begin{bmatrix} -1 \\ 2 \end{bmatrix} + c_{2}\begin{bmatrix} -3 \\ 0 \end{bmatrix} + c_{3}\begin{bmatrix} -1 \\ 2 \end{bmatrix}$ for some scalars $c_1, c_2, c_3$.
This is the linear system $\left[\begin{array}{ccc|c} -1 & -3 & -1 & -2 \\ 2 & 0 & 2 & -2 \end{array}\right]$, which row reduces to $\left[\begin{array}{ccc|c} 1 & 0 & 1 & -1 \\ 0 & 1 & 0 & 1 \end{array}\right]$.
$c_{1} = -1$, $c_{2} = 1$, $c_{3} = 0$ is a solution.
So, $\begin{bmatrix} -2 \\ -2 \end{bmatrix} = -\begin{bmatrix} -1 \\ 2 \end{bmatrix} \begin{bmatrix} -3 \\ 0 \end{bmatrix}$.

Q: Is $\begin{bmatrix} -6 \\ 0 \end{bmatrix}$ a linear combination of $\begin{bmatrix} 3 \\ -3 \end{bmatrix}$, $\begin{bmatrix} -3 \\ -3 \end{bmatrix}$, and $\begin{bmatrix} 3 \\ 9 \end{bmatrix}$?
A: $\begin{bmatrix} -6 \\ 0 \end{bmatrix}$ is a linear combination of $\begin{bmatrix} 3 \\ -3 \end{bmatrix}$, $\begin{bmatrix} -3 \\ -3 \end{bmatrix}$, and $\begin{bmatrix} 3 \\ 9 \end{bmatrix}$ if $\begin{bmatrix} -6 \\ 0 \end{bmatrix} = c_{1}\begin{bmatrix} 3 \\ -3 \end{bmatrix} + c_{2}\begin{bmatrix} -3 \\ -3 \end{bmatrix} + c_{3}\begin{bmatrix} 3 \\ 9 \end{bmatrix}$ for some scalars $c_1, c_2, c_3$.
This is the linear system $\left[\begin{array}{ccc|c} 3 & -3 & 3 & -6 \\ -3 & -3 & 9 & 0 \end{array}\right]$, which row reduces to $\left[\begin{array}{ccc|c} 1 & 0 & -1 & -1 \\ 0 & 1 & -2 & 1 \end{array}\right]$.
$c_{1} = -1$, $c_{2} = 1$, $c_{3} = 0$ is a solution.
So, $\begin{bmatrix} -6 \\ 0 \end{bmatrix} = -\begin{bmatrix} 3 \\ -3 \end{bmatrix} \begin{bmatrix} -3 \\ -3 \end{bmatrix}$.

Q: Is the vector $\begin{bmatrix} 5 \\ 0 \end{bmatrix}$ in the span of $\left\{ \begin{bmatrix} 1 \\ 2 \end{bmatrix}, \begin{bmatrix} 3 \\ 1 \end{bmatrix}, \begin{bmatrix} 0 \\ 0 \end{bmatrix} \right\}$?
A: $\begin{bmatrix} 5 \\ 0 \end{bmatrix}$ is in the span of $\begin{bmatrix} 1 \\ 2 \end{bmatrix}$, $\begin{bmatrix} 3 \\ 1 \end{bmatrix}$, and $\begin{bmatrix} 0 \\ 0 \end{bmatrix}$ if $\begin{bmatrix} 5 \\ 0 \end{bmatrix} = c_{1}\begin{bmatrix} 1 \\ 2 \end{bmatrix} + c_{2}\begin{bmatrix} 3 \\ 1 \end{bmatrix} + c_{3}\begin{bmatrix} 0 \\ 0 \end{bmatrix}$ for some scalars $c_1, c_2, c_3$.
This is the linear system $\left[\begin{array}{ccc|c} 1 & 3 & 0 & 5 \\ 2 & 1 & 0 & 0 \end{array}\right]$, which row reduces to $\left[\begin{array}{ccc|c} 1 & 0 & 0 & -1 \\ 0 & 1 & 0 & 2 \end{array}\right]$.
$c_{1} = -1$, $c_{2} = 2$, $c_{3} = 0$ is a solution.
So, $\begin{bmatrix} 5 \\ 0 \end{bmatrix} = -\begin{bmatrix} 1 \\ 2 \end{bmatrix} + 2\begin{bmatrix} 3 \\ 1 \end{bmatrix}$.

Q: Is $\begin{bmatrix} 3 \\ 6 \end{bmatrix}$ a linear combination of $\begin{bmatrix} 3 \\ 2 \end{bmatrix}$, $\begin{bmatrix} -3 \\ 2 \end{bmatrix}$, and $\begin{bmatrix} 6 \\ 4 \end{bmatrix}$?
A: $\begin{bmatrix} 3 \\ 6 \end{bmatrix}$ is a linear combination of $\begin{bmatrix} 3 \\ 2 \end{bmatrix}$, $\begin{bmatrix} -3 \\ 2 \end{bmatrix}$, and $\begin{bmatrix} 6 \\ 4 \end{bmatrix}$ if $\begin{bmatrix} 3 \\ 6 \end{bmatrix} = c_{1}\begin{bmatrix} 3 \\ 2 \end{bmatrix} + c_{2}\begin{bmatrix} -3 \\ 2 \end{bmatrix} + c_{3}\begin{bmatrix} 6 \\ 4 \end{bmatrix}$ for some scalars $c_1, c_2, c_3$.
This is the linear system $\left[\begin{array}{ccc|c} 3 & -3 & 6 & 3 \\ 2 & 2 & 4 & 6 \end{array}\right]$, which row reduces to $\left[\begin{array}{ccc|c} 1 & 0 & 2 & 2 \\ 0 & 1 & 0 & 1 \end{array}\right]$.
$c_{1} = 2$, $c_{2} = 1$, $c_{3} = 0$ is a solution.
So, $\begin{bmatrix} 3 \\ 6 \end{bmatrix} = 2\begin{bmatrix} 3 \\ 2 \end{bmatrix} \begin{bmatrix} -3 \\ 2 \end{bmatrix}$.

Q: Is the vector $\begin{bmatrix} 3 \\ 4 \end{bmatrix}$ in the span of $\left\{ \begin{bmatrix} 3 \\ -2 \end{bmatrix}, \begin{bmatrix} 0 \\ -3 \end{bmatrix}, \begin{bmatrix} 3 \\ -8 \end{bmatrix} \right\}$?
A: $\begin{bmatrix} 3 \\ 4 \end{bmatrix}$ is in the span of $\begin{bmatrix} 3 \\ -2 \end{bmatrix}$, $\begin{bmatrix} 0 \\ -3 \end{bmatrix}$, and $\begin{bmatrix} 3 \\ -8 \end{bmatrix}$ if $\begin{bmatrix} 3 \\ 4 \end{bmatrix} = c_{1}\begin{bmatrix} 3 \\ -2 \end{bmatrix} + c_{2}\begin{bmatrix} 0 \\ -3 \end{bmatrix} + c_{3}\begin{bmatrix} 3 \\ -8 \end{bmatrix}$ for some scalars $c_1, c_2, c_3$.
This is the linear system $\left[\begin{array}{ccc|c} 3 & 0 & 3 & 3 \\ -2 & -3 & -8 & 4 \end{array}\right]$, which row reduces to $\left[\begin{array}{ccc|c} 1 & 0 & 1 & 1 \\ 0 & 1 & 2 & -2 \end{array}\right]$.
$c_{1} = 1$, $c_{2} = -2$, $c_{3} = 0$ is a solution.
So, $\begin{bmatrix} 3 \\ 4 \end{bmatrix} = \begin{bmatrix} 3 \\ -2 \end{bmatrix} -2\begin{bmatrix} 0 \\ -3 \end{bmatrix}$.

Q: Is $\begin{bmatrix} 5 \\ 1 \end{bmatrix}$ a linear combination of $\begin{bmatrix} -2 \\ 0 \end{bmatrix}$, $\begin{bmatrix} 1 \\ 1 \end{bmatrix}$, and $\begin{bmatrix} 0 \\ -2 \end{bmatrix}$?
A: $\begin{bmatrix} 5 \\ 1 \end{bmatrix}$ is a linear combination of $\begin{bmatrix} -2 \\ 0 \end{bmatrix}$, $\begin{bmatrix} 1 \\ 1 \end{bmatrix}$, and $\begin{bmatrix} 0 \\ -2 \end{bmatrix}$ if $\begin{bmatrix} 5 \\ 1 \end{bmatrix} = c_{1}\begin{bmatrix} -2 \\ 0 \end{bmatrix} + c_{2}\begin{bmatrix} 1 \\ 1 \end{bmatrix} + c_{3}\begin{bmatrix} 0 \\ -2 \end{bmatrix}$ for some scalars $c_1, c_2, c_3$.
This is the linear system $\left[\begin{array}{ccc|c} -2 & 1 & 0 & 5 \\ 0 & 1 & -2 & 1 \end{array}\right]$, which row reduces to $\left[\begin{array}{ccc|c} 1 & 0 & -1 & -2 \\ 0 & 1 & -2 & 1 \end{array}\right]$.
$c_{1} = -2$, $c_{2} = 1$, $c_{3} = 0$ is a solution.
So, $\begin{bmatrix} 5 \\ 1 \end{bmatrix} = -2\begin{bmatrix} -2 \\ 0 \end{bmatrix} \begin{bmatrix} 1 \\ 1 \end{bmatrix}$.

Q: Is $\begin{bmatrix} 2 \\ 4 \\ -1 \end{bmatrix}$ a linear combination of $\begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix}$ and $\begin{bmatrix} 1 \\ 3 \\ -2 \end{bmatrix}$?
A: $\begin{bmatrix} 2 \\ 4 \\ -1 \end{bmatrix}$ is a linear combination of $\begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix}$ and $\begin{bmatrix} 1 \\ 3 \\ -2 \end{bmatrix}$ if $\begin{bmatrix} 2 \\ 4 \\ -1 \end{bmatrix} = c_{1}\begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix} + c_{2}\begin{bmatrix} 1 \\ 3 \\ -2 \end{bmatrix}$ for some scalars $c_1, c_2$.
This is the linear system $\left[\begin{array}{cc|c} 1 & 1 & 2 \\ 1 & 3 & 4 \\ 1 & -2 & -1 \end{array}\right]$, which row reduces to $\left[\begin{array}{cc|c} 1 & 0 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 0 \end{array}\right]$.
$c_{1} = 1$, $c_{2} = 1$ is a solution.
So, $\begin{bmatrix} 2 \\ 4 \\ -1 \end{bmatrix} = \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix} \begin{bmatrix} 1 \\ 3 \\ -2 \end{bmatrix}$.

Q: Is the vector $\begin{bmatrix} 0 \\ 2 \\ -6 \end{bmatrix}$ in the span of $\left\{ \begin{bmatrix} 1 \\ 1 \\ 3 \end{bmatrix}, \begin{bmatrix} -1 \\ -2 \\ 0 \end{bmatrix} \right\}$?
A: $\begin{bmatrix} 0 \\ 2 \\ -6 \end{bmatrix}$ is in the span of $\begin{bmatrix} 1 \\ 1 \\ 3 \end{bmatrix}$ and $\begin{bmatrix} -1 \\ -2 \\ 0 \end{bmatrix}$ if $\begin{bmatrix} 0 \\ 2 \\ -6 \end{bmatrix} = c_{1}\begin{bmatrix} 1 \\ 1 \\ 3 \end{bmatrix} + c_{2}\begin{bmatrix} -1 \\ -2 \\ 0 \end{bmatrix}$ for some scalars $c_1, c_2$.
This is the linear system $\left[\begin{array}{cc|c} 1 & -1 & 0 \\ 1 & -2 & 2 \\ 3 & 0 & -6 \end{array}\right]$, which row reduces to $\left[\begin{array}{cc|c} 1 & 0 & -2 \\ 0 & 1 & -2 \\ 0 & 0 & 0 \end{array}\right]$.
$c_{1} = -2$, $c_{2} = -2$ is a solution.
So, $\begin{bmatrix} 0 \\ 2 \\ -6 \end{bmatrix} = -2\begin{bmatrix} 1 \\ 1 \\ 3 \end{bmatrix} -2\begin{bmatrix} -1 \\ -2 \\ 0 \end{bmatrix}$.

Q: Is $\begin{bmatrix} 3 \\ 5 \\ 7 \end{bmatrix}$ a linear combination of $\begin{bmatrix} 3 \\ -3 \\ -1 \end{bmatrix}$ and $\begin{bmatrix} -3 \\ -1 \\ -3 \end{bmatrix}$?
A: $\begin{bmatrix} 3 \\ 5 \\ 7 \end{bmatrix}$ is a linear combination of $\begin{bmatrix} 3 \\ -3 \\ -1 \end{bmatrix}$ and $\begin{bmatrix} -3 \\ -1 \\ -3 \end{bmatrix}$ if $\begin{bmatrix} 3 \\ 5 \\ 7 \end{bmatrix} = c_{1}\begin{bmatrix} 3 \\ -3 \\ -1 \end{bmatrix} + c_{2}\begin{bmatrix} -3 \\ -1 \\ -3 \end{bmatrix}$ for some scalars $c_1, c_2$.
This is the linear system $\left[\begin{array}{cc|c} 3 & -3 & 3 \\ -3 & -1 & 5 \\ -1 & -3 & 7 \end{array}\right]$, which row reduces to $\left[\begin{array}{cc|c} 1 & 0 & -1 \\ 0 & 1 & -2 \\ 0 & 0 & 0 \end{array}\right]$.
$c_{1} = -1$, $c_{2} = -2$ is a solution.
So, $\begin{bmatrix} 3 \\ 5 \\ 7 \end{bmatrix} = -\begin{bmatrix} 3 \\ -3 \\ -1 \end{bmatrix} -2\begin{bmatrix} -3 \\ -1 \\ -3 \end{bmatrix}$.

Q: Is the vector $\begin{bmatrix} 1 \\ 1 \\ -5 \end{bmatrix}$ in the span of $\left\{ \begin{bmatrix} 0 \\ -2 \\ 3 \end{bmatrix}, \begin{bmatrix} 1 \\ -3 \\ 1 \end{bmatrix} \right\}$?
A: $\begin{bmatrix} 1 \\ 1 \\ -5 \end{bmatrix}$ is in the span of $\begin{bmatrix} 0 \\ -2 \\ 3 \end{bmatrix}$ and $\begin{bmatrix} 1 \\ -3 \\ 1 \end{bmatrix}$ if $\begin{bmatrix} 1 \\ 1 \\ -5 \end{bmatrix} = c_{1}\begin{bmatrix} 0 \\ -2 \\ 3 \end{bmatrix} + c_{2}\begin{bmatrix} 1 \\ -3 \\ 1 \end{bmatrix}$ for some scalars $c_1, c_2$.
This is the linear system $\left[\begin{array}{cc|c} 0 & 1 & 1 \\ -2 & -3 & 1 \\ 3 & 1 & -5 \end{array}\right]$, which row reduces to $\left[\begin{array}{cc|c} 1 & 0 & -2 \\ 0 & 1 & 1 \\ 0 & 0 & 0 \end{array}\right]$.
$c_{1} = -2$, $c_{2} = 1$ is a solution.
So, $\begin{bmatrix} 1 \\ 1 \\ -5 \end{bmatrix} = -2\begin{bmatrix} 0 \\ -2 \\ 3 \end{bmatrix} \begin{bmatrix} 1 \\ -3 \\ 1 \end{bmatrix}$.

Q: Is $\begin{bmatrix} 4 \\ -3 \\ -5 \end{bmatrix}$ a linear combination of $\begin{bmatrix} 2 \\ -3 \\ -1 \end{bmatrix}$ and $\begin{bmatrix} -3 \\ 3 \\ 3 \end{bmatrix}$?
A: $\begin{bmatrix} 4 \\ -3 \\ -5 \end{bmatrix}$ is a linear combination of $\begin{bmatrix} 2 \\ -3 \\ -1 \end{bmatrix}$ and $\begin{bmatrix} -3 \\ 3 \\ 3 \end{bmatrix}$ if $\begin{bmatrix} 4 \\ -3 \\ -5 \end{bmatrix} = c_{1}\begin{bmatrix} 2 \\ -3 \\ -1 \end{bmatrix} + c_{2}\begin{bmatrix} -3 \\ 3 \\ 3 \end{bmatrix}$ for some scalars $c_1, c_2$.
This is the linear system $\left[\begin{array}{cc|c} 2 & -3 & 4 \\ -3 & 3 & -3 \\ -1 & 3 & -5 \end{array}\right]$, which row reduces to $\left[\begin{array}{cc|c} 1 & 0 & -1 \\ 0 & 1 & -2 \\ 0 & 0 & 0 \end{array}\right]$.
$c_{1} = -1$, $c_{2} = -2$ is a solution.
So, $\begin{bmatrix} 4 \\ -3 \\ -5 \end{bmatrix} = -\begin{bmatrix} 2 \\ -3 \\ -1 \end{bmatrix} -2\begin{bmatrix} -3 \\ 3 \\ 3 \end{bmatrix}$.

Q: Is the vector $\begin{bmatrix} 2 \\ 3 \\ 0 \end{bmatrix}$ in the span of $\left\{ \begin{bmatrix} -3 \\ -1 \\ -3 \end{bmatrix}, \begin{bmatrix} -1 \\ 2 \\ -3 \end{bmatrix} \right\}$?
A: $\begin{bmatrix} 2 \\ 3 \\ 0 \end{bmatrix}$ is in the span of $\begin{bmatrix} -3 \\ -1 \\ -3 \end{bmatrix}$ and $\begin{bmatrix} -1 \\ 2 \\ -3 \end{bmatrix}$ if $\begin{bmatrix} 2 \\ 3 \\ 0 \end{bmatrix} = c_{1}\begin{bmatrix} -3 \\ -1 \\ -3 \end{bmatrix} + c_{2}\begin{bmatrix} -1 \\ 2 \\ -3 \end{bmatrix}$ for some scalars $c_1, c_2$.
This is the linear system $\left[\begin{array}{cc|c} -3 & -1 & 2 \\ -1 & 2 & 3 \\ -3 & -3 & 0 \end{array}\right]$, which row reduces to $\left[\begin{array}{cc|c} 1 & 0 & -1 \\ 0 & 1 & 1 \\ 0 & 0 & 0 \end{array}\right]$.
$c_{1} = -1$, $c_{2} = 1$ is a solution.
So, $\begin{bmatrix} 2 \\ 3 \\ 0 \end{bmatrix} = -\begin{bmatrix} -3 \\ -1 \\ -3 \end{bmatrix} \begin{bmatrix} -1 \\ 2 \\ -3 \end{bmatrix}$.

Q: Is $\begin{bmatrix} 6 \\ -5 \\ 5 \end{bmatrix}$ a linear combination of $\begin{bmatrix} 3 \\ -1 \\ 3 \end{bmatrix}$ and $\begin{bmatrix} 0 \\ 3 \\ 1 \end{bmatrix}$?
A: $\begin{bmatrix} 6 \\ -5 \\ 5 \end{bmatrix}$ is a linear combination of $\begin{bmatrix} 3 \\ -1 \\ 3 \end{bmatrix}$ and $\begin{bmatrix} 0 \\ 3 \\ 1 \end{bmatrix}$ if $\begin{bmatrix} 6 \\ -5 \\ 5 \end{bmatrix} = c_{1}\begin{bmatrix} 3 \\ -1 \\ 3 \end{bmatrix} + c_{2}\begin{bmatrix} 0 \\ 3 \\ 1 \end{bmatrix}$ for some scalars $c_1, c_2$.
This is the linear system $\left[\begin{array}{cc|c} 3 & 0 & 6 \\ -1 & 3 & -5 \\ 3 & 1 & 5 \end{array}\right]$, which row reduces to $\left[\begin{array}{cc|c} 1 & 0 & 2 \\ 0 & 1 & -1 \\ 0 & 0 & 0 \end{array}\right]$.
$c_{1} = 2$, $c_{2} = -1$ is a solution.
So, $\begin{bmatrix} 6 \\ -5 \\ 5 \end{bmatrix} = 2\begin{bmatrix} 3 \\ -1 \\ 3 \end{bmatrix} -\begin{bmatrix} 0 \\ 3 \\ 1 \end{bmatrix}$.

Q: Is $\begin{bmatrix} 4 \\ -4 \\ 2 \end{bmatrix}$ a linear combination of $\begin{bmatrix} 1 \\ 2 \\ -3 \end{bmatrix}$ and $\begin{bmatrix} 3 \\ 0 \\ -2 \end{bmatrix}$?
A: $\begin{bmatrix} 4 \\ -4 \\ 2 \end{bmatrix}$ is a linear combination of $\begin{bmatrix} 1 \\ 2 \\ -3 \end{bmatrix}$ and $\begin{bmatrix} 3 \\ 0 \\ -2 \end{bmatrix}$ if $\begin{bmatrix} 4 \\ -4 \\ 2 \end{bmatrix} = c_{1}\begin{bmatrix} 1 \\ 2 \\ -3 \end{bmatrix} + c_{2}\begin{bmatrix} 3 \\ 0 \\ -2 \end{bmatrix}$ for some scalars $c_1, c_2$.
This is the linear system $\left[\begin{array}{cc|c} 1 & 3 & 4 \\ 2 & 0 & -4 \\ -3 & -2 & 2 \end{array}\right]$, which row reduces to $\left[\begin{array}{cc|c} 1 & 0 & -2 \\ 0 & 1 & 2 \\ 0 & 0 & 0 \end{array}\right]$.
$c_{1} = -2$, $c_{2} = 2$ is a solution.
So, $\begin{bmatrix} 4 \\ -4 \\ 2 \end{bmatrix} = -2\begin{bmatrix} 1 \\ 2 \\ -3 \end{bmatrix} + 2\begin{bmatrix} 3 \\ 0 \\ -2 \end{bmatrix}$.

Q: Is the vector $\begin{bmatrix} -5 \\ 0 \\ 8 \end{bmatrix}$ in the span of $\left\{ \begin{bmatrix} 1 \\ 0 \\ 2 \end{bmatrix}, \begin{bmatrix} 3 \\ 0 \\ -3 \end{bmatrix} \right\}$?
A: $\begin{bmatrix} -5 \\ 0 \\ 8 \end{bmatrix}$ is in the span of $\begin{bmatrix} 1 \\ 0 \\ 2 \end{bmatrix}$ and $\begin{bmatrix} 3 \\ 0 \\ -3 \end{bmatrix}$ if $\begin{bmatrix} -5 \\ 0 \\ 8 \end{bmatrix} = c_{1}\begin{bmatrix} 1 \\ 0 \\ 2 \end{bmatrix} + c_{2}\begin{bmatrix} 3 \\ 0 \\ -3 \end{bmatrix}$ for some scalars $c_1, c_2$.
This is the linear system $\left[\begin{array}{cc|c} 1 & 3 & -5 \\ 0 & 0 & 0 \\ 2 & -3 & 8 \end{array}\right]$, which row reduces to $\left[\begin{array}{cc|c} 1 & 0 & 1 \\ 0 & 1 & -2 \\ 0 & 0 & 0 \end{array}\right]$.
$c_{1} = 1$, $c_{2} = -2$ is a solution.
So, $\begin{bmatrix} -5 \\ 0 \\ 8 \end{bmatrix} = \begin{bmatrix} 1 \\ 0 \\ 2 \end{bmatrix} -2\begin{bmatrix} 3 \\ 0 \\ -3 \end{bmatrix}$.

Q: Is the vector $\begin{bmatrix} 0 \\ -5 \\ 5 \end{bmatrix}$ in the span of $\left\{ \begin{bmatrix} 0 \\ -3 \\ 3 \end{bmatrix}, \begin{bmatrix} 0 \\ -2 \\ 2 \end{bmatrix} \right\}$?
A: $\begin{bmatrix} 0 \\ -5 \\ 5 \end{bmatrix}$ is in the span of $\begin{bmatrix} 0 \\ -3 \\ 3 \end{bmatrix}$ and $\begin{bmatrix} 0 \\ -2 \\ 2 \end{bmatrix}$ if $\begin{bmatrix} 0 \\ -5 \\ 5 \end{bmatrix} = c_{1}\begin{bmatrix} 0 \\ -3 \\ 3 \end{bmatrix} + c_{2}\begin{bmatrix} 0 \\ -2 \\ 2 \end{bmatrix}$ for some scalars $c_1, c_2$.
This is the linear system $\left[\begin{array}{cc|c} 0 & 0 & 0 \\ -3 & -2 & -5 \\ 3 & 2 & 5 \end{array}\right]$, which row reduces to $\left[\begin{array}{cc|c} 1 & 0.67 & 1.67 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right]$.
$c_{1} = 1.6666666667$, $c_{2} = 0$ is a solution.
So, $\begin{bmatrix} 0 \\ -5 \\ 5 \end{bmatrix} = 1.6666666667\begin{bmatrix} 0 \\ -3 \\ 3 \end{bmatrix}$.

Q: Is the vector $\begin{bmatrix} 1 \\ -1 \\ 5 \end{bmatrix}$ in the span of $\left\{ \begin{bmatrix} 3 \\ -3 \\ -3 \end{bmatrix}, \begin{bmatrix} -2 \\ 2 \\ -1 \end{bmatrix} \right\}$?
A: $\begin{bmatrix} 1 \\ -1 \\ 5 \end{bmatrix}$ is in the span of $\begin{bmatrix} 3 \\ -3 \\ -3 \end{bmatrix}$ and $\begin{bmatrix} -2 \\ 2 \\ -1 \end{bmatrix}$ if $\begin{bmatrix} 1 \\ -1 \\ 5 \end{bmatrix} = c_{1}\begin{bmatrix} 3 \\ -3 \\ -3 \end{bmatrix} + c_{2}\begin{bmatrix} -2 \\ 2 \\ -1 \end{bmatrix}$ for some scalars $c_1, c_2$.
This is the linear system $\left[\begin{array}{cc|c} 3 & -2 & 1 \\ -3 & 2 & -1 \\ -3 & -1 & 5 \end{array}\right]$, which row reduces to $\left[\begin{array}{cc|c} 1 & 0 & -1 \\ 0 & 1 & -2 \\ 0 & 0 & 0 \end{array}\right]$.
$c_{1} = -1$, $c_{2} = -2$ is a solution.
So, $\begin{bmatrix} 1 \\ -1 \\ 5 \end{bmatrix} = -\begin{bmatrix} 3 \\ -3 \\ -3 \end{bmatrix} -2\begin{bmatrix} -2 \\ 2 \\ -1 \end{bmatrix}$.

Q: Is $\begin{bmatrix} 3 \\ 5 \\ 5 \end{bmatrix}$ a linear combination of $\begin{bmatrix} -3 \\ 1 \\ 1 \end{bmatrix}$ and $\begin{bmatrix} 3 \\ 2 \\ 2 \end{bmatrix}$?
A: $\begin{bmatrix} 3 \\ 5 \\ 5 \end{bmatrix}$ is a linear combination of $\begin{bmatrix} -3 \\ 1 \\ 1 \end{bmatrix}$ and $\begin{bmatrix} 3 \\ 2 \\ 2 \end{bmatrix}$ if $\begin{bmatrix} 3 \\ 5 \\ 5 \end{bmatrix} = c_{1}\begin{bmatrix} -3 \\ 1 \\ 1 \end{bmatrix} + c_{2}\begin{bmatrix} 3 \\ 2 \\ 2 \end{bmatrix}$ for some scalars $c_1, c_2$.
This is the linear system $\left[\begin{array}{cc|c} -3 & 3 & 3 \\ 1 & 2 & 5 \\ 1 & 2 & 5 \end{array}\right]$, which row reduces to $\left[\begin{array}{cc|c} 1 & 0 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & 0 \end{array}\right]$.
$c_{1} = 1$, $c_{2} = 2$ is a solution.
So, $\begin{bmatrix} 3 \\ 5 \\ 5 \end{bmatrix} = \begin{bmatrix} -3 \\ 1 \\ 1 \end{bmatrix} + 2\begin{bmatrix} 3 \\ 2 \\ 2 \end{bmatrix}$.

Q: Is $\begin{bmatrix} 0 \\ 5 \\ -4 \end{bmatrix}$ a linear combination of $\begin{bmatrix} -3 \\ 2 \\ -2 \end{bmatrix}$ and $\begin{bmatrix} 3 \\ 3 \\ -2 \end{bmatrix}$?
A: $\begin{bmatrix} 0 \\ 5 \\ -4 \end{bmatrix}$ is a linear combination of $\begin{bmatrix} -3 \\ 2 \\ -2 \end{bmatrix}$ and $\begin{bmatrix} 3 \\ 3 \\ -2 \end{bmatrix}$ if $\begin{bmatrix} 0 \\ 5 \\ -4 \end{bmatrix} = c_{1}\begin{bmatrix} -3 \\ 2 \\ -2 \end{bmatrix} + c_{2}\begin{bmatrix} 3 \\ 3 \\ -2 \end{bmatrix}$ for some scalars $c_1, c_2$.
This is the linear system $\left[\begin{array}{cc|c} -3 & 3 & 0 \\ 2 & 3 & 5 \\ -2 & -2 & -4 \end{array}\right]$, which row reduces to $\left[\begin{array}{cc|c} 1 & 0 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 0 \end{array}\right]$.
$c_{1} = 1$, $c_{2} = 1$ is a solution.
So, $\begin{bmatrix} 0 \\ 5 \\ -4 \end{bmatrix} = \begin{bmatrix} -3 \\ 2 \\ -2 \end{bmatrix} \begin{bmatrix} 3 \\ 3 \\ -2 \end{bmatrix}$.

Q: Is the vector $\begin{bmatrix} 4 \\ -2 \\ 0 \end{bmatrix}$ in the span of $\left\{ \begin{bmatrix} -3 \\ 2 \\ 0 \end{bmatrix}, \begin{bmatrix} 1 \\ -1 \\ 0 \end{bmatrix} \right\}$?
A: $\begin{bmatrix} 4 \\ -2 \\ 0 \end{bmatrix}$ is in the span of $\begin{bmatrix} -3 \\ 2 \\ 0 \end{bmatrix}$ and $\begin{bmatrix} 1 \\ -1 \\ 0 \end{bmatrix}$ if $\begin{bmatrix} 4 \\ -2 \\ 0 \end{bmatrix} = c_{1}\begin{bmatrix} -3 \\ 2 \\ 0 \end{bmatrix} + c_{2}\begin{bmatrix} 1 \\ -1 \\ 0 \end{bmatrix}$ for some scalars $c_1, c_2$.
This is the linear system $\left[\begin{array}{cc|c} -3 & 1 & 4 \\ 2 & -1 & -2 \\ 0 & 0 & 0 \end{array}\right]$, which row reduces to $\left[\begin{array}{cc|c} 1 & 0 & -2 \\ 0 & 1 & -2 \\ 0 & 0 & 0 \end{array}\right]$.
$c_{1} = -2$, $c_{2} = -2$ is a solution.
So, $\begin{bmatrix} 4 \\ -2 \\ 0 \end{bmatrix} = -2\begin{bmatrix} -3 \\ 2 \\ 0 \end{bmatrix} -2\begin{bmatrix} 1 \\ -1 \\ 0 \end{bmatrix}$.

Q: Is the vector $\begin{bmatrix} 0 \\ 2 \\ -6 \end{bmatrix}$ in the span of $\left\{ \begin{bmatrix} 2 \\ 1 \\ -2 \end{bmatrix}, \begin{bmatrix} -2 \\ 0 \\ -1 \end{bmatrix} \right\}$?
A: $\begin{bmatrix} 0 \\ 2 \\ -6 \end{bmatrix}$ is in the span of $\begin{bmatrix} 2 \\ 1 \\ -2 \end{bmatrix}$ and $\begin{bmatrix} -2 \\ 0 \\ -1 \end{bmatrix}$ if $\begin{bmatrix} 0 \\ 2 \\ -6 \end{bmatrix} = c_{1}\begin{bmatrix} 2 \\ 1 \\ -2 \end{bmatrix} + c_{2}\begin{bmatrix} -2 \\ 0 \\ -1 \end{bmatrix}$ for some scalars $c_1, c_2$.
This is the linear system $\left[\begin{array}{cc|c} 2 & -2 & 0 \\ 1 & 0 & 2 \\ -2 & -1 & -6 \end{array}\right]$, which row reduces to $\left[\begin{array}{cc|c} 1 & 0 & 2 \\ 0 & 1 & 2 \\ 0 & 0 & 0 \end{array}\right]$.
$c_{1} = 2$, $c_{2} = 2$ is a solution.
So, $\begin{bmatrix} 0 \\ 2 \\ -6 \end{bmatrix} = 2\begin{bmatrix} 2 \\ 1 \\ -2 \end{bmatrix} + 2\begin{bmatrix} -2 \\ 0 \\ -1 \end{bmatrix}$.

Q: Is $\begin{bmatrix} 1 \\ 0 \\ 2 \end{bmatrix}$ a linear combination of $\begin{bmatrix} 1 \\ 1 \\ 2 \end{bmatrix}$ and $\begin{bmatrix} 0 \\ -1 \\ 0 \end{bmatrix}$?
A: $\begin{bmatrix} 1 \\ 0 \\ 2 \end{bmatrix}$ is a linear combination of $\begin{bmatrix} 1 \\ 1 \\ 2 \end{bmatrix}$ and $\begin{bmatrix} 0 \\ -1 \\ 0 \end{bmatrix}$ if $\begin{bmatrix} 1 \\ 0 \\ 2 \end{bmatrix} = c_{1}\begin{bmatrix} 1 \\ 1 \\ 2 \end{bmatrix} + c_{2}\begin{bmatrix} 0 \\ -1 \\ 0 \end{bmatrix}$ for some scalars $c_1, c_2$.
This is the linear system $\left[\begin{array}{cc|c} 1 & 0 & 1 \\ 1 & -1 & 0 \\ 2 & 0 & 2 \end{array}\right]$, which row reduces to $\left[\begin{array}{cc|c} 1 & 0 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 0 \end{array}\right]$.
$c_{1} = 1$, $c_{2} = 1$ is a solution.
So, $\begin{bmatrix} 1 \\ 0 \\ 2 \end{bmatrix} = \begin{bmatrix} 1 \\ 1 \\ 2 \end{bmatrix} \begin{bmatrix} 0 \\ -1 \\ 0 \end{bmatrix}$.

Q: Is $\begin{bmatrix} 4 \\ 0 \\ 1 \end{bmatrix}$ a linear combination of $\begin{bmatrix} -3 \\ 3 \\ -3 \end{bmatrix}$ and $\begin{bmatrix} 1 \\ 3 \\ -2 \end{bmatrix}$?
A: $\begin{bmatrix} 4 \\ 0 \\ 1 \end{bmatrix}$ is a linear combination of $\begin{bmatrix} -3 \\ 3 \\ -3 \end{bmatrix}$ and $\begin{bmatrix} 1 \\ 3 \\ -2 \end{bmatrix}$ if $\begin{bmatrix} 4 \\ 0 \\ 1 \end{bmatrix} = c_{1}\begin{bmatrix} -3 \\ 3 \\ -3 \end{bmatrix} + c_{2}\begin{bmatrix} 1 \\ 3 \\ -2 \end{bmatrix}$ for some scalars $c_1, c_2$.
This is the linear system $\left[\begin{array}{cc|c} -3 & 1 & 4 \\ 3 & 3 & 0 \\ -3 & -2 & 1 \end{array}\right]$, which row reduces to $\left[\begin{array}{cc|c} 1 & 0 & -1 \\ 0 & 1 & 1 \\ 0 & 0 & 0 \end{array}\right]$.
$c_{1} = -1$, $c_{2} = 1$ is a solution.
So, $\begin{bmatrix} 4 \\ 0 \\ 1 \end{bmatrix} = -\begin{bmatrix} -3 \\ 3 \\ -3 \end{bmatrix} \begin{bmatrix} 1 \\ 3 \\ -2 \end{bmatrix}$.

Q: Is $\begin{bmatrix} -4 \\ -6 \\ -3 \end{bmatrix}$ a linear combination of $\begin{bmatrix} 0 \\ 2 \\ -1 \end{bmatrix}$ and $\begin{bmatrix} 2 \\ 2 \\ 2 \end{bmatrix}$?
A: $\begin{bmatrix} -4 \\ -6 \\ -3 \end{bmatrix}$ is a linear combination of $\begin{bmatrix} 0 \\ 2 \\ -1 \end{bmatrix}$ and $\begin{bmatrix} 2 \\ 2 \\ 2 \end{bmatrix}$ if $\begin{bmatrix} -4 \\ -6 \\ -3 \end{bmatrix} = c_{1}\begin{bmatrix} 0 \\ 2 \\ -1 \end{bmatrix} + c_{2}\begin{bmatrix} 2 \\ 2 \\ 2 \end{bmatrix}$ for some scalars $c_1, c_2$.
This is the linear system $\left[\begin{array}{cc|c} 0 & 2 & -4 \\ 2 & 2 & -6 \\ -1 & 2 & -3 \end{array}\right]$, which row reduces to $\left[\begin{array}{cc|c} 1 & 0 & -1 \\ 0 & 1 & -2 \\ 0 & 0 & 0 \end{array}\right]$.
$c_{1} = -1$, $c_{2} = -2$ is a solution.
So, $\begin{bmatrix} -4 \\ -6 \\ -3 \end{bmatrix} = -\begin{bmatrix} 0 \\ 2 \\ -1 \end{bmatrix} -2\begin{bmatrix} 2 \\ 2 \\ 2 \end{bmatrix}$.

Q: Is $\begin{bmatrix} 5 \\ 3 \\ -2 \end{bmatrix}$ a linear combination of $\begin{bmatrix} -1 \\ -3 \\ 0 \end{bmatrix}$ and $\begin{bmatrix} -3 \\ -3 \\ 1 \end{bmatrix}$?
A: $\begin{bmatrix} 5 \\ 3 \\ -2 \end{bmatrix}$ is a linear combination of $\begin{bmatrix} -1 \\ -3 \\ 0 \end{bmatrix}$ and $\begin{bmatrix} -3 \\ -3 \\ 1 \end{bmatrix}$ if $\begin{bmatrix} 5 \\ 3 \\ -2 \end{bmatrix} = c_{1}\begin{bmatrix} -1 \\ -3 \\ 0 \end{bmatrix} + c_{2}\begin{bmatrix} -3 \\ -3 \\ 1 \end{bmatrix}$ for some scalars $c_1, c_2$.
This is the linear system $\left[\begin{array}{cc|c} -1 & -3 & 5 \\ -3 & -3 & 3 \\ 0 & 1 & -2 \end{array}\right]$, which row reduces to $\left[\begin{array}{cc|c} 1 & 0 & 1 \\ 0 & 1 & -2 \\ 0 & 0 & 0 \end{array}\right]$.
$c_{1} = 1$, $c_{2} = -2$ is a solution.
So, $\begin{bmatrix} 5 \\ 3 \\ -2 \end{bmatrix} = \begin{bmatrix} -1 \\ -3 \\ 0 \end{bmatrix} -2\begin{bmatrix} -3 \\ -3 \\ 1 \end{bmatrix}$.

Q: Is the vector $\begin{bmatrix} -2 \\ 6 \\ -2 \end{bmatrix}$ in the span of $\left\{ \begin{bmatrix} 1 \\ 3 \\ -1 \end{bmatrix}, \begin{bmatrix} -2 \\ 0 \\ 0 \end{bmatrix} \right\}$?
A: $\begin{bmatrix} -2 \\ 6 \\ -2 \end{bmatrix}$ is in the span of $\begin{bmatrix} 1 \\ 3 \\ -1 \end{bmatrix}$ and $\begin{bmatrix} -2 \\ 0 \\ 0 \end{bmatrix}$ if $\begin{bmatrix} -2 \\ 6 \\ -2 \end{bmatrix} = c_{1}\begin{bmatrix} 1 \\ 3 \\ -1 \end{bmatrix} + c_{2}\begin{bmatrix} -2 \\ 0 \\ 0 \end{bmatrix}$ for some scalars $c_1, c_2$.
This is the linear system $\left[\begin{array}{cc|c} 1 & -2 & -2 \\ 3 & 0 & 6 \\ -1 & 0 & -2 \end{array}\right]$, which row reduces to $\left[\begin{array}{cc|c} 1 & 0 & 2 \\ 0 & 1 & 2 \\ 0 & 0 & 0 \end{array}\right]$.
$c_{1} = 2$, $c_{2} = 2$ is a solution.
So, $\begin{bmatrix} -2 \\ 6 \\ -2 \end{bmatrix} = 2\begin{bmatrix} 1 \\ 3 \\ -1 \end{bmatrix} + 2\begin{bmatrix} -2 \\ 0 \\ 0 \end{bmatrix}$.

Q: Is the vector $\begin{bmatrix} -4 \\ 1 \\ -8 \end{bmatrix}$ in the span of $\left\{ \begin{bmatrix} -1 \\ -2 \\ 2 \end{bmatrix}, \begin{bmatrix} 3 \\ 2 \\ 3 \end{bmatrix}, \begin{bmatrix} -1 \\ -3 \\ 0 \end{bmatrix} \right\}$?
A: $\begin{bmatrix} -4 \\ 1 \\ -8 \end{bmatrix}$ is in the span of $\begin{bmatrix} -1 \\ -2 \\ 2 \end{bmatrix}$, $\begin{bmatrix} 3 \\ 2 \\ 3 \end{bmatrix}$, and $\begin{bmatrix} -1 \\ -3 \\ 0 \end{bmatrix}$ if $\begin{bmatrix} -4 \\ 1 \\ -8 \end{bmatrix} = c_{1}\begin{bmatrix} -1 \\ -2 \\ 2 \end{bmatrix} + c_{2}\begin{bmatrix} 3 \\ 2 \\ 3 \end{bmatrix} + c_{3}\begin{bmatrix} -1 \\ -3 \\ 0 \end{bmatrix}$ for some scalars $c_1, c_2, c_3$.
This is the linear system $\left[\begin{array}{ccc|c} -1 & 3 & -1 & -4 \\ -2 & 2 & -3 & 1 \\ 2 & 3 & 0 & -8 \end{array}\right]$, which row reduces to $\left[\begin{array}{ccc|c} 1 & 0 & 0 & -1 \\ 0 & 1 & 0 & -2 \\ 0 & 0 & 1 & -1 \end{array}\right]$.
$c_{1} = -1$, $c_{2} = -2$, $c_{3} = -1$ is a solution.
So, $\begin{bmatrix} -4 \\ 1 \\ -8 \end{bmatrix} = -\begin{bmatrix} -1 \\ -2 \\ 2 \end{bmatrix} -2\begin{bmatrix} 3 \\ 2 \\ 3 \end{bmatrix} -\begin{bmatrix} -1 \\ -3 \\ 0 \end{bmatrix}$.

Q: Is the vector $\begin{bmatrix} 2 \\ 0 \\ 4 \end{bmatrix}$ in the span of $\left\{ \begin{bmatrix} 1 \\ -2 \\ -2 \end{bmatrix}, \begin{bmatrix} 2 \\ -3 \\ 0 \end{bmatrix}, \begin{bmatrix} -1 \\ -1 \\ -2 \end{bmatrix} \right\}$?
A: $\begin{bmatrix} 2 \\ 0 \\ 4 \end{bmatrix}$ is in the span of $\begin{bmatrix} 1 \\ -2 \\ -2 \end{bmatrix}$, $\begin{bmatrix} 2 \\ -3 \\ 0 \end{bmatrix}$, and $\begin{bmatrix} -1 \\ -1 \\ -2 \end{bmatrix}$ if $\begin{bmatrix} 2 \\ 0 \\ 4 \end{bmatrix} = c_{1}\begin{bmatrix} 1 \\ -2 \\ -2 \end{bmatrix} + c_{2}\begin{bmatrix} 2 \\ -3 \\ 0 \end{bmatrix} + c_{3}\begin{bmatrix} -1 \\ -1 \\ -2 \end{bmatrix}$ for some scalars $c_1, c_2, c_3$.
This is the linear system $\left[\begin{array}{ccc|c} 1 & 2 & -1 & 2 \\ -2 & -3 & -1 & 0 \\ -2 & 0 & -2 & 4 \end{array}\right]$, which row reduces to $\left[\begin{array}{ccc|c} 1 & 0 & 0 & -1 \\ 0 & 1 & 0 & 1 \\ 0 & 0 & 1 & -1 \end{array}\right]$.
$c_{1} = -1$, $c_{2} = 1$, $c_{3} = -1$ is a solution.
So, $\begin{bmatrix} 2 \\ 0 \\ 4 \end{bmatrix} = -\begin{bmatrix} 1 \\ -2 \\ -2 \end{bmatrix} \begin{bmatrix} 2 \\ -3 \\ 0 \end{bmatrix} -\begin{bmatrix} -1 \\ -1 \\ -2 \end{bmatrix}$.

Q: Is the vector $\begin{bmatrix} -5 \\ -5 \\ -2 \end{bmatrix}$ in the span of $\left\{ \begin{bmatrix} 0 \\ -1 \\ 1 \end{bmatrix}, \begin{bmatrix} 1 \\ 1 \\ 3 \end{bmatrix}, \begin{bmatrix} -3 \\ -1 \\ 2 \end{bmatrix} \right\}$?
A: $\begin{bmatrix} -5 \\ -5 \\ -2 \end{bmatrix}$ is in the span of $\begin{bmatrix} 0 \\ -1 \\ 1 \end{bmatrix}$, $\begin{bmatrix} 1 \\ 1 \\ 3 \end{bmatrix}$, and $\begin{bmatrix} -3 \\ -1 \\ 2 \end{bmatrix}$ if $\begin{bmatrix} -5 \\ -5 \\ -2 \end{bmatrix} = c_{1}\begin{bmatrix} 0 \\ -1 \\ 1 \end{bmatrix} + c_{2}\begin{bmatrix} 1 \\ 1 \\ 3 \end{bmatrix} + c_{3}\begin{bmatrix} -3 \\ -1 \\ 2 \end{bmatrix}$ for some scalars $c_1, c_2, c_3$.
This is the linear system $\left[\begin{array}{ccc|c} 0 & 1 & -3 & -5 \\ -1 & 1 & -1 & -5 \\ 1 & 3 & 2 & -2 \end{array}\right]$, which row reduces to $\left[\begin{array}{ccc|c} 1 & 0 & 0 & 2 \\ 0 & 1 & 0 & -2 \\ 0 & 0 & 1 & 1 \end{array}\right]$.
$c_{1} = 2$, $c_{2} = -2$, $c_{3} = 1$ is a solution.
So, $\begin{bmatrix} -5 \\ -5 \\ -2 \end{bmatrix} = 2\begin{bmatrix} 0 \\ -1 \\ 1 \end{bmatrix} -2\begin{bmatrix} 1 \\ 1 \\ 3 \end{bmatrix} \begin{bmatrix} -3 \\ -1 \\ 2 \end{bmatrix}$.

Q: Is $\begin{bmatrix} 0 \\ 2 \\ 5 \end{bmatrix}$ a linear combination of $\begin{bmatrix} 2 \\ 1 \\ 2 \end{bmatrix}$, $\begin{bmatrix} 3 \\ 2 \\ -3 \end{bmatrix}$, and $\begin{bmatrix} -1 \\ -3 \\ 0 \end{bmatrix}$?
A: $\begin{bmatrix} 0 \\ 2 \\ 5 \end{bmatrix}$ is a linear combination of $\begin{bmatrix} 2 \\ 1 \\ 2 \end{bmatrix}$, $\begin{bmatrix} 3 \\ 2 \\ -3 \end{bmatrix}$, and $\begin{bmatrix} -1 \\ -3 \\ 0 \end{bmatrix}$ if $\begin{bmatrix} 0 \\ 2 \\ 5 \end{bmatrix} = c_{1}\begin{bmatrix} 2 \\ 1 \\ 2 \end{bmatrix} + c_{2}\begin{bmatrix} 3 \\ 2 \\ -3 \end{bmatrix} + c_{3}\begin{bmatrix} -1 \\ -3 \\ 0 \end{bmatrix}$ for some scalars $c_1, c_2, c_3$.
This is the linear system $\left[\begin{array}{ccc|c} 2 & 3 & -1 & 0 \\ 1 & 2 & -3 & 2 \\ 2 & -3 & 0 & 5 \end{array}\right]$, which row reduces to $\left[\begin{array}{ccc|c} 1 & 0 & 0 & 1 \\ 0 & 1 & 0 & -1 \\ 0 & 0 & 1 & -1 \end{array}\right]$.
$c_{1} = 1$, $c_{2} = -1$, $c_{3} = -1$ is a solution.
So, $\begin{bmatrix} 0 \\ 2 \\ 5 \end{bmatrix} = \begin{bmatrix} 2 \\ 1 \\ 2 \end{bmatrix} -\begin{bmatrix} 3 \\ 2 \\ -3 \end{bmatrix} -\begin{bmatrix} -1 \\ -3 \\ 0 \end{bmatrix}$.

Q: Is the vector $\begin{bmatrix} -4 \\ 6 \\ -1 \end{bmatrix}$ in the span of $\left\{ \begin{bmatrix} -3 \\ 1 \\ -2 \end{bmatrix}, \begin{bmatrix} -2 \\ 0 \\ 3 \end{bmatrix}, \begin{bmatrix} 0 \\ 2 \\ 3 \end{bmatrix} \right\}$?
A: $\begin{bmatrix} -4 \\ 6 \\ -1 \end{bmatrix}$ is in the span of $\begin{bmatrix} -3 \\ 1 \\ -2 \end{bmatrix}$, $\begin{bmatrix} -2 \\ 0 \\ 3 \end{bmatrix}$, and $\begin{bmatrix} 0 \\ 2 \\ 3 \end{bmatrix}$ if $\begin{bmatrix} -4 \\ 6 \\ -1 \end{bmatrix} = c_{1}\begin{bmatrix} -3 \\ 1 \\ -2 \end{bmatrix} + c_{2}\begin{bmatrix} -2 \\ 0 \\ 3 \end{bmatrix} + c_{3}\begin{bmatrix} 0 \\ 2 \\ 3 \end{bmatrix}$ for some scalars $c_1, c_2, c_3$.
This is the linear system $\left[\begin{array}{ccc|c} -3 & -2 & 0 & -4 \\ 1 & 0 & 2 & 6 \\ -2 & 3 & 3 & -1 \end{array}\right]$, which row reduces to $\left[\begin{array}{ccc|c} 1 & 0 & 0 & 2 \\ 0 & 1 & 0 & -1 \\ 0 & 0 & 1 & 2 \end{array}\right]$.
$c_{1} = 2$, $c_{2} = -1$, $c_{3} = 2$ is a solution.
So, $\begin{bmatrix} -4 \\ 6 \\ -1 \end{bmatrix} = 2\begin{bmatrix} -3 \\ 1 \\ -2 \end{bmatrix} -\begin{bmatrix} -2 \\ 0 \\ 3 \end{bmatrix} + 2\begin{bmatrix} 0 \\ 2 \\ 3 \end{bmatrix}$.

Q: Is the vector $\begin{bmatrix} -9 \\ -3 \\ 5 \end{bmatrix}$ in the span of $\left\{ \begin{bmatrix} -3 \\ -3 \\ -3 \end{bmatrix}, \begin{bmatrix} 0 \\ -3 \\ 2 \end{bmatrix}, \begin{bmatrix} -3 \\ 3 \\ 2 \end{bmatrix} \right\}$?
A: $\begin{bmatrix} -9 \\ -3 \\ 5 \end{bmatrix}$ is in the span of $\begin{bmatrix} -3 \\ -3 \\ -3 \end{bmatrix}$, $\begin{bmatrix} 0 \\ -3 \\ 2 \end{bmatrix}$, and $\begin{bmatrix} -3 \\ 3 \\ 2 \end{bmatrix}$ if $\begin{bmatrix} -9 \\ -3 \\ 5 \end{bmatrix} = c_{1}\begin{bmatrix} -3 \\ -3 \\ -3 \end{bmatrix} + c_{2}\begin{bmatrix} 0 \\ -3 \\ 2 \end{bmatrix} + c_{3}\begin{bmatrix} -3 \\ 3 \\ 2 \end{bmatrix}$ for some scalars $c_1, c_2, c_3$.
This is the linear system $\left[\begin{array}{ccc|c} -3 & 0 & -3 & -9 \\ -3 & -3 & 3 & -3 \\ -3 & 2 & 2 & 5 \end{array}\right]$, which row reduces to $\left[\begin{array}{ccc|c} 1 & 0 & 0 & 1 \\ 0 & 1 & 0 & 2 \\ 0 & 0 & 1 & 2 \end{array}\right]$.
$c_{1} = 1$, $c_{2} = 2$, $c_{3} = 2$ is a solution.
So, $\begin{bmatrix} -9 \\ -3 \\ 5 \end{bmatrix} = \begin{bmatrix} -3 \\ -3 \\ -3 \end{bmatrix} + 2\begin{bmatrix} 0 \\ -3 \\ 2 \end{bmatrix} + 2\begin{bmatrix} -3 \\ 3 \\ 2 \end{bmatrix}$.

Q: Is the vector $\begin{bmatrix} 4 \\ 4 \\ -12 \end{bmatrix}$ in the span of $\left\{ \begin{bmatrix} 2 \\ 2 \\ -2 \end{bmatrix}, \begin{bmatrix} 3 \\ -3 \\ -2 \end{bmatrix}, \begin{bmatrix} 3 \\ -3 \\ 2 \end{bmatrix} \right\}$?
A: $\begin{bmatrix} 4 \\ 4 \\ -12 \end{bmatrix}$ is in the span of $\begin{bmatrix} 2 \\ 2 \\ -2 \end{bmatrix}$, $\begin{bmatrix} 3 \\ -3 \\ -2 \end{bmatrix}$, and $\begin{bmatrix} 3 \\ -3 \\ 2 \end{bmatrix}$ if $\begin{bmatrix} 4 \\ 4 \\ -12 \end{bmatrix} = c_{1}\begin{bmatrix} 2 \\ 2 \\ -2 \end{bmatrix} + c_{2}\begin{bmatrix} 3 \\ -3 \\ -2 \end{bmatrix} + c_{3}\begin{bmatrix} 3 \\ -3 \\ 2 \end{bmatrix}$ for some scalars $c_1, c_2, c_3$.
This is the linear system $\left[\begin{array}{ccc|c} 2 & 3 & 3 & 4 \\ 2 & -3 & -3 & 4 \\ -2 & -2 & 2 & -12 \end{array}\right]$, which row reduces to $\left[\begin{array}{ccc|c} 1 & 0 & 0 & 2 \\ 0 & 1 & 0 & 2 \\ 0 & 0 & 1 & -2 \end{array}\right]$.
$c_{1} = 2$, $c_{2} = 2$, $c_{3} = -2$ is a solution.
So, $\begin{bmatrix} 4 \\ 4 \\ -12 \end{bmatrix} = 2\begin{bmatrix} 2 \\ 2 \\ -2 \end{bmatrix} + 2\begin{bmatrix} 3 \\ -3 \\ -2 \end{bmatrix} -2\begin{bmatrix} 3 \\ -3 \\ 2 \end{bmatrix}$.

Q: Is the vector $\begin{bmatrix} 5 \\ 2 \\ 3 \end{bmatrix}$ in the span of $\left\{ \begin{bmatrix} 1 \\ -1 \\ 0 \end{bmatrix}, \begin{bmatrix} -2 \\ -3 \\ -3 \end{bmatrix}, \begin{bmatrix} 2 \\ -1 \\ 0 \end{bmatrix} \right\}$?
A: $\begin{bmatrix} 5 \\ 2 \\ 3 \end{bmatrix}$ is in the span of $\begin{bmatrix} 1 \\ -1 \\ 0 \end{bmatrix}$, $\begin{bmatrix} -2 \\ -3 \\ -3 \end{bmatrix}$, and $\begin{bmatrix} 2 \\ -1 \\ 0 \end{bmatrix}$ if $\begin{bmatrix} 5 \\ 2 \\ 3 \end{bmatrix} = c_{1}\begin{bmatrix} 1 \\ -1 \\ 0 \end{bmatrix} + c_{2}\begin{bmatrix} -2 \\ -3 \\ -3 \end{bmatrix} + c_{3}\begin{bmatrix} 2 \\ -1 \\ 0 \end{bmatrix}$ for some scalars $c_1, c_2, c_3$.
This is the linear system $\left[\begin{array}{ccc|c} 1 & -2 & 2 & 5 \\ -1 & -3 & -1 & 2 \\ 0 & -3 & 0 & 3 \end{array}\right]$, which row reduces to $\left[\begin{array}{ccc|c} 1 & 0 & 0 & -1 \\ 0 & 1 & 0 & -1 \\ 0 & 0 & 1 & 2 \end{array}\right]$.
$c_{1} = -1$, $c_{2} = -1$, $c_{3} = 2$ is a solution.
So, $\begin{bmatrix} 5 \\ 2 \\ 3 \end{bmatrix} = -\begin{bmatrix} 1 \\ -1 \\ 0 \end{bmatrix} -\begin{bmatrix} -2 \\ -3 \\ -3 \end{bmatrix} + 2\begin{bmatrix} 2 \\ -1 \\ 0 \end{bmatrix}$.

Q: Is $\begin{bmatrix} 7 \\ -4 \\ -3 \end{bmatrix}$ a linear combination of $\begin{bmatrix} -3 \\ 2 \\ 0 \end{bmatrix}$, $\begin{bmatrix} -1 \\ 0 \\ 2 \end{bmatrix}$, and $\begin{bmatrix} -2 \\ 2 \\ -1 \end{bmatrix}$?
A: $\begin{bmatrix} 7 \\ -4 \\ -3 \end{bmatrix}$ is a linear combination of $\begin{bmatrix} -3 \\ 2 \\ 0 \end{bmatrix}$, $\begin{bmatrix} -1 \\ 0 \\ 2 \end{bmatrix}$, and $\begin{bmatrix} -2 \\ 2 \\ -1 \end{bmatrix}$ if $\begin{bmatrix} 7 \\ -4 \\ -3 \end{bmatrix} = c_{1}\begin{bmatrix} -3 \\ 2 \\ 0 \end{bmatrix} + c_{2}\begin{bmatrix} -1 \\ 0 \\ 2 \end{bmatrix} + c_{3}\begin{bmatrix} -2 \\ 2 \\ -1 \end{bmatrix}$ for some scalars $c_1, c_2, c_3$.
This is the linear system $\left[\begin{array}{ccc|c} -3 & -1 & -2 & 7 \\ 2 & 0 & 2 & -4 \\ 0 & 2 & -1 & -3 \end{array}\right]$, which row reduces to $\left[\begin{array}{ccc|c} 1 & 0 & 0 & -1 \\ 0 & 1 & 0 & -2 \\ 0 & 0 & 1 & -1 \end{array}\right]$.
$c_{1} = -1$, $c_{2} = -2$, $c_{3} = -1$ is a solution.
So, $\begin{bmatrix} 7 \\ -4 \\ -3 \end{bmatrix} = -\begin{bmatrix} -3 \\ 2 \\ 0 \end{bmatrix} -2\begin{bmatrix} -1 \\ 0 \\ 2 \end{bmatrix} -\begin{bmatrix} -2 \\ 2 \\ -1 \end{bmatrix}$.

Q: Is $\begin{bmatrix} 3 \\ -3 \\ -2 \end{bmatrix}$ a linear combination of $\begin{bmatrix} 1 \\ -3 \\ -2 \end{bmatrix}$, $\begin{bmatrix} 2 \\ -1 \\ 2 \end{bmatrix}$, and $\begin{bmatrix} 0 \\ 1 \\ -2 \end{bmatrix}$?
A: $\begin{bmatrix} 3 \\ -3 \\ -2 \end{bmatrix}$ is a linear combination of $\begin{bmatrix} 1 \\ -3 \\ -2 \end{bmatrix}$, $\begin{bmatrix} 2 \\ -1 \\ 2 \end{bmatrix}$, and $\begin{bmatrix} 0 \\ 1 \\ -2 \end{bmatrix}$ if $\begin{bmatrix} 3 \\ -3 \\ -2 \end{bmatrix} = c_{1}\begin{bmatrix} 1 \\ -3 \\ -2 \end{bmatrix} + c_{2}\begin{bmatrix} 2 \\ -1 \\ 2 \end{bmatrix} + c_{3}\begin{bmatrix} 0 \\ 1 \\ -2 \end{bmatrix}$ for some scalars $c_1, c_2, c_3$.
This is the linear system $\left[\begin{array}{ccc|c} 1 & 2 & 0 & 3 \\ -3 & -1 & 1 & -3 \\ -2 & 2 & -2 & -2 \end{array}\right]$, which row reduces to $\left[\begin{array}{ccc|c} 1 & 0 & 0 & 1 \\ 0 & 1 & 0 & 1 \\ 0 & 0 & 1 & 1 \end{array}\right]$.
$c_{1} = 1$, $c_{2} = 1$, $c_{3} = 1$ is a solution.
So, $\begin{bmatrix} 3 \\ -3 \\ -2 \end{bmatrix} = \begin{bmatrix} 1 \\ -3 \\ -2 \end{bmatrix} \begin{bmatrix} 2 \\ -1 \\ 2 \end{bmatrix} \begin{bmatrix} 0 \\ 1 \\ -2 \end{bmatrix}$.

Q: Is $\begin{bmatrix} 5 \\ -2 \\ 3 \end{bmatrix}$ a linear combination of $\begin{bmatrix} -2 \\ 0 \\ 0 \end{bmatrix}$, $\begin{bmatrix} 3 \\ 2 \\ -3 \end{bmatrix}$, and $\begin{bmatrix} 3 \\ -2 \\ 3 \end{bmatrix}$?
A: $\begin{bmatrix} 5 \\ -2 \\ 3 \end{bmatrix}$ is a linear combination of $\begin{bmatrix} -2 \\ 0 \\ 0 \end{bmatrix}$, $\begin{bmatrix} 3 \\ 2 \\ -3 \end{bmatrix}$, and $\begin{bmatrix} 3 \\ -2 \\ 3 \end{bmatrix}$ if $\begin{bmatrix} 5 \\ -2 \\ 3 \end{bmatrix} = c_{1}\begin{bmatrix} -2 \\ 0 \\ 0 \end{bmatrix} + c_{2}\begin{bmatrix} 3 \\ 2 \\ -3 \end{bmatrix} + c_{3}\begin{bmatrix} 3 \\ -2 \\ 3 \end{bmatrix}$ for some scalars $c_1, c_2, c_3$.
This is the linear system $\left[\begin{array}{ccc|c} -2 & 3 & 3 & 5 \\ 0 & 2 & -2 & -2 \\ 0 & -3 & 3 & 3 \end{array}\right]$, which row reduces to $\left[\begin{array}{ccc|c} 1 & 0 & -3 & -4 \\ 0 & 1 & -1 & -1 \\ 0 & 0 & 0 & 0 \end{array}\right]$.
$c_{1} = -4$, $c_{2} = -1$, $c_{3} = 0$ is a solution.
So, $\begin{bmatrix} 5 \\ -2 \\ 3 \end{bmatrix} = -4\begin{bmatrix} -2 \\ 0 \\ 0 \end{bmatrix} -\begin{bmatrix} 3 \\ 2 \\ -3 \end{bmatrix}$.

Q: Is $\begin{bmatrix} 0 \\ -6 \\ 0 \end{bmatrix}$ a linear combination of $\begin{bmatrix} -2 \\ 2 \\ -2 \end{bmatrix}$, $\begin{bmatrix} -1 \\ -3 \\ -2 \end{bmatrix}$, and $\begin{bmatrix} 1 \\ -2 \\ 0 \end{bmatrix}$?
A: $\begin{bmatrix} 0 \\ -6 \\ 0 \end{bmatrix}$ is a linear combination of $\begin{bmatrix} -2 \\ 2 \\ -2 \end{bmatrix}$, $\begin{bmatrix} -1 \\ -3 \\ -2 \end{bmatrix}$, and $\begin{bmatrix} 1 \\ -2 \\ 0 \end{bmatrix}$ if $\begin{bmatrix} 0 \\ -6 \\ 0 \end{bmatrix} = c_{1}\begin{bmatrix} -2 \\ 2 \\ -2 \end{bmatrix} + c_{2}\begin{bmatrix} -1 \\ -3 \\ -2 \end{bmatrix} + c_{3}\begin{bmatrix} 1 \\ -2 \\ 0 \end{bmatrix}$ for some scalars $c_1, c_2, c_3$.
This is the linear system $\left[\begin{array}{ccc|c} -2 & -1 & 1 & 0 \\ 2 & -3 & -2 & -6 \\ -2 & -2 & 0 & 0 \end{array}\right]$, which row reduces to $\left[\begin{array}{ccc|c} 1 & 0 & 0 & -2 \\ 0 & 1 & 0 & 2 \\ 0 & 0 & 1 & -2 \end{array}\right]$.
$c_{1} = -2$, $c_{2} = 2$, $c_{3} = -2$ is a solution.
So, $\begin{bmatrix} 0 \\ -6 \\ 0 \end{bmatrix} = -2\begin{bmatrix} -2 \\ 2 \\ -2 \end{bmatrix} + 2\begin{bmatrix} -1 \\ -3 \\ -2 \end{bmatrix} -2\begin{bmatrix} 1 \\ -2 \\ 0 \end{bmatrix}$.

Q: Is $\begin{bmatrix} 1 \\ -4 \\ -3 \end{bmatrix}$ a linear combination of $\begin{bmatrix} 2 \\ -1 \\ 0 \end{bmatrix}$, $\begin{bmatrix} 0 \\ 0 \\ -1 \end{bmatrix}$, and $\begin{bmatrix} 3 \\ 2 \\ 1 \end{bmatrix}$?
A: $\begin{bmatrix} 1 \\ -4 \\ -3 \end{bmatrix}$ is a linear combination of $\begin{bmatrix} 2 \\ -1 \\ 0 \end{bmatrix}$, $\begin{bmatrix} 0 \\ 0 \\ -1 \end{bmatrix}$, and $\begin{bmatrix} 3 \\ 2 \\ 1 \end{bmatrix}$ if $\begin{bmatrix} 1 \\ -4 \\ -3 \end{bmatrix} = c_{1}\begin{bmatrix} 2 \\ -1 \\ 0 \end{bmatrix} + c_{2}\begin{bmatrix} 0 \\ 0 \\ -1 \end{bmatrix} + c_{3}\begin{bmatrix} 3 \\ 2 \\ 1 \end{bmatrix}$ for some scalars $c_1, c_2, c_3$.
This is the linear system $\left[\begin{array}{ccc|c} 2 & 0 & 3 & 1 \\ -1 & 0 & 2 & -4 \\ 0 & -1 & 1 & -3 \end{array}\right]$, which row reduces to $\left[\begin{array}{ccc|c} 1 & 0 & 0 & 2 \\ 0 & 1 & 0 & 2 \\ 0 & 0 & 1 & -1 \end{array}\right]$.
$c_{1} = 2$, $c_{2} = 2$, $c_{3} = -1$ is a solution.
So, $\begin{bmatrix} 1 \\ -4 \\ -3 \end{bmatrix} = 2\begin{bmatrix} 2 \\ -1 \\ 0 \end{bmatrix} + 2\begin{bmatrix} 0 \\ 0 \\ -1 \end{bmatrix} -\begin{bmatrix} 3 \\ 2 \\ 1 \end{bmatrix}$.

Q: Is the vector $\begin{bmatrix} -9 \\ 9 \\ 3 \end{bmatrix}$ in the span of $\left\{ \begin{bmatrix} -2 \\ 3 \\ 2 \end{bmatrix}, \begin{bmatrix} 1 \\ -1 \\ -3 \end{bmatrix}, \begin{bmatrix} 2 \\ -1 \\ 2 \end{bmatrix} \right\}$?
A: $\begin{bmatrix} -9 \\ 9 \\ 3 \end{bmatrix}$ is in the span of $\begin{bmatrix} -2 \\ 3 \\ 2 \end{bmatrix}$, $\begin{bmatrix} 1 \\ -1 \\ -3 \end{bmatrix}$, and $\begin{bmatrix} 2 \\ -1 \\ 2 \end{bmatrix}$ if $\begin{bmatrix} -9 \\ 9 \\ 3 \end{bmatrix} = c_{1}\begin{bmatrix} -2 \\ 3 \\ 2 \end{bmatrix} + c_{2}\begin{bmatrix} 1 \\ -1 \\ -3 \end{bmatrix} + c_{3}\begin{bmatrix} 2 \\ -1 \\ 2 \end{bmatrix}$ for some scalars $c_1, c_2, c_3$.
This is the linear system $\left[\begin{array}{ccc|c} -2 & 1 & 2 & -9 \\ 3 & -1 & -1 & 9 \\ 2 & -3 & 2 & 3 \end{array}\right]$, which row reduces to $\left[\begin{array}{ccc|c} 1 & 0 & 0 & 2 \\ 0 & 1 & 0 & -1 \\ 0 & 0 & 1 & -2 \end{array}\right]$.
$c_{1} = 2$, $c_{2} = -1$, $c_{3} = -2$ is a solution.
So, $\begin{bmatrix} -9 \\ 9 \\ 3 \end{bmatrix} = 2\begin{bmatrix} -2 \\ 3 \\ 2 \end{bmatrix} -\begin{bmatrix} 1 \\ -1 \\ -3 \end{bmatrix} -2\begin{bmatrix} 2 \\ -1 \\ 2 \end{bmatrix}$.

Q: Is $\begin{bmatrix} 3 \\ 2 \\ -4 \end{bmatrix}$ a linear combination of $\begin{bmatrix} 2 \\ 0 \\ -2 \end{bmatrix}$, $\begin{bmatrix} -2 \\ -2 \\ 0 \end{bmatrix}$, and $\begin{bmatrix} 3 \\ 2 \\ 2 \end{bmatrix}$?
A: $\begin{bmatrix} 3 \\ 2 \\ -4 \end{bmatrix}$ is a linear combination of $\begin{bmatrix} 2 \\ 0 \\ -2 \end{bmatrix}$, $\begin{bmatrix} -2 \\ -2 \\ 0 \end{bmatrix}$, and $\begin{bmatrix} 3 \\ 2 \\ 2 \end{bmatrix}$ if $\begin{bmatrix} 3 \\ 2 \\ -4 \end{bmatrix} = c_{1}\begin{bmatrix} 2 \\ 0 \\ -2 \end{bmatrix} + c_{2}\begin{bmatrix} -2 \\ -2 \\ 0 \end{bmatrix} + c_{3}\begin{bmatrix} 3 \\ 2 \\ 2 \end{bmatrix}$ for some scalars $c_1, c_2, c_3$.
This is the linear system $\left[\begin{array}{ccc|c} 2 & -2 & 3 & 3 \\ 0 & -2 & 2 & 2 \\ -2 & 0 & 2 & -4 \end{array}\right]$, which row reduces to $\left[\begin{array}{ccc|c} 1 & 0 & 0 & 1 \\ 0 & 1 & 0 & -2 \\ 0 & 0 & 1 & -1 \end{array}\right]$.
$c_{1} = 1$, $c_{2} = -2$, $c_{3} = -1$ is a solution.
So, $\begin{bmatrix} 3 \\ 2 \\ -4 \end{bmatrix} = \begin{bmatrix} 2 \\ 0 \\ -2 \end{bmatrix} -2\begin{bmatrix} -2 \\ -2 \\ 0 \end{bmatrix} -\begin{bmatrix} 3 \\ 2 \\ 2 \end{bmatrix}$.

Q: Is the vector $\begin{bmatrix} 5 \\ 5 \\ -6 \end{bmatrix}$ in the span of $\left\{ \begin{bmatrix} 3 \\ 1 \\ 2 \end{bmatrix}, \begin{bmatrix} 2 \\ 3 \\ -3 \end{bmatrix}, \begin{bmatrix} 2 \\ 0 \\ 1 \end{bmatrix} \right\}$?
A: $\begin{bmatrix} 5 \\ 5 \\ -6 \end{bmatrix}$ is in the span of $\begin{bmatrix} 3 \\ 1 \\ 2 \end{bmatrix}$, $\begin{bmatrix} 2 \\ 3 \\ -3 \end{bmatrix}$, and $\begin{bmatrix} 2 \\ 0 \\ 1 \end{bmatrix}$ if $\begin{bmatrix} 5 \\ 5 \\ -6 \end{bmatrix} = c_{1}\begin{bmatrix} 3 \\ 1 \\ 2 \end{bmatrix} + c_{2}\begin{bmatrix} 2 \\ 3 \\ -3 \end{bmatrix} + c_{3}\begin{bmatrix} 2 \\ 0 \\ 1 \end{bmatrix}$ for some scalars $c_1, c_2, c_3$.
This is the linear system $\left[\begin{array}{ccc|c} 3 & 2 & 2 & 5 \\ 1 & 3 & 0 & 5 \\ 2 & -3 & 1 & -6 \end{array}\right]$, which row reduces to $\left[\begin{array}{ccc|c} 1 & 0 & 0 & -1 \\ 0 & 1 & 0 & 2 \\ 0 & 0 & 1 & 2 \end{array}\right]$.
$c_{1} = -1$, $c_{2} = 2$, $c_{3} = 2$ is a solution.
So, $\begin{bmatrix} 5 \\ 5 \\ -6 \end{bmatrix} = -\begin{bmatrix} 3 \\ 1 \\ 2 \end{bmatrix} + 2\begin{bmatrix} 2 \\ 3 \\ -3 \end{bmatrix} + 2\begin{bmatrix} 2 \\ 0 \\ 1 \end{bmatrix}$.

Q: Is the vector $\begin{bmatrix} -2 \\ 2 \\ -2 \end{bmatrix}$ in the span of $\left\{ \begin{bmatrix} -3 \\ 1 \\ 3 \end{bmatrix}, \begin{bmatrix} 1 \\ 3 \\ -3 \end{bmatrix}, \begin{bmatrix} 0 \\ 2 \\ 2 \end{bmatrix} \right\}$?
A: $\begin{bmatrix} -2 \\ 2 \\ -2 \end{bmatrix}$ is in the span of $\begin{bmatrix} -3 \\ 1 \\ 3 \end{bmatrix}$, $\begin{bmatrix} 1 \\ 3 \\ -3 \end{bmatrix}$, and $\begin{bmatrix} 0 \\ 2 \\ 2 \end{bmatrix}$ if $\begin{bmatrix} -2 \\ 2 \\ -2 \end{bmatrix} = c_{1}\begin{bmatrix} -3 \\ 1 \\ 3 \end{bmatrix} + c_{2}\begin{bmatrix} 1 \\ 3 \\ -3 \end{bmatrix} + c_{3}\begin{bmatrix} 0 \\ 2 \\ 2 \end{bmatrix}$ for some scalars $c_1, c_2, c_3$.
This is the linear system $\left[\begin{array}{ccc|c} -3 & 1 & 0 & -2 \\ 1 & 3 & 2 & 2 \\ 3 & -3 & 2 & -2 \end{array}\right]$, which row reduces to $\left[\begin{array}{ccc|c} 1 & 0 & 0 & 1 \\ 0 & 1 & 0 & 1 \\ 0 & 0 & 1 & -1 \end{array}\right]$.
$c_{1} = 1$, $c_{2} = 1$, $c_{3} = -1$ is a solution.
So, $\begin{bmatrix} -2 \\ 2 \\ -2 \end{bmatrix} = \begin{bmatrix} -3 \\ 1 \\ 3 \end{bmatrix} \begin{bmatrix} 1 \\ 3 \\ -3 \end{bmatrix} -\begin{bmatrix} 0 \\ 2 \\ 2 \end{bmatrix}$.

Q: Is the vector $\begin{bmatrix} 4 \\ -6 \\ 6 \end{bmatrix}$ in the span of $\left\{ \begin{bmatrix} 3 \\ -2 \\ 2 \end{bmatrix}, \begin{bmatrix} 1 \\ -1 \\ -1 \end{bmatrix}, \begin{bmatrix} 2 \\ 0 \\ -2 \end{bmatrix} \right\}$?
A: $\begin{bmatrix} 4 \\ -6 \\ 6 \end{bmatrix}$ is in the span of $\begin{bmatrix} 3 \\ -2 \\ 2 \end{bmatrix}$, $\begin{bmatrix} 1 \\ -1 \\ -1 \end{bmatrix}$, and $\begin{bmatrix} 2 \\ 0 \\ -2 \end{bmatrix}$ if $\begin{bmatrix} 4 \\ -6 \\ 6 \end{bmatrix} = c_{1}\begin{bmatrix} 3 \\ -2 \\ 2 \end{bmatrix} + c_{2}\begin{bmatrix} 1 \\ -1 \\ -1 \end{bmatrix} + c_{3}\begin{bmatrix} 2 \\ 0 \\ -2 \end{bmatrix}$ for some scalars $c_1, c_2, c_3$.
This is the linear system $\left[\begin{array}{ccc|c} 3 & 1 & 2 & 4 \\ -2 & -1 & 0 & -6 \\ 2 & -1 & -2 & 6 \end{array}\right]$, which row reduces to $\left[\begin{array}{ccc|c} 1 & 0 & 0 & 2 \\ 0 & 1 & 0 & 2 \\ 0 & 0 & 1 & -2 \end{array}\right]$.
$c_{1} = 2$, $c_{2} = 2$, $c_{3} = -2$ is a solution.
So, $\begin{bmatrix} 4 \\ -6 \\ 6 \end{bmatrix} = 2\begin{bmatrix} 3 \\ -2 \\ 2 \end{bmatrix} + 2\begin{bmatrix} 1 \\ -1 \\ -1 \end{bmatrix} -2\begin{bmatrix} 2 \\ 0 \\ -2 \end{bmatrix}$.

Q: Is the vector $\begin{bmatrix} 2 \\ -5 \\ -2 \end{bmatrix}$ in the span of $\left\{ \begin{bmatrix} 2 \\ 3 \\ 0 \end{bmatrix}, \begin{bmatrix} -3 \\ 0 \\ 1 \end{bmatrix}, \begin{bmatrix} -3 \\ -1 \\ 1 \end{bmatrix} \right\}$?
A: $\begin{bmatrix} 2 \\ -5 \\ -2 \end{bmatrix}$ is in the span of $\begin{bmatrix} 2 \\ 3 \\ 0 \end{bmatrix}$, $\begin{bmatrix} -3 \\ 0 \\ 1 \end{bmatrix}$, and $\begin{bmatrix} -3 \\ -1 \\ 1 \end{bmatrix}$ if $\begin{bmatrix} 2 \\ -5 \\ -2 \end{bmatrix} = c_{1}\begin{bmatrix} 2 \\ 3 \\ 0 \end{bmatrix} + c_{2}\begin{bmatrix} -3 \\ 0 \\ 1 \end{bmatrix} + c_{3}\begin{bmatrix} -3 \\ -1 \\ 1 \end{bmatrix}$ for some scalars $c_1, c_2, c_3$.
This is the linear system $\left[\begin{array}{ccc|c} 2 & -3 & -3 & 2 \\ 3 & 0 & -1 & -5 \\ 0 & 1 & 1 & -2 \end{array}\right]$, which row reduces to $\left[\begin{array}{ccc|c} 1 & 0 & 0 & -2 \\ 0 & 1 & 0 & -1 \\ 0 & 0 & 1 & -1 \end{array}\right]$.
$c_{1} = -2$, $c_{2} = -1$, $c_{3} = -1$ is a solution.
So, $\begin{bmatrix} 2 \\ -5 \\ -2 \end{bmatrix} = -2\begin{bmatrix} 2 \\ 3 \\ 0 \end{bmatrix} -\begin{bmatrix} -3 \\ 0 \\ 1 \end{bmatrix} -\begin{bmatrix} -3 \\ -1 \\ 1 \end{bmatrix}$.

Q: Is $\begin{bmatrix} -4 \\ 6 \\ -10 \end{bmatrix}$ a linear combination of $\begin{bmatrix} 3 \\ -3 \\ 3 \end{bmatrix}$, $\begin{bmatrix} 1 \\ 2 \\ 1 \end{bmatrix}$, and $\begin{bmatrix} -3 \\ -2 \\ 3 \end{bmatrix}$?
A: $\begin{bmatrix} -4 \\ 6 \\ -10 \end{bmatrix}$ is a linear combination of $\begin{bmatrix} 3 \\ -3 \\ 3 \end{bmatrix}$, $\begin{bmatrix} 1 \\ 2 \\ 1 \end{bmatrix}$, and $\begin{bmatrix} -3 \\ -2 \\ 3 \end{bmatrix}$ if $\begin{bmatrix} -4 \\ 6 \\ -10 \end{bmatrix} = c_{1}\begin{bmatrix} 3 \\ -3 \\ 3 \end{bmatrix} + c_{2}\begin{bmatrix} 1 \\ 2 \\ 1 \end{bmatrix} + c_{3}\begin{bmatrix} -3 \\ -2 \\ 3 \end{bmatrix}$ for some scalars $c_1, c_2, c_3$.
This is the linear system $\left[\begin{array}{ccc|c} 3 & 1 & -3 & -4 \\ -3 & 2 & -2 & 6 \\ 3 & 1 & 3 & -10 \end{array}\right]$, which row reduces to $\left[\begin{array}{ccc|c} 1 & 0 & 0 & -2 \\ 0 & 1 & 0 & -1 \\ 0 & 0 & 1 & -1 \end{array}\right]$.
$c_{1} = -2$, $c_{2} = -1$, $c_{3} = -1$ is a solution.
So, $\begin{bmatrix} -4 \\ 6 \\ -10 \end{bmatrix} = -2\begin{bmatrix} 3 \\ -3 \\ 3 \end{bmatrix} -\begin{bmatrix} 1 \\ 2 \\ 1 \end{bmatrix} -\begin{bmatrix} -3 \\ -2 \\ 3 \end{bmatrix}$.

Q: Is $\begin{bmatrix} 2 \\ 0 \\ 2 \end{bmatrix}$ a linear combination of $\begin{bmatrix} -1 \\ 1 \\ 1 \end{bmatrix}$, $\begin{bmatrix} 0 \\ 0 \\ -1 \end{bmatrix}$, $\begin{bmatrix} 0 \\ 2 \\ 3 \end{bmatrix}$, and $\begin{bmatrix} -1 \\ -1 \\ -1 \end{bmatrix}$?
A: $\begin{bmatrix} 2 \\ 0 \\ 2 \end{bmatrix}$ is a linear combination of $\begin{bmatrix} -1 \\ 1 \\ 1 \end{bmatrix}$, $\begin{bmatrix} 0 \\ 0 \\ -1 \end{bmatrix}$, $\begin{bmatrix} 0 \\ 2 \\ 3 \end{bmatrix}$, $\begin{bmatrix} -1 \\ -1 \\ -1 \end{bmatrix}$ if $\begin{bmatrix} 2 \\ 0 \\ 2 \end{bmatrix} = c_{1}\begin{bmatrix} -1 \\ 1 \\ 1 \end{bmatrix} + c_{2}\begin{bmatrix} 0 \\ 0 \\ -1 \end{bmatrix} + c_{3}\begin{bmatrix} 0 \\ 2 \\ 3 \end{bmatrix} + c_{4}\begin{bmatrix} -1 \\ -1 \\ -1 \end{bmatrix}$ for some scalars $c_1, c_2, c_3, c_4$.
This is the linear system $\left[\begin{array}{cccc|c} -1 & 0 & 0 & -1 & 2 \\ 1 & 0 & 2 & -1 & 0 \\ 1 & -1 & 3 & -1 & 2 \end{array}\right]$, which row reduces to $\left[\begin{array}{cccc|c} 1 & 0 & 0 & 1 & -2 \\ 0 & 1 & 0 & -1 & -1 \\ 0 & 0 & 1 & -1 & 1 \end{array}\right]$.
$c_{1} = -2$, $c_{2} = -1$, $c_{3} = 1$, $c_{4} = 0$ is a solution.
So, $\begin{bmatrix} 2 \\ 0 \\ 2 \end{bmatrix} = -2\begin{bmatrix} -1 \\ 1 \\ 1 \end{bmatrix} -\begin{bmatrix} 0 \\ 0 \\ -1 \end{bmatrix} \begin{bmatrix} 0 \\ 2 \\ 3 \end{bmatrix}$.

Q: Is the vector $\begin{bmatrix} 5 \\ -2 \\ -5 \end{bmatrix}$ in the span of $\left\{ \begin{bmatrix} -1 \\ -3 \\ 1 \end{bmatrix}, \begin{bmatrix} -3 \\ -2 \\ 3 \end{bmatrix}, \begin{bmatrix} 1 \\ 0 \\ -1 \end{bmatrix}, \begin{bmatrix} 1 \\ -1 \\ -1 \end{bmatrix} \right\}$?
A: $\begin{bmatrix} 5 \\ -2 \\ -5 \end{bmatrix}$ is in the span of $\begin{bmatrix} -1 \\ -3 \\ 1 \end{bmatrix}$, $\begin{bmatrix} -3 \\ -2 \\ 3 \end{bmatrix}$, $\begin{bmatrix} 1 \\ 0 \\ -1 \end{bmatrix}$, $\begin{bmatrix} 1 \\ -1 \\ -1 \end{bmatrix}$ if $\begin{bmatrix} 5 \\ -2 \\ -5 \end{bmatrix} = c_{1}\begin{bmatrix} -1 \\ -3 \\ 1 \end{bmatrix} + c_{2}\begin{bmatrix} -3 \\ -2 \\ 3 \end{bmatrix} + c_{3}\begin{bmatrix} 1 \\ 0 \\ -1 \end{bmatrix} + c_{4}\begin{bmatrix} 1 \\ -1 \\ -1 \end{bmatrix}$ for some scalars $c_1, c_2, c_3, c_4$.
This is the linear system $\left[\begin{array}{cccc|c} -1 & -3 & 1 & 1 & 5 \\ -3 & -2 & 0 & -1 & -2 \\ 1 & 3 & -1 & -1 & -5 \end{array}\right]$, which row reduces to $\left[\begin{array}{cccc|c} 1 & 0 & 0.29 & 0.71 & 2.29 \\ 0 & 1 & -0.43 & -0.57 & -2.43 \\ 0 & 0 & 0 & 0 & 0 \end{array}\right]$.
$c_{1} = 2.2857142857$, $c_{2} = -2.4285714286$, $c_{3} = 0$, $c_{4} = 0$ is a solution.
So, $\begin{bmatrix} 5 \\ -2 \\ -5 \end{bmatrix} = 2.2857142857\begin{bmatrix} -1 \\ -3 \\ 1 \end{bmatrix} -2.4285714286\begin{bmatrix} -3 \\ -2 \\ 3 \end{bmatrix}$.

Q: Is $\begin{bmatrix} -4 \\ 3 \\ -6 \end{bmatrix}$ a linear combination of $\begin{bmatrix} 3 \\ 0 \\ 3 \end{bmatrix}$, $\begin{bmatrix} 0 \\ -2 \\ 3 \end{bmatrix}$, $\begin{bmatrix} -2 \\ -3 \\ 0 \end{bmatrix}$, and $\begin{bmatrix} 3 \\ 2 \\ 0 \end{bmatrix}$?
A: $\begin{bmatrix} -4 \\ 3 \\ -6 \end{bmatrix}$ is a linear combination of $\begin{bmatrix} 3 \\ 0 \\ 3 \end{bmatrix}$, $\begin{bmatrix} 0 \\ -2 \\ 3 \end{bmatrix}$, $\begin{bmatrix} -2 \\ -3 \\ 0 \end{bmatrix}$, $\begin{bmatrix} 3 \\ 2 \\ 0 \end{bmatrix}$ if $\begin{bmatrix} -4 \\ 3 \\ -6 \end{bmatrix} = c_{1}\begin{bmatrix} 3 \\ 0 \\ 3 \end{bmatrix} + c_{2}\begin{bmatrix} 0 \\ -2 \\ 3 \end{bmatrix} + c_{3}\begin{bmatrix} -2 \\ -3 \\ 0 \end{bmatrix} + c_{4}\begin{bmatrix} 3 \\ 2 \\ 0 \end{bmatrix}$ for some scalars $c_1, c_2, c_3, c_4$.
This is the linear system $\left[\begin{array}{cccc|c} 3 & 0 & -2 & 3 & -4 \\ 0 & -2 & -3 & 2 & 3 \\ 3 & 3 & 0 & 0 & -6 \end{array}\right]$, which row reduces to $\left[\begin{array}{cccc|c} 1 & 0 & 0 & 1 & -2 \\ 0 & 1 & 0 & -1 & 0 \\ 0 & 0 & 1 & 0 & -1 \end{array}\right]$.
$c_{1} = -2$, $c_{2} = 0$, $c_{3} = -1$, $c_{4} = 0$ is a solution.
So, $\begin{bmatrix} -4 \\ 3 \\ -6 \end{bmatrix} = -2\begin{bmatrix} 3 \\ 0 \\ 3 \end{bmatrix} -\begin{bmatrix} -2 \\ -3 \\ 0 \end{bmatrix}$.

Q: Is the vector $\begin{bmatrix} -1 \\ 2 \\ 0 \end{bmatrix}$ in the span of $\left\{ \begin{bmatrix} -3 \\ 3 \\ 3 \end{bmatrix}, \begin{bmatrix} -2 \\ -3 \\ -1 \end{bmatrix}, \begin{bmatrix} 3 \\ 1 \\ 1 \end{bmatrix}, \begin{bmatrix} -3 \\ 7 \\ 7 \end{bmatrix} \right\}$?
A: $\begin{bmatrix} -1 \\ 2 \\ 0 \end{bmatrix}$ is in the span of $\begin{bmatrix} -3 \\ 3 \\ 3 \end{bmatrix}$, $\begin{bmatrix} -2 \\ -3 \\ -1 \end{bmatrix}$, $\begin{bmatrix} 3 \\ 1 \\ 1 \end{bmatrix}$, $\begin{bmatrix} -3 \\ 7 \\ 7 \end{bmatrix}$ if $\begin{bmatrix} -1 \\ 2 \\ 0 \end{bmatrix} = c_{1}\begin{bmatrix} -3 \\ 3 \\ 3 \end{bmatrix} + c_{2}\begin{bmatrix} -2 \\ -3 \\ -1 \end{bmatrix} + c_{3}\begin{bmatrix} 3 \\ 1 \\ 1 \end{bmatrix} + c_{4}\begin{bmatrix} -3 \\ 7 \\ 7 \end{bmatrix}$ for some scalars $c_1, c_2, c_3, c_4$.
This is the linear system $\left[\begin{array}{cccc|c} -3 & -2 & 3 & -3 & -1 \\ 3 & -3 & 1 & 7 & 2 \\ 3 & -1 & 1 & 7 & 0 \end{array}\right]$, which row reduces to $\left[\begin{array}{cccc|c} 1 & 0 & 0 & 2 & 0 \\ 0 & 1 & 0 & 0 & -1 \\ 0 & 0 & 1 & 1 & -1 \end{array}\right]$.
$c_{1} = 0$, $c_{2} = -1$, $c_{3} = -1$, $c_{4} = 0$ is a solution.
So, $\begin{bmatrix} -1 \\ 2 \\ 0 \end{bmatrix} = -\begin{bmatrix} -2 \\ -3 \\ -1 \end{bmatrix} -\begin{bmatrix} 3 \\ 1 \\ 1 \end{bmatrix}$.

Q: Is $\begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix}$ a linear combination of $\begin{bmatrix} -1 \\ 0 \\ 1 \end{bmatrix}$, $\begin{bmatrix} -2 \\ 0 \\ 2 \end{bmatrix}$, $\begin{bmatrix} -1 \\ -1 \\ -1 \end{bmatrix}$, and $\begin{bmatrix} 0 \\ -1 \\ -2 \end{bmatrix}$?
A: $\begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix}$ is a linear combination of $\begin{bmatrix} -1 \\ 0 \\ 1 \end{bmatrix}$, $\begin{bmatrix} -2 \\ 0 \\ 2 \end{bmatrix}$, $\begin{bmatrix} -1 \\ -1 \\ -1 \end{bmatrix}$, $\begin{bmatrix} 0 \\ -1 \\ -2 \end{bmatrix}$ if $\begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix} = c_{1}\begin{bmatrix} -1 \\ 0 \\ 1 \end{bmatrix} + c_{2}\begin{bmatrix} -2 \\ 0 \\ 2 \end{bmatrix} + c_{3}\begin{bmatrix} -1 \\ -1 \\ -1 \end{bmatrix} + c_{4}\begin{bmatrix} 0 \\ -1 \\ -2 \end{bmatrix}$ for some scalars $c_1, c_2, c_3, c_4$.
This is the linear system $\left[\begin{array}{cccc|c} -1 & -2 & -1 & 0 & 1 \\ 0 & 0 & -1 & -1 & 1 \\ 1 & 2 & -1 & -2 & 1 \end{array}\right]$, which row reduces to $\left[\begin{array}{cccc|c} 1 & 2 & 0 & -1 & 0 \\ 0 & 0 & 1 & 1 & -1 \\ 0 & 0 & 0 & 0 & 0 \end{array}\right]$.
$c_{1} = 0$, $c_{2} = 0$, $c_{3} = -1$, $c_{4} = 0$ is a solution.
So, $\begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix} = -\begin{bmatrix} -1 \\ -1 \\ -1 \end{bmatrix}$.

Q: Is the vector $\begin{bmatrix} 1 \\ 6 \\ -1 \end{bmatrix}$ in the span of $\left\{ \begin{bmatrix} 2 \\ 2 \\ -2 \end{bmatrix}, \begin{bmatrix} 1 \\ 2 \\ -1 \end{bmatrix}, \begin{bmatrix} -2 \\ 2 \\ 2 \end{bmatrix}, \begin{bmatrix} 2 \\ 2 \\ -2 \end{bmatrix} \right\}$?
A: $\begin{bmatrix} 1 \\ 6 \\ -1 \end{bmatrix}$ is in the span of $\begin{bmatrix} 2 \\ 2 \\ -2 \end{bmatrix}$, $\begin{bmatrix} 1 \\ 2 \\ -1 \end{bmatrix}$, $\begin{bmatrix} -2 \\ 2 \\ 2 \end{bmatrix}$, $\begin{bmatrix} 2 \\ 2 \\ -2 \end{bmatrix}$ if $\begin{bmatrix} 1 \\ 6 \\ -1 \end{bmatrix} = c_{1}\begin{bmatrix} 2 \\ 2 \\ -2 \end{bmatrix} + c_{2}\begin{bmatrix} 1 \\ 2 \\ -1 \end{bmatrix} + c_{3}\begin{bmatrix} -2 \\ 2 \\ 2 \end{bmatrix} + c_{4}\begin{bmatrix} 2 \\ 2 \\ -2 \end{bmatrix}$ for some scalars $c_1, c_2, c_3, c_4$.
This is the linear system $\left[\begin{array}{cccc|c} 2 & 1 & -2 & 2 & 1 \\ 2 & 2 & 2 & 2 & 6 \\ -2 & -1 & 2 & -2 & -1 \end{array}\right]$, which row reduces to $\left[\begin{array}{cccc|c} 1 & 0 & -3 & 1 & -2 \\ 0 & 1 & 4 & 0 & 5 \\ 0 & 0 & 0 & 0 & 0 \end{array}\right]$.
$c_{1} = -2$, $c_{2} = 5$, $c_{3} = 0$, $c_{4} = 0$ is a solution.
So, $\begin{bmatrix} 1 \\ 6 \\ -1 \end{bmatrix} = -2\begin{bmatrix} 2 \\ 2 \\ -2 \end{bmatrix} + 5\begin{bmatrix} 1 \\ 2 \\ -1 \end{bmatrix}$.

Q: Is the vector $\begin{bmatrix} 3 \\ 7 \\ 6 \end{bmatrix}$ in the span of $\left\{ \begin{bmatrix} 3 \\ 2 \\ 3 \end{bmatrix}, \begin{bmatrix} -1 \\ -3 \\ -2 \end{bmatrix}, \begin{bmatrix} 1 \\ 1 \\ 2 \end{bmatrix}, \begin{bmatrix} -8 \\ -10 \\ -10 \end{bmatrix} \right\}$?
A: $\begin{bmatrix} 3 \\ 7 \\ 6 \end{bmatrix}$ is in the span of $\begin{bmatrix} 3 \\ 2 \\ 3 \end{bmatrix}$, $\begin{bmatrix} -1 \\ -3 \\ -2 \end{bmatrix}$, $\begin{bmatrix} 1 \\ 1 \\ 2 \end{bmatrix}$, $\begin{bmatrix} -8 \\ -10 \\ -10 \end{bmatrix}$ if $\begin{bmatrix} 3 \\ 7 \\ 6 \end{bmatrix} = c_{1}\begin{bmatrix} 3 \\ 2 \\ 3 \end{bmatrix} + c_{2}\begin{bmatrix} -1 \\ -3 \\ -2 \end{bmatrix} + c_{3}\begin{bmatrix} 1 \\ 1 \\ 2 \end{bmatrix} + c_{4}\begin{bmatrix} -8 \\ -10 \\ -10 \end{bmatrix}$ for some scalars $c_1, c_2, c_3, c_4$.
This is the linear system $\left[\begin{array}{cccc|c} 3 & -1 & 1 & -8 & 3 \\ 2 & -3 & 1 & -10 & 7 \\ 3 & -2 & 2 & -10 & 6 \end{array}\right]$, which row reduces to $\left[\begin{array}{cccc|c} 1 & 0 & 0 & -2 & 0 \\ 0 & 1 & 0 & 2 & -2 \\ 0 & 0 & 1 & 0 & 1 \end{array}\right]$.
$c_{1} = 0$, $c_{2} = -2$, $c_{3} = 1$, $c_{4} = 0$ is a solution.
So, $\begin{bmatrix} 3 \\ 7 \\ 6 \end{bmatrix} = -2\begin{bmatrix} -1 \\ -3 \\ -2 \end{bmatrix} \begin{bmatrix} 1 \\ 1 \\ 2 \end{bmatrix}$.

Q: Is $\begin{bmatrix} 2 \\ 8 \\ 2 \end{bmatrix}$ a linear combination of $\begin{bmatrix} 3 \\ 1 \\ 2 \end{bmatrix}$, $\begin{bmatrix} 2 \\ -3 \\ 1 \end{bmatrix}$, $\begin{bmatrix} 0 \\ 0 \\ 2 \end{bmatrix}$, and $\begin{bmatrix} 3 \\ 1 \\ 0 \end{bmatrix}$?
A: $\begin{bmatrix} 2 \\ 8 \\ 2 \end{bmatrix}$ is a linear combination of $\begin{bmatrix} 3 \\ 1 \\ 2 \end{bmatrix}$, $\begin{bmatrix} 2 \\ -3 \\ 1 \end{bmatrix}$, $\begin{bmatrix} 0 \\ 0 \\ 2 \end{bmatrix}$, $\begin{bmatrix} 3 \\ 1 \\ 0 \end{bmatrix}$ if $\begin{bmatrix} 2 \\ 8 \\ 2 \end{bmatrix} = c_{1}\begin{bmatrix} 3 \\ 1 \\ 2 \end{bmatrix} + c_{2}\begin{bmatrix} 2 \\ -3 \\ 1 \end{bmatrix} + c_{3}\begin{bmatrix} 0 \\ 0 \\ 2 \end{bmatrix} + c_{4}\begin{bmatrix} 3 \\ 1 \\ 0 \end{bmatrix}$ for some scalars $c_1, c_2, c_3, c_4$.
This is the linear system $\left[\begin{array}{cccc|c} 3 & 2 & 0 & 3 & 2 \\ 1 & -3 & 0 & 1 & 8 \\ 2 & 1 & 2 & 0 & 2 \end{array}\right]$, which row reduces to $\left[\begin{array}{cccc|c} 1 & 0 & 0 & 1 & 2 \\ 0 & 1 & 0 & 0 & -2 \\ 0 & 0 & 1 & -1 & 0 \end{array}\right]$.
$c_{1} = 2$, $c_{2} = -2$, $c_{3} = 0$, $c_{4} = 0$ is a solution.
So, $\begin{bmatrix} 2 \\ 8 \\ 2 \end{bmatrix} = 2\begin{bmatrix} 3 \\ 1 \\ 2 \end{bmatrix} -2\begin{bmatrix} 2 \\ -3 \\ 1 \end{bmatrix}$.

Q: Is $\begin{bmatrix} -4 \\ 6 \\ -2 \end{bmatrix}$ a linear combination of $\begin{bmatrix} 0 \\ 3 \\ -3 \end{bmatrix}$, $\begin{bmatrix} 2 \\ 0 \\ 1 \end{bmatrix}$, $\begin{bmatrix} 2 \\ 0 \\ -2 \end{bmatrix}$, and $\begin{bmatrix} 6 \\ 6 \\ -6 \end{bmatrix}$?
A: $\begin{bmatrix} -4 \\ 6 \\ -2 \end{bmatrix}$ is a linear combination of $\begin{bmatrix} 0 \\ 3 \\ -3 \end{bmatrix}$, $\begin{bmatrix} 2 \\ 0 \\ 1 \end{bmatrix}$, $\begin{bmatrix} 2 \\ 0 \\ -2 \end{bmatrix}$, $\begin{bmatrix} 6 \\ 6 \\ -6 \end{bmatrix}$ if $\begin{bmatrix} -4 \\ 6 \\ -2 \end{bmatrix} = c_{1}\begin{bmatrix} 0 \\ 3 \\ -3 \end{bmatrix} + c_{2}\begin{bmatrix} 2 \\ 0 \\ 1 \end{bmatrix} + c_{3}\begin{bmatrix} 2 \\ 0 \\ -2 \end{bmatrix} + c_{4}\begin{bmatrix} 6 \\ 6 \\ -6 \end{bmatrix}$ for some scalars $c_1, c_2, c_3, c_4$.
This is the linear system $\left[\begin{array}{cccc|c} 0 & 2 & 2 & 6 & -4 \\ 3 & 0 & 0 & 6 & 6 \\ -3 & 1 & -2 & -6 & -2 \end{array}\right]$, which row reduces to $\left[\begin{array}{cccc|c} 1 & 0 & 0 & 2 & 2 \\ 0 & 1 & 0 & 2 & 0 \\ 0 & 0 & 1 & 1 & -2 \end{array}\right]$.
$c_{1} = 2$, $c_{2} = 0$, $c_{3} = -2$, $c_{4} = 0$ is a solution.
So, $\begin{bmatrix} -4 \\ 6 \\ -2 \end{bmatrix} = 2\begin{bmatrix} 0 \\ 3 \\ -3 \end{bmatrix} -2\begin{bmatrix} 2 \\ 0 \\ -2 \end{bmatrix}$.

Q: Is the vector $\begin{bmatrix} 4 \\ 6 \\ -5 \end{bmatrix}$ in the span of $\left\{ \begin{bmatrix} -1 \\ -2 \\ 0 \end{bmatrix}, \begin{bmatrix} 3 \\ 0 \\ -2 \end{bmatrix}, \begin{bmatrix} -1 \\ 2 \\ -3 \end{bmatrix}, \begin{bmatrix} -3 \\ 8 \\ -4 \end{bmatrix} \right\}$?
A: $\begin{bmatrix} 4 \\ 6 \\ -5 \end{bmatrix}$ is in the span of $\begin{bmatrix} -1 \\ -2 \\ 0 \end{bmatrix}$, $\begin{bmatrix} 3 \\ 0 \\ -2 \end{bmatrix}$, $\begin{bmatrix} -1 \\ 2 \\ -3 \end{bmatrix}$, $\begin{bmatrix} -3 \\ 8 \\ -4 \end{bmatrix}$ if $\begin{bmatrix} 4 \\ 6 \\ -5 \end{bmatrix} = c_{1}\begin{bmatrix} -1 \\ -2 \\ 0 \end{bmatrix} + c_{2}\begin{bmatrix} 3 \\ 0 \\ -2 \end{bmatrix} + c_{3}\begin{bmatrix} -1 \\ 2 \\ -3 \end{bmatrix} + c_{4}\begin{bmatrix} -3 \\ 8 \\ -4 \end{bmatrix}$ for some scalars $c_1, c_2, c_3, c_4$.
This is the linear system $\left[\begin{array}{cccc|c} -1 & 3 & -1 & -3 & 4 \\ -2 & 0 & 2 & 8 & 6 \\ 0 & -2 & -3 & -4 & -5 \end{array}\right]$, which row reduces to $\left[\begin{array}{cccc|c} 1 & 0 & 0 & -2 & -2 \\ 0 & 1 & 0 & -1 & 1 \\ 0 & 0 & 1 & 2 & 1 \end{array}\right]$.
$c_{1} = -2$, $c_{2} = 1$, $c_{3} = 1$, $c_{4} = 0$ is a solution.
So, $\begin{bmatrix} 4 \\ 6 \\ -5 \end{bmatrix} = -2\begin{bmatrix} -1 \\ -2 \\ 0 \end{bmatrix} \begin{bmatrix} 3 \\ 0 \\ -2 \end{bmatrix} \begin{bmatrix} -1 \\ 2 \\ -3 \end{bmatrix}$.

Q: Is $\begin{bmatrix} -1 \\ 5 \\ -10 \end{bmatrix}$ a linear combination of $\begin{bmatrix} 2 \\ 2 \\ 2 \end{bmatrix}$, $\begin{bmatrix} -1 \\ -3 \\ 2 \end{bmatrix}$, $\begin{bmatrix} -1 \\ -3 \\ 2 \end{bmatrix}$, and $\begin{bmatrix} -1 \\ 5 \\ -10 \end{bmatrix}$?
A: $\begin{bmatrix} -1 \\ 5 \\ -10 \end{bmatrix}$ is a linear combination of $\begin{bmatrix} 2 \\ 2 \\ 2 \end{bmatrix}$, $\begin{bmatrix} -1 \\ -3 \\ 2 \end{bmatrix}$, $\begin{bmatrix} -1 \\ -3 \\ 2 \end{bmatrix}$, $\begin{bmatrix} -1 \\ 5 \\ -10 \end{bmatrix}$ if $\begin{bmatrix} -1 \\ 5 \\ -10 \end{bmatrix} = c_{1}\begin{bmatrix} 2 \\ 2 \\ 2 \end{bmatrix} + c_{2}\begin{bmatrix} -1 \\ -3 \\ 2 \end{bmatrix} + c_{3}\begin{bmatrix} -1 \\ -3 \\ 2 \end{bmatrix} + c_{4}\begin{bmatrix} -1 \\ 5 \\ -10 \end{bmatrix}$ for some scalars $c_1, c_2, c_3, c_4$.
This is the linear system $\left[\begin{array}{cccc|c} 2 & -1 & -1 & -1 & -1 \\ 2 & -3 & -3 & 5 & 5 \\ 2 & 2 & 2 & -10 & -10 \end{array}\right]$, which row reduces to $\left[\begin{array}{cccc|c} 1 & 0 & 0 & -2 & -2 \\ 0 & 1 & 1 & -3 & -3 \\ 0 & 0 & 0 & 0 & 0 \end{array}\right]$.
$c_{1} = -2$, $c_{2} = -3$, $c_{3} = 0$, $c_{4} = 0$ is a solution.
So, $\begin{bmatrix} -1 \\ 5 \\ -10 \end{bmatrix} = -2\begin{bmatrix} 2 \\ 2 \\ 2 \end{bmatrix} -3\begin{bmatrix} -1 \\ -3 \\ 2 \end{bmatrix}$.

Q: Is the vector $\begin{bmatrix} -2 \\ -2 \\ -2 \end{bmatrix}$ in the span of $\left\{ \begin{bmatrix} 0 \\ 1 \\ -1 \end{bmatrix}, \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix}, \begin{bmatrix} -1 \\ 0 \\ -2 \end{bmatrix}, \begin{bmatrix} -2 \\ -3 \\ -1 \end{bmatrix} \right\}$?
A: $\begin{bmatrix} -2 \\ -2 \\ -2 \end{bmatrix}$ is in the span of $\begin{bmatrix} 0 \\ 1 \\ -1 \end{bmatrix}$, $\begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix}$, $\begin{bmatrix} -1 \\ 0 \\ -2 \end{bmatrix}$, $\begin{bmatrix} -2 \\ -3 \\ -1 \end{bmatrix}$ if $\begin{bmatrix} -2 \\ -2 \\ -2 \end{bmatrix} = c_{1}\begin{bmatrix} 0 \\ 1 \\ -1 \end{bmatrix} + c_{2}\begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix} + c_{3}\begin{bmatrix} -1 \\ 0 \\ -2 \end{bmatrix} + c_{4}\begin{bmatrix} -2 \\ -3 \\ -1 \end{bmatrix}$ for some scalars $c_1, c_2, c_3, c_4$.
This is the linear system $\left[\begin{array}{cccc|c} 0 & 1 & -1 & -2 & -2 \\ 1 & 1 & 0 & -3 & -2 \\ -1 & 1 & -2 & -1 & -2 \end{array}\right]$, which row reduces to $\left[\begin{array}{cccc|c} 1 & 0 & 1 & -1 & 0 \\ 0 & 1 & -1 & -2 & -2 \\ 0 & 0 & 0 & 0 & 0 \end{array}\right]$.
$c_{1} = 0$, $c_{2} = -2$, $c_{3} = 0$, $c_{4} = 0$ is a solution.
So, $\begin{bmatrix} -2 \\ -2 \\ -2 \end{bmatrix} = -2\begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix}$.

Q: Is the vector $\begin{bmatrix} 10 \\ -1 \\ 2 \end{bmatrix}$ in the span of $\left\{ \begin{bmatrix} 2 \\ 1 \\ -1 \end{bmatrix}, \begin{bmatrix} -3 \\ 1 \\ -3 \end{bmatrix}, \begin{bmatrix} 0 \\ -1 \\ -2 \end{bmatrix}, \begin{bmatrix} -4 \\ -2 \\ 2 \end{bmatrix} \right\}$?
A: $\begin{bmatrix} 10 \\ -1 \\ 2 \end{bmatrix}$ is in the span of $\begin{bmatrix} 2 \\ 1 \\ -1 \end{bmatrix}$, $\begin{bmatrix} -3 \\ 1 \\ -3 \end{bmatrix}$, $\begin{bmatrix} 0 \\ -1 \\ -2 \end{bmatrix}$, $\begin{bmatrix} -4 \\ -2 \\ 2 \end{bmatrix}$ if $\begin{bmatrix} 10 \\ -1 \\ 2 \end{bmatrix} = c_{1}\begin{bmatrix} 2 \\ 1 \\ -1 \end{bmatrix} + c_{2}\begin{bmatrix} -3 \\ 1 \\ -3 \end{bmatrix} + c_{3}\begin{bmatrix} 0 \\ -1 \\ -2 \end{bmatrix} + c_{4}\begin{bmatrix} -4 \\ -2 \\ 2 \end{bmatrix}$ for some scalars $c_1, c_2, c_3, c_4$.
This is the linear system $\left[\begin{array}{cccc|c} 2 & -3 & 0 & -4 & 10 \\ 1 & 1 & -1 & -2 & -1 \\ -1 & -3 & -2 & 2 & 2 \end{array}\right]$, which row reduces to $\left[\begin{array}{cccc|c} 1 & 0 & 0 & -2 & 2 \\ 0 & 1 & 0 & 0 & -2 \\ 0 & 0 & 1 & 0 & 1 \end{array}\right]$.
$c_{1} = 2$, $c_{2} = -2$, $c_{3} = 1$, $c_{4} = 0$ is a solution.
So, $\begin{bmatrix} 10 \\ -1 \\ 2 \end{bmatrix} = 2\begin{bmatrix} 2 \\ 1 \\ -1 \end{bmatrix} -2\begin{bmatrix} -3 \\ 1 \\ -3 \end{bmatrix} \begin{bmatrix} 0 \\ -1 \\ -2 \end{bmatrix}$.

Q: Is the vector $\begin{bmatrix} -2 \\ -2 \\ 6 \end{bmatrix}$ in the span of $\left\{ \begin{bmatrix} -3 \\ 2 \\ -1 \end{bmatrix}, \begin{bmatrix} -2 \\ 1 \\ 2 \end{bmatrix}, \begin{bmatrix} -1 \\ 2 \\ -1 \end{bmatrix}, \begin{bmatrix} 2 \\ -3 \\ 4 \end{bmatrix} \right\}$?
A: $\begin{bmatrix} -2 \\ -2 \\ 6 \end{bmatrix}$ is in the span of $\begin{bmatrix} -3 \\ 2 \\ -1 \end{bmatrix}$, $\begin{bmatrix} -2 \\ 1 \\ 2 \end{bmatrix}$, $\begin{bmatrix} -1 \\ 2 \\ -1 \end{bmatrix}$, $\begin{bmatrix} 2 \\ -3 \\ 4 \end{bmatrix}$ if $\begin{bmatrix} -2 \\ -2 \\ 6 \end{bmatrix} = c_{1}\begin{bmatrix} -3 \\ 2 \\ -1 \end{bmatrix} + c_{2}\begin{bmatrix} -2 \\ 1 \\ 2 \end{bmatrix} + c_{3}\begin{bmatrix} -1 \\ 2 \\ -1 \end{bmatrix} + c_{4}\begin{bmatrix} 2 \\ -3 \\ 4 \end{bmatrix}$ for some scalars $c_1, c_2, c_3, c_4$.
This is the linear system $\left[\begin{array}{cccc|c} -3 & -2 & -1 & 2 & -2 \\ 2 & 1 & 2 & -3 & -2 \\ -1 & 2 & -1 & 4 & 6 \end{array}\right]$, which row reduces to $\left[\begin{array}{cccc|c} 1 & 0 & 0 & -1 & 0 \\ 0 & 1 & 0 & 1 & 2 \\ 0 & 0 & 1 & -1 & -2 \end{array}\right]$.
$c_{1} = 0$, $c_{2} = 2$, $c_{3} = -2$, $c_{4} = 0$ is a solution.
So, $\begin{bmatrix} -2 \\ -2 \\ 6 \end{bmatrix} = 2\begin{bmatrix} -2 \\ 1 \\ 2 \end{bmatrix} -2\begin{bmatrix} -1 \\ 2 \\ -1 \end{bmatrix}$.

Q: Is the vector $\begin{bmatrix} -6 \\ -5 \\ -3 \end{bmatrix}$ in the span of $\left\{ \begin{bmatrix} -1 \\ -1 \\ 1 \end{bmatrix}, \begin{bmatrix} -3 \\ -1 \\ -3 \end{bmatrix}, \begin{bmatrix} -2 \\ 1 \\ -1 \end{bmatrix}, \begin{bmatrix} -4 \\ 2 \\ -2 \end{bmatrix} \right\}$?
A: $\begin{bmatrix} -6 \\ -5 \\ -3 \end{bmatrix}$ is in the span of $\begin{bmatrix} -1 \\ -1 \\ 1 \end{bmatrix}$, $\begin{bmatrix} -3 \\ -1 \\ -3 \end{bmatrix}$, $\begin{bmatrix} -2 \\ 1 \\ -1 \end{bmatrix}$, $\begin{bmatrix} -4 \\ 2 \\ -2 \end{bmatrix}$ if $\begin{bmatrix} -6 \\ -5 \\ -3 \end{bmatrix} = c_{1}\begin{bmatrix} -1 \\ -1 \\ 1 \end{bmatrix} + c_{2}\begin{bmatrix} -3 \\ -1 \\ -3 \end{bmatrix} + c_{3}\begin{bmatrix} -2 \\ 1 \\ -1 \end{bmatrix} + c_{4}\begin{bmatrix} -4 \\ 2 \\ -2 \end{bmatrix}$ for some scalars $c_1, c_2, c_3, c_4$.
This is the linear system $\left[\begin{array}{cccc|c} -1 & -3 & -2 & -4 & -6 \\ -1 & -1 & 1 & 2 & -5 \\ 1 & -3 & -1 & -2 & -3 \end{array}\right]$, which row reduces to $\left[\begin{array}{cccc|c} 1 & 0 & 0 & 0 & 2 \\ 0 & 1 & 0 & 0 & 2 \\ 0 & 0 & 1 & 2 & -1 \end{array}\right]$.
$c_{1} = 2$, $c_{2} = 2$, $c_{3} = -1$, $c_{4} = 0$ is a solution.
So, $\begin{bmatrix} -6 \\ -5 \\ -3 \end{bmatrix} = 2\begin{bmatrix} -1 \\ -1 \\ 1 \end{bmatrix} + 2\begin{bmatrix} -3 \\ -1 \\ -3 \end{bmatrix} -\begin{bmatrix} -2 \\ 1 \\ -1 \end{bmatrix}$.

Q: Is $\begin{bmatrix} 0 \\ -2 \end{bmatrix}$ a linear combination of $\begin{bmatrix} -2 \\ 0 \end{bmatrix}$ and $\begin{bmatrix} -1 \\ 0 \end{bmatrix}$?
A: No. $\begin{bmatrix} 0 \\ -2 \end{bmatrix}$ is NOT in the span of $\begin{bmatrix} -2 \\ 0 \end{bmatrix}$ and $\begin{bmatrix} -1 \\ 0 \end{bmatrix}$.
For $\begin{bmatrix} 0 \\ -2 \end{bmatrix}$ to be in the span, we need $\begin{bmatrix} 0 \\ -2 \end{bmatrix} = c_{1}\begin{bmatrix} -2 \\ 0 \end{bmatrix} + c_{2}\begin{bmatrix} -1 \\ 0 \end{bmatrix}$ for some scalars $c_1, c_2$.
This gives us the linear system $\left[\begin{array}{cc|c} -2 & -1 & 0 \\ 0 & 0 & -2 \end{array}\right]$, which row reduces to $\left[\begin{array}{cc|c} 1 & 0.50 & 0 \\ 0 & 0 & 1 \end{array}\right]$.
Row 2 has the form [0 ... 0 | 1], which represents the equation 0 = 1, indicating no solution exists for the system.

Q: Is the vector $\begin{bmatrix} 0 \\ 1 \end{bmatrix}$ in the span of $\left\{ \begin{bmatrix} 1 \\ 0 \end{bmatrix}, \begin{bmatrix} 2 \\ 0 \end{bmatrix} \right\}$?
A: No. $\begin{bmatrix} 0 \\ 1 \end{bmatrix}$ is NOT in the span of $\begin{bmatrix} 1 \\ 0 \end{bmatrix}$ and $\begin{bmatrix} 2 \\ 0 \end{bmatrix}$.
For $\begin{bmatrix} 0 \\ 1 \end{bmatrix}$ to be in the span, we need $\begin{bmatrix} 0 \\ 1 \end{bmatrix} = c_{1}\begin{bmatrix} 1 \\ 0 \end{bmatrix} + c_{2}\begin{bmatrix} 2 \\ 0 \end{bmatrix}$ for some scalars $c_1, c_2$.
This gives us the linear system $\left[\begin{array}{cc|c} 1 & 2 & 0 \\ 0 & 0 & 1 \end{array}\right]$, which row reduces to $\left[\begin{array}{cc|c} 1 & 2 & 0 \\ 0 & 0 & 1 \end{array}\right]$.
Row 2 has the form [0 ... 0 | 1], which represents the equation 0 = 1, indicating no solution exists for the system.

Q: Is the vector $\begin{bmatrix} -1 \\ 0 \end{bmatrix}$ in the span of $\left\{ \begin{bmatrix} 0 \\ 1 \end{bmatrix}, \begin{bmatrix} 0 \\ 4 \end{bmatrix} \right\}$?
A: No. $\begin{bmatrix} -1 \\ 0 \end{bmatrix}$ is NOT in the span of $\begin{bmatrix} 0 \\ 1 \end{bmatrix}$ and $\begin{bmatrix} 0 \\ 4 \end{bmatrix}$.
For $\begin{bmatrix} -1 \\ 0 \end{bmatrix}$ to be in the span, we need $\begin{bmatrix} -1 \\ 0 \end{bmatrix} = c_{1}\begin{bmatrix} 0 \\ 1 \end{bmatrix} + c_{2}\begin{bmatrix} 0 \\ 4 \end{bmatrix}$ for some scalars $c_1, c_2$.
This gives us the linear system $\left[\begin{array}{cc|c} 0 & 0 & -1 \\ 1 & 4 & 0 \end{array}\right]$, which row reduces to $\left[\begin{array}{cc|c} 1 & 4 & 0 \\ 0 & 0 & 1 \end{array}\right]$.
Row 2 has the form [0 ... 0 | 1], which represents the equation 0 = 1, indicating no solution exists for the system.

Q: Is the vector $\begin{bmatrix} 0 \\ -1 \end{bmatrix}$ in the span of $\left\{ \begin{bmatrix} -1 \\ 0 \end{bmatrix}, \begin{bmatrix} -3 \\ 0 \end{bmatrix} \right\}$?
A: No. $\begin{bmatrix} 0 \\ -1 \end{bmatrix}$ is NOT in the span of $\begin{bmatrix} -1 \\ 0 \end{bmatrix}$ and $\begin{bmatrix} -3 \\ 0 \end{bmatrix}$.
For $\begin{bmatrix} 0 \\ -1 \end{bmatrix}$ to be in the span, we need $\begin{bmatrix} 0 \\ -1 \end{bmatrix} = c_{1}\begin{bmatrix} -1 \\ 0 \end{bmatrix} + c_{2}\begin{bmatrix} -3 \\ 0 \end{bmatrix}$ for some scalars $c_1, c_2$.
This gives us the linear system $\left[\begin{array}{cc|c} -1 & -3 & 0 \\ 0 & 0 & -1 \end{array}\right]$, which row reduces to $\left[\begin{array}{cc|c} 1 & 3 & 0 \\ 0 & 0 & 1 \end{array}\right]$.
Row 2 has the form [0 ... 0 | 1], which represents the equation 0 = 1, indicating no solution exists for the system.

Q: Is $\begin{bmatrix} -2 \\ 0 \end{bmatrix}$ a linear combination of $\begin{bmatrix} 0 \\ 2 \end{bmatrix}$ and $\begin{bmatrix} 0 \\ 5 \end{bmatrix}$?
A: No. $\begin{bmatrix} -2 \\ 0 \end{bmatrix}$ is NOT in the span of $\begin{bmatrix} 0 \\ 2 \end{bmatrix}$ and $\begin{bmatrix} 0 \\ 5 \end{bmatrix}$.
For $\begin{bmatrix} -2 \\ 0 \end{bmatrix}$ to be in the span, we need $\begin{bmatrix} -2 \\ 0 \end{bmatrix} = c_{1}\begin{bmatrix} 0 \\ 2 \end{bmatrix} + c_{2}\begin{bmatrix} 0 \\ 5 \end{bmatrix}$ for some scalars $c_1, c_2$.
This gives us the linear system $\left[\begin{array}{cc|c} 0 & 0 & -2 \\ 2 & 5 & 0 \end{array}\right]$, which row reduces to $\left[\begin{array}{cc|c} 1 & 2.50 & 0 \\ 0 & 0 & 1 \end{array}\right]$.
Row 2 has the form [0 ... 0 | 1], which represents the equation 0 = 1, indicating no solution exists for the system.

Q: Is $\begin{bmatrix} 0 \\ 1 \end{bmatrix}$ a linear combination of $\begin{bmatrix} 1 \\ 0 \end{bmatrix}$ and $\begin{bmatrix} 2 \\ 0 \end{bmatrix}$?
A: No. $\begin{bmatrix} 0 \\ 1 \end{bmatrix}$ is NOT in the span of $\begin{bmatrix} 1 \\ 0 \end{bmatrix}$ and $\begin{bmatrix} 2 \\ 0 \end{bmatrix}$.
For $\begin{bmatrix} 0 \\ 1 \end{bmatrix}$ to be in the span, we need $\begin{bmatrix} 0 \\ 1 \end{bmatrix} = c_{1}\begin{bmatrix} 1 \\ 0 \end{bmatrix} + c_{2}\begin{bmatrix} 2 \\ 0 \end{bmatrix}$ for some scalars $c_1, c_2$.
This gives us the linear system $\left[\begin{array}{cc|c} 1 & 2 & 0 \\ 0 & 0 & 1 \end{array}\right]$, which row reduces to $\left[\begin{array}{cc|c} 1 & 2 & 0 \\ 0 & 0 & 1 \end{array}\right]$.
Row 2 has the form [0 ... 0 | 1], which represents the equation 0 = 1, indicating no solution exists for the system.

Q: Is the vector $\begin{bmatrix} 0 \\ 1 \end{bmatrix}$ in the span of $\left\{ \begin{bmatrix} 1 \\ 0 \end{bmatrix}, \begin{bmatrix} 1 \\ 0 \end{bmatrix} \right\}$?
A: No. $\begin{bmatrix} 0 \\ 1 \end{bmatrix}$ is NOT in the span of $\begin{bmatrix} 1 \\ 0 \end{bmatrix}$ and $\begin{bmatrix} 1 \\ 0 \end{bmatrix}$.
For $\begin{bmatrix} 0 \\ 1 \end{bmatrix}$ to be in the span, we need $\begin{bmatrix} 0 \\ 1 \end{bmatrix} = c_{1}\begin{bmatrix} 1 \\ 0 \end{bmatrix} + c_{2}\begin{bmatrix} 1 \\ 0 \end{bmatrix}$ for some scalars $c_1, c_2$.
This gives us the linear system $\left[\begin{array}{cc|c} 1 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]$, which row reduces to $\left[\begin{array}{cc|c} 1 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]$.
Row 2 has the form [0 ... 0 | 1], which represents the equation 0 = 1, indicating no solution exists for the system.

Q: Is the vector $\begin{bmatrix} 3 \\ 0 \end{bmatrix}$ in the span of $\left\{ \begin{bmatrix} 0 \\ -3 \end{bmatrix}, \begin{bmatrix} 0 \\ -8 \end{bmatrix} \right\}$?
A: No. $\begin{bmatrix} 3 \\ 0 \end{bmatrix}$ is NOT in the span of $\begin{bmatrix} 0 \\ -3 \end{bmatrix}$ and $\begin{bmatrix} 0 \\ -8 \end{bmatrix}$.
For $\begin{bmatrix} 3 \\ 0 \end{bmatrix}$ to be in the span, we need $\begin{bmatrix} 3 \\ 0 \end{bmatrix} = c_{1}\begin{bmatrix} 0 \\ -3 \end{bmatrix} + c_{2}\begin{bmatrix} 0 \\ -8 \end{bmatrix}$ for some scalars $c_1, c_2$.
This gives us the linear system $\left[\begin{array}{cc|c} 0 & 0 & 3 \\ -3 & -8 & 0 \end{array}\right]$, which row reduces to $\left[\begin{array}{cc|c} 1 & 2.67 & 0 \\ 0 & 0 & 1 \end{array}\right]$.
Row 2 has the form [0 ... 0 | 1], which represents the equation 0 = 1, indicating no solution exists for the system.

Q: Is the vector $\begin{bmatrix} -1 \\ 0 \end{bmatrix}$ in the span of $\left\{ \begin{bmatrix} 0 \\ 1 \end{bmatrix}, \begin{bmatrix} 0 \\ 1 \end{bmatrix} \right\}$?
A: No. $\begin{bmatrix} -1 \\ 0 \end{bmatrix}$ is NOT in the span of $\begin{bmatrix} 0 \\ 1 \end{bmatrix}$ and $\begin{bmatrix} 0 \\ 1 \end{bmatrix}$.
For $\begin{bmatrix} -1 \\ 0 \end{bmatrix}$ to be in the span, we need $\begin{bmatrix} -1 \\ 0 \end{bmatrix} = c_{1}\begin{bmatrix} 0 \\ 1 \end{bmatrix} + c_{2}\begin{bmatrix} 0 \\ 1 \end{bmatrix}$ for some scalars $c_1, c_2$.
This gives us the linear system $\left[\begin{array}{cc|c} 0 & 0 & -1 \\ 1 & 1 & 0 \end{array}\right]$, which row reduces to $\left[\begin{array}{cc|c} 1 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]$.
Row 2 has the form [0 ... 0 | 1], which represents the equation 0 = 1, indicating no solution exists for the system.

Q: Is $\begin{bmatrix} 0 \\ -2 \end{bmatrix}$ a linear combination of $\begin{bmatrix} -2 \\ 0 \end{bmatrix}$ and $\begin{bmatrix} -5 \\ 0 \end{bmatrix}$?
A: No. $\begin{bmatrix} 0 \\ -2 \end{bmatrix}$ is NOT in the span of $\begin{bmatrix} -2 \\ 0 \end{bmatrix}$ and $\begin{bmatrix} -5 \\ 0 \end{bmatrix}$.
For $\begin{bmatrix} 0 \\ -2 \end{bmatrix}$ to be in the span, we need $\begin{bmatrix} 0 \\ -2 \end{bmatrix} = c_{1}\begin{bmatrix} -2 \\ 0 \end{bmatrix} + c_{2}\begin{bmatrix} -5 \\ 0 \end{bmatrix}$ for some scalars $c_1, c_2$.
This gives us the linear system $\left[\begin{array}{cc|c} -2 & -5 & 0 \\ 0 & 0 & -2 \end{array}\right]$, which row reduces to $\left[\begin{array}{cc|c} 1 & 2.50 & 0 \\ 0 & 0 & 1 \end{array}\right]$.
Row 2 has the form [0 ... 0 | 1], which represents the equation 0 = 1, indicating no solution exists for the system.

Q: Is $\begin{bmatrix} 3 \\ 0 \end{bmatrix}$ a linear combination of $\begin{bmatrix} 0 \\ -3 \end{bmatrix}$ and $\begin{bmatrix} 0 \\ -4 \end{bmatrix}$?
A: No. $\begin{bmatrix} 3 \\ 0 \end{bmatrix}$ is NOT in the span of $\begin{bmatrix} 0 \\ -3 \end{bmatrix}$ and $\begin{bmatrix} 0 \\ -4 \end{bmatrix}$.
For $\begin{bmatrix} 3 \\ 0 \end{bmatrix}$ to be in the span, we need $\begin{bmatrix} 3 \\ 0 \end{bmatrix} = c_{1}\begin{bmatrix} 0 \\ -3 \end{bmatrix} + c_{2}\begin{bmatrix} 0 \\ -4 \end{bmatrix}$ for some scalars $c_1, c_2$.
This gives us the linear system $\left[\begin{array}{cc|c} 0 & 0 & 3 \\ -3 & -4 & 0 \end{array}\right]$, which row reduces to $\left[\begin{array}{cc|c} 1 & 1.33 & 0 \\ 0 & 0 & 1 \end{array}\right]$.
Row 2 has the form [0 ... 0 | 1], which represents the equation 0 = 1, indicating no solution exists for the system.

Q: Is $\begin{bmatrix} 0 \\ -2 \end{bmatrix}$ a linear combination of $\begin{bmatrix} -2 \\ 0 \end{bmatrix}$ and $\begin{bmatrix} -7 \\ 0 \end{bmatrix}$?
A: No. $\begin{bmatrix} 0 \\ -2 \end{bmatrix}$ is NOT in the span of $\begin{bmatrix} -2 \\ 0 \end{bmatrix}$ and $\begin{bmatrix} -7 \\ 0 \end{bmatrix}$.
For $\begin{bmatrix} 0 \\ -2 \end{bmatrix}$ to be in the span, we need $\begin{bmatrix} 0 \\ -2 \end{bmatrix} = c_{1}\begin{bmatrix} -2 \\ 0 \end{bmatrix} + c_{2}\begin{bmatrix} -7 \\ 0 \end{bmatrix}$ for some scalars $c_1, c_2$.
This gives us the linear system $\left[\begin{array}{cc|c} -2 & -7 & 0 \\ 0 & 0 & -2 \end{array}\right]$, which row reduces to $\left[\begin{array}{cc|c} 1 & 3.50 & 0 \\ 0 & 0 & 1 \end{array}\right]$.
Row 2 has the form [0 ... 0 | 1], which represents the equation 0 = 1, indicating no solution exists for the system.

Q: Is the vector $\begin{bmatrix} 0 \\ 1 \end{bmatrix}$ in the span of $\left\{ \begin{bmatrix} 1 \\ 0 \end{bmatrix}, \begin{bmatrix} 2 \\ 0 \end{bmatrix}, \begin{bmatrix} -1 \\ 0 \end{bmatrix} \right\}$?
A: No. $\begin{bmatrix} 0 \\ 1 \end{bmatrix}$ is NOT in the span of $\begin{bmatrix} 1 \\ 0 \end{bmatrix}$, $\begin{bmatrix} 2 \\ 0 \end{bmatrix}$, and $\begin{bmatrix} -1 \\ 0 \end{bmatrix}$.
For $\begin{bmatrix} 0 \\ 1 \end{bmatrix}$ to be in the span, we need $\begin{bmatrix} 0 \\ 1 \end{bmatrix} = c_{1}\begin{bmatrix} 1 \\ 0 \end{bmatrix} + c_{2}\begin{bmatrix} 2 \\ 0 \end{bmatrix} + c_{3}\begin{bmatrix} -1 \\ 0 \end{bmatrix}$ for some scalars $c_1, c_2, c_3$.
This gives us the linear system $\left[\begin{array}{ccc|c} 1 & 2 & -1 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right]$, which row reduces to $\left[\begin{array}{ccc|c} 1 & 2 & -1 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right]$.
Row 2 has the form [0 ... 0 | 1], which represents the equation 0 = 1, indicating no solution exists for the system.

Q: Is the vector $\begin{bmatrix} -2 \\ 3 \end{bmatrix}$ in the span of $\left\{ \begin{bmatrix} 3 \\ 2 \end{bmatrix}, \begin{bmatrix} 9 \\ 6 \end{bmatrix}, \begin{bmatrix} -9 \\ -6 \end{bmatrix} \right\}$?
A: No. $\begin{bmatrix} -2 \\ 3 \end{bmatrix}$ is NOT in the span of $\begin{bmatrix} 3 \\ 2 \end{bmatrix}$, $\begin{bmatrix} 9 \\ 6 \end{bmatrix}$, and $\begin{bmatrix} -9 \\ -6 \end{bmatrix}$.
For $\begin{bmatrix} -2 \\ 3 \end{bmatrix}$ to be in the span, we need $\begin{bmatrix} -2 \\ 3 \end{bmatrix} = c_{1}\begin{bmatrix} 3 \\ 2 \end{bmatrix} + c_{2}\begin{bmatrix} 9 \\ 6 \end{bmatrix} + c_{3}\begin{bmatrix} -9 \\ -6 \end{bmatrix}$ for some scalars $c_1, c_2, c_3$.
This gives us the linear system $\left[\begin{array}{ccc|c} 3 & 9 & -9 & -2 \\ 2 & 6 & -6 & 3 \end{array}\right]$, which row reduces to $\left[\begin{array}{ccc|c} 1 & 3 & -3 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right]$.
Row 2 has the form [0 ... 0 | 1], which represents the equation 0 = 1, indicating no solution exists for the system.

Q: Is $\begin{bmatrix} 1 \\ 1 \end{bmatrix}$ a linear combination of $\begin{bmatrix} 1 \\ -1 \end{bmatrix}$, $\begin{bmatrix} 3 \\ -3 \end{bmatrix}$, and $\begin{bmatrix} -2 \\ 2 \end{bmatrix}$?
A: No. $\begin{bmatrix} 1 \\ 1 \end{bmatrix}$ is NOT in the span of $\begin{bmatrix} 1 \\ -1 \end{bmatrix}$, $\begin{bmatrix} 3 \\ -3 \end{bmatrix}$, and $\begin{bmatrix} -2 \\ 2 \end{bmatrix}$.
For $\begin{bmatrix} 1 \\ 1 \end{bmatrix}$ to be in the span, we need $\begin{bmatrix} 1 \\ 1 \end{bmatrix} = c_{1}\begin{bmatrix} 1 \\ -1 \end{bmatrix} + c_{2}\begin{bmatrix} 3 \\ -3 \end{bmatrix} + c_{3}\begin{bmatrix} -2 \\ 2 \end{bmatrix}$ for some scalars $c_1, c_2, c_3$.
This gives us the linear system $\left[\begin{array}{ccc|c} 1 & 3 & -2 & 1 \\ -1 & -3 & 2 & 1 \end{array}\right]$, which row reduces to $\left[\begin{array}{ccc|c} 1 & 3 & -2 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right]$.
Row 2 has the form [0 ... 0 | 1], which represents the equation 0 = 1, indicating no solution exists for the system.

Q: Is $\begin{bmatrix} -3 \\ -2 \end{bmatrix}$ a linear combination of $\begin{bmatrix} -2 \\ 3 \end{bmatrix}$, $\begin{bmatrix} -2 \\ 3 \end{bmatrix}$, and $\begin{bmatrix} 6 \\ -9 \end{bmatrix}$?
A: No. $\begin{bmatrix} -3 \\ -2 \end{bmatrix}$ is NOT in the span of $\begin{bmatrix} -2 \\ 3 \end{bmatrix}$, $\begin{bmatrix} -2 \\ 3 \end{bmatrix}$, and $\begin{bmatrix} 6 \\ -9 \end{bmatrix}$.
For $\begin{bmatrix} -3 \\ -2 \end{bmatrix}$ to be in the span, we need $\begin{bmatrix} -3 \\ -2 \end{bmatrix} = c_{1}\begin{bmatrix} -2 \\ 3 \end{bmatrix} + c_{2}\begin{bmatrix} -2 \\ 3 \end{bmatrix} + c_{3}\begin{bmatrix} 6 \\ -9 \end{bmatrix}$ for some scalars $c_1, c_2, c_3$.
This gives us the linear system $\left[\begin{array}{ccc|c} -2 & -2 & 6 & -3 \\ 3 & 3 & -9 & -2 \end{array}\right]$, which row reduces to $\left[\begin{array}{ccc|c} 1 & 1 & -3 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right]$.
Row 2 has the form [0 ... 0 | 1], which represents the equation 0 = 1, indicating no solution exists for the system.

Q: Is $\begin{bmatrix} -3 \\ -2 \end{bmatrix}$ a linear combination of $\begin{bmatrix} -2 \\ 3 \end{bmatrix}$, $\begin{bmatrix} -2 \\ 3 \end{bmatrix}$, and $\begin{bmatrix} 2 \\ -3 \end{bmatrix}$?
A: No. $\begin{bmatrix} -3 \\ -2 \end{bmatrix}$ is NOT in the span of $\begin{bmatrix} -2 \\ 3 \end{bmatrix}$, $\begin{bmatrix} -2 \\ 3 \end{bmatrix}$, and $\begin{bmatrix} 2 \\ -3 \end{bmatrix}$.
For $\begin{bmatrix} -3 \\ -2 \end{bmatrix}$ to be in the span, we need $\begin{bmatrix} -3 \\ -2 \end{bmatrix} = c_{1}\begin{bmatrix} -2 \\ 3 \end{bmatrix} + c_{2}\begin{bmatrix} -2 \\ 3 \end{bmatrix} + c_{3}\begin{bmatrix} 2 \\ -3 \end{bmatrix}$ for some scalars $c_1, c_2, c_3$.
This gives us the linear system $\left[\begin{array}{ccc|c} -2 & -2 & 2 & -3 \\ 3 & 3 & -3 & -2 \end{array}\right]$, which row reduces to $\left[\begin{array}{ccc|c} 1 & 1 & -1 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right]$.
Row 2 has the form [0 ... 0 | 1], which represents the equation 0 = 1, indicating no solution exists for the system.

Q: Is $\begin{bmatrix} 3 \\ -3 \end{bmatrix}$ a linear combination of $\begin{bmatrix} -3 \\ -3 \end{bmatrix}$, $\begin{bmatrix} -6 \\ -6 \end{bmatrix}$, and $\begin{bmatrix} 9 \\ 9 \end{bmatrix}$?
A: No. $\begin{bmatrix} 3 \\ -3 \end{bmatrix}$ is NOT in the span of $\begin{bmatrix} -3 \\ -3 \end{bmatrix}$, $\begin{bmatrix} -6 \\ -6 \end{bmatrix}$, and $\begin{bmatrix} 9 \\ 9 \end{bmatrix}$.
For $\begin{bmatrix} 3 \\ -3 \end{bmatrix}$ to be in the span, we need $\begin{bmatrix} 3 \\ -3 \end{bmatrix} = c_{1}\begin{bmatrix} -3 \\ -3 \end{bmatrix} + c_{2}\begin{bmatrix} -6 \\ -6 \end{bmatrix} + c_{3}\begin{bmatrix} 9 \\ 9 \end{bmatrix}$ for some scalars $c_1, c_2, c_3$.
This gives us the linear system $\left[\begin{array}{ccc|c} -3 & -6 & 9 & 3 \\ -3 & -6 & 9 & -3 \end{array}\right]$, which row reduces to $\left[\begin{array}{ccc|c} 1 & 2 & -3 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right]$.
Row 2 has the form [0 ... 0 | 1], which represents the equation 0 = 1, indicating no solution exists for the system.

Q: Is the vector $\begin{bmatrix} -3 \\ 2 \end{bmatrix}$ in the span of $\left\{ \begin{bmatrix} 2 \\ 3 \end{bmatrix}, \begin{bmatrix} 2 \\ 3 \end{bmatrix}, \begin{bmatrix} -2 \\ -3 \end{bmatrix} \right\}$?
A: No. $\begin{bmatrix} -3 \\ 2 \end{bmatrix}$ is NOT in the span of $\begin{bmatrix} 2 \\ 3 \end{bmatrix}$, $\begin{bmatrix} 2 \\ 3 \end{bmatrix}$, and $\begin{bmatrix} -2 \\ -3 \end{bmatrix}$.
For $\begin{bmatrix} -3 \\ 2 \end{bmatrix}$ to be in the span, we need $\begin{bmatrix} -3 \\ 2 \end{bmatrix} = c_{1}\begin{bmatrix} 2 \\ 3 \end{bmatrix} + c_{2}\begin{bmatrix} 2 \\ 3 \end{bmatrix} + c_{3}\begin{bmatrix} -2 \\ -3 \end{bmatrix}$ for some scalars $c_1, c_2, c_3$.
This gives us the linear system $\left[\begin{array}{ccc|c} 2 & 2 & -2 & -3 \\ 3 & 3 & -3 & 2 \end{array}\right]$, which row reduces to $\left[\begin{array}{ccc|c} 1 & 1 & -1 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right]$.
Row 2 has the form [0 ... 0 | 1], which represents the equation 0 = 1, indicating no solution exists for the system.

Q: Is $\begin{bmatrix} 0 \\ 3 \end{bmatrix}$ a linear combination of $\begin{bmatrix} 3 \\ 0 \end{bmatrix}$, $\begin{bmatrix} 9 \\ 0 \end{bmatrix}$, and $\begin{bmatrix} -9 \\ 0 \end{bmatrix}$?
A: No. $\begin{bmatrix} 0 \\ 3 \end{bmatrix}$ is NOT in the span of $\begin{bmatrix} 3 \\ 0 \end{bmatrix}$, $\begin{bmatrix} 9 \\ 0 \end{bmatrix}$, and $\begin{bmatrix} -9 \\ 0 \end{bmatrix}$.
For $\begin{bmatrix} 0 \\ 3 \end{bmatrix}$ to be in the span, we need $\begin{bmatrix} 0 \\ 3 \end{bmatrix} = c_{1}\begin{bmatrix} 3 \\ 0 \end{bmatrix} + c_{2}\begin{bmatrix} 9 \\ 0 \end{bmatrix} + c_{3}\begin{bmatrix} -9 \\ 0 \end{bmatrix}$ for some scalars $c_1, c_2, c_3$.
This gives us the linear system $\left[\begin{array}{ccc|c} 3 & 9 & -9 & 0 \\ 0 & 0 & 0 & 3 \end{array}\right]$, which row reduces to $\left[\begin{array}{ccc|c} 1 & 3 & -3 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right]$.
Row 2 has the form [0 ... 0 | 1], which represents the equation 0 = 1, indicating no solution exists for the system.

Q: Is $\begin{bmatrix} 0 \\ 1 \end{bmatrix}$ a linear combination of $\begin{bmatrix} 1 \\ 0 \end{bmatrix}$, $\begin{bmatrix} 3 \\ 0 \end{bmatrix}$, and $\begin{bmatrix} -3 \\ 0 \end{bmatrix}$?
A: No. $\begin{bmatrix} 0 \\ 1 \end{bmatrix}$ is NOT in the span of $\begin{bmatrix} 1 \\ 0 \end{bmatrix}$, $\begin{bmatrix} 3 \\ 0 \end{bmatrix}$, and $\begin{bmatrix} -3 \\ 0 \end{bmatrix}$.
For $\begin{bmatrix} 0 \\ 1 \end{bmatrix}$ to be in the span, we need $\begin{bmatrix} 0 \\ 1 \end{bmatrix} = c_{1}\begin{bmatrix} 1 \\ 0 \end{bmatrix} + c_{2}\begin{bmatrix} 3 \\ 0 \end{bmatrix} + c_{3}\begin{bmatrix} -3 \\ 0 \end{bmatrix}$ for some scalars $c_1, c_2, c_3$.
This gives us the linear system $\left[\begin{array}{ccc|c} 1 & 3 & -3 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right]$, which row reduces to $\left[\begin{array}{ccc|c} 1 & 3 & -3 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right]$.
Row 2 has the form [0 ... 0 | 1], which represents the equation 0 = 1, indicating no solution exists for the system.

Q: Is $\begin{bmatrix} -3 \\ 0 \end{bmatrix}$ a linear combination of $\begin{bmatrix} 0 \\ 3 \end{bmatrix}$, $\begin{bmatrix} 0 \\ 9 \end{bmatrix}$, and $\begin{bmatrix} 0 \\ -6 \end{bmatrix}$?
A: No. $\begin{bmatrix} -3 \\ 0 \end{bmatrix}$ is NOT in the span of $\begin{bmatrix} 0 \\ 3 \end{bmatrix}$, $\begin{bmatrix} 0 \\ 9 \end{bmatrix}$, and $\begin{bmatrix} 0 \\ -6 \end{bmatrix}$.
For $\begin{bmatrix} -3 \\ 0 \end{bmatrix}$ to be in the span, we need $\begin{bmatrix} -3 \\ 0 \end{bmatrix} = c_{1}\begin{bmatrix} 0 \\ 3 \end{bmatrix} + c_{2}\begin{bmatrix} 0 \\ 9 \end{bmatrix} + c_{3}\begin{bmatrix} 0 \\ -6 \end{bmatrix}$ for some scalars $c_1, c_2, c_3$.
This gives us the linear system $\left[\begin{array}{ccc|c} 0 & 0 & 0 & -3 \\ 3 & 9 & -6 & 0 \end{array}\right]$, which row reduces to $\left[\begin{array}{ccc|c} 1 & 3 & -2 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right]$.
Row 2 has the form [0 ... 0 | 1], which represents the equation 0 = 1, indicating no solution exists for the system.

Q: Is $\begin{bmatrix} -3 \\ 2 \end{bmatrix}$ a linear combination of $\begin{bmatrix} 2 \\ 3 \end{bmatrix}$, $\begin{bmatrix} 4 \\ 6 \end{bmatrix}$, and $\begin{bmatrix} -4 \\ -6 \end{bmatrix}$?
A: No. $\begin{bmatrix} -3 \\ 2 \end{bmatrix}$ is NOT in the span of $\begin{bmatrix} 2 \\ 3 \end{bmatrix}$, $\begin{bmatrix} 4 \\ 6 \end{bmatrix}$, and $\begin{bmatrix} -4 \\ -6 \end{bmatrix}$.
For $\begin{bmatrix} -3 \\ 2 \end{bmatrix}$ to be in the span, we need $\begin{bmatrix} -3 \\ 2 \end{bmatrix} = c_{1}\begin{bmatrix} 2 \\ 3 \end{bmatrix} + c_{2}\begin{bmatrix} 4 \\ 6 \end{bmatrix} + c_{3}\begin{bmatrix} -4 \\ -6 \end{bmatrix}$ for some scalars $c_1, c_2, c_3$.
This gives us the linear system $\left[\begin{array}{ccc|c} 2 & 4 & -4 & -3 \\ 3 & 6 & -6 & 2 \end{array}\right]$, which row reduces to $\left[\begin{array}{ccc|c} 1 & 2 & -2 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right]$.
Row 2 has the form [0 ... 0 | 1], which represents the equation 0 = 1, indicating no solution exists for the system.

Q: Is the vector $\begin{bmatrix} -3 \\ 0 \end{bmatrix}$ in the span of $\left\{ \begin{bmatrix} 0 \\ 3 \end{bmatrix}, \begin{bmatrix} 0 \\ 9 \end{bmatrix}, \begin{bmatrix} 0 \\ -3 \end{bmatrix} \right\}$?
A: No. $\begin{bmatrix} -3 \\ 0 \end{bmatrix}$ is NOT in the span of $\begin{bmatrix} 0 \\ 3 \end{bmatrix}$, $\begin{bmatrix} 0 \\ 9 \end{bmatrix}$, and $\begin{bmatrix} 0 \\ -3 \end{bmatrix}$.
For $\begin{bmatrix} -3 \\ 0 \end{bmatrix}$ to be in the span, we need $\begin{bmatrix} -3 \\ 0 \end{bmatrix} = c_{1}\begin{bmatrix} 0 \\ 3 \end{bmatrix} + c_{2}\begin{bmatrix} 0 \\ 9 \end{bmatrix} + c_{3}\begin{bmatrix} 0 \\ -3 \end{bmatrix}$ for some scalars $c_1, c_2, c_3$.
This gives us the linear system $\left[\begin{array}{ccc|c} 0 & 0 & 0 & -3 \\ 3 & 9 & -3 & 0 \end{array}\right]$, which row reduces to $\left[\begin{array}{ccc|c} 1 & 3 & -1 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right]$.
Row 2 has the form [0 ... 0 | 1], which represents the equation 0 = 1, indicating no solution exists for the system.

Q: Is the vector $\begin{bmatrix} 0 \\ 1 \end{bmatrix}$ in the span of $\left\{ \begin{bmatrix} 1 \\ 0 \end{bmatrix}, \begin{bmatrix} 3 \\ 0 \end{bmatrix}, \begin{bmatrix} -2 \\ 0 \end{bmatrix} \right\}$?
A: No. $\begin{bmatrix} 0 \\ 1 \end{bmatrix}$ is NOT in the span of $\begin{bmatrix} 1 \\ 0 \end{bmatrix}$, $\begin{bmatrix} 3 \\ 0 \end{bmatrix}$, and $\begin{bmatrix} -2 \\ 0 \end{bmatrix}$.
For $\begin{bmatrix} 0 \\ 1 \end{bmatrix}$ to be in the span, we need $\begin{bmatrix} 0 \\ 1 \end{bmatrix} = c_{1}\begin{bmatrix} 1 \\ 0 \end{bmatrix} + c_{2}\begin{bmatrix} 3 \\ 0 \end{bmatrix} + c_{3}\begin{bmatrix} -2 \\ 0 \end{bmatrix}$ for some scalars $c_1, c_2, c_3$.
This gives us the linear system $\left[\begin{array}{ccc|c} 1 & 3 & -2 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right]$, which row reduces to $\left[\begin{array}{ccc|c} 1 & 3 & -2 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right]$.
Row 2 has the form [0 ... 0 | 1], which represents the equation 0 = 1, indicating no solution exists for the system.

Q: Is the vector $\begin{bmatrix} 3 \\ 1 \end{bmatrix}$ in the span of $\left\{ \begin{bmatrix} 1 \\ -3 \end{bmatrix}, \begin{bmatrix} 2 \\ -6 \end{bmatrix}, \begin{bmatrix} -3 \\ 9 \end{bmatrix} \right\}$?
A: No. $\begin{bmatrix} 3 \\ 1 \end{bmatrix}$ is NOT in the span of $\begin{bmatrix} 1 \\ -3 \end{bmatrix}$, $\begin{bmatrix} 2 \\ -6 \end{bmatrix}$, and $\begin{bmatrix} -3 \\ 9 \end{bmatrix}$.
For $\begin{bmatrix} 3 \\ 1 \end{bmatrix}$ to be in the span, we need $\begin{bmatrix} 3 \\ 1 \end{bmatrix} = c_{1}\begin{bmatrix} 1 \\ -3 \end{bmatrix} + c_{2}\begin{bmatrix} 2 \\ -6 \end{bmatrix} + c_{3}\begin{bmatrix} -3 \\ 9 \end{bmatrix}$ for some scalars $c_1, c_2, c_3$.
This gives us the linear system $\left[\begin{array}{ccc|c} 1 & 2 & -3 & 3 \\ -3 & -6 & 9 & 1 \end{array}\right]$, which row reduces to $\left[\begin{array}{ccc|c} 1 & 2 & -3 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right]$.
Row 2 has the form [0 ... 0 | 1], which represents the equation 0 = 1, indicating no solution exists for the system.

Q: Is the vector $\begin{bmatrix} 2 \\ 0 \end{bmatrix}$ in the span of $\left\{ \begin{bmatrix} 0 \\ -2 \end{bmatrix}, \begin{bmatrix} 0 \\ -2 \end{bmatrix}, \begin{bmatrix} 0 \\ 4 \end{bmatrix} \right\}$?
A: No. $\begin{bmatrix} 2 \\ 0 \end{bmatrix}$ is NOT in the span of $\begin{bmatrix} 0 \\ -2 \end{bmatrix}$, $\begin{bmatrix} 0 \\ -2 \end{bmatrix}$, and $\begin{bmatrix} 0 \\ 4 \end{bmatrix}$.
For $\begin{bmatrix} 2 \\ 0 \end{bmatrix}$ to be in the span, we need $\begin{bmatrix} 2 \\ 0 \end{bmatrix} = c_{1}\begin{bmatrix} 0 \\ -2 \end{bmatrix} + c_{2}\begin{bmatrix} 0 \\ -2 \end{bmatrix} + c_{3}\begin{bmatrix} 0 \\ 4 \end{bmatrix}$ for some scalars $c_1, c_2, c_3$.
This gives us the linear system $\left[\begin{array}{ccc|c} 0 & 0 & 0 & 2 \\ -2 & -2 & 4 & 0 \end{array}\right]$, which row reduces to $\left[\begin{array}{ccc|c} 1 & 1 & -2 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right]$.
Row 2 has the form [0 ... 0 | 1], which represents the equation 0 = 1, indicating no solution exists for the system.

Q: Is $\begin{bmatrix} -1 \\ -2 \end{bmatrix}$ a linear combination of $\begin{bmatrix} -2 \\ 1 \end{bmatrix}$, $\begin{bmatrix} -4 \\ 2 \end{bmatrix}$, and $\begin{bmatrix} 4 \\ -2 \end{bmatrix}$?
A: No. $\begin{bmatrix} -1 \\ -2 \end{bmatrix}$ is NOT in the span of $\begin{bmatrix} -2 \\ 1 \end{bmatrix}$, $\begin{bmatrix} -4 \\ 2 \end{bmatrix}$, and $\begin{bmatrix} 4 \\ -2 \end{bmatrix}$.
For $\begin{bmatrix} -1 \\ -2 \end{bmatrix}$ to be in the span, we need $\begin{bmatrix} -1 \\ -2 \end{bmatrix} = c_{1}\begin{bmatrix} -2 \\ 1 \end{bmatrix} + c_{2}\begin{bmatrix} -4 \\ 2 \end{bmatrix} + c_{3}\begin{bmatrix} 4 \\ -2 \end{bmatrix}$ for some scalars $c_1, c_2, c_3$.
This gives us the linear system $\left[\begin{array}{ccc|c} -2 & -4 & 4 & -1 \\ 1 & 2 & -2 & -2 \end{array}\right]$, which row reduces to $\left[\begin{array}{ccc|c} 1 & 2 & -2 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right]$.
Row 2 has the form [0 ... 0 | 1], which represents the equation 0 = 1, indicating no solution exists for the system.

Q: Is $\begin{bmatrix} 1 \\ 3 \end{bmatrix}$ a linear combination of $\begin{bmatrix} 3 \\ -1 \end{bmatrix}$, $\begin{bmatrix} 6 \\ -2 \end{bmatrix}$, and $\begin{bmatrix} -6 \\ 2 \end{bmatrix}$?
A: No. $\begin{bmatrix} 1 \\ 3 \end{bmatrix}$ is NOT in the span of $\begin{bmatrix} 3 \\ -1 \end{bmatrix}$, $\begin{bmatrix} 6 \\ -2 \end{bmatrix}$, and $\begin{bmatrix} -6 \\ 2 \end{bmatrix}$.
For $\begin{bmatrix} 1 \\ 3 \end{bmatrix}$ to be in the span, we need $\begin{bmatrix} 1 \\ 3 \end{bmatrix} = c_{1}\begin{bmatrix} 3 \\ -1 \end{bmatrix} + c_{2}\begin{bmatrix} 6 \\ -2 \end{bmatrix} + c_{3}\begin{bmatrix} -6 \\ 2 \end{bmatrix}$ for some scalars $c_1, c_2, c_3$.
This gives us the linear system $\left[\begin{array}{ccc|c} 3 & 6 & -6 & 1 \\ -1 & -2 & 2 & 3 \end{array}\right]$, which row reduces to $\left[\begin{array}{ccc|c} 1 & 2 & -2 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right]$.
Row 2 has the form [0 ... 0 | 1], which represents the equation 0 = 1, indicating no solution exists for the system.

Q: Is $\begin{bmatrix} 2 \\ -3 \end{bmatrix}$ a linear combination of $\begin{bmatrix} -3 \\ -2 \end{bmatrix}$, $\begin{bmatrix} -6 \\ -4 \end{bmatrix}$, and $\begin{bmatrix} 3 \\ 2 \end{bmatrix}$?
A: No. $\begin{bmatrix} 2 \\ -3 \end{bmatrix}$ is NOT in the span of $\begin{bmatrix} -3 \\ -2 \end{bmatrix}$, $\begin{bmatrix} -6 \\ -4 \end{bmatrix}$, and $\begin{bmatrix} 3 \\ 2 \end{bmatrix}$.
For $\begin{bmatrix} 2 \\ -3 \end{bmatrix}$ to be in the span, we need $\begin{bmatrix} 2 \\ -3 \end{bmatrix} = c_{1}\begin{bmatrix} -3 \\ -2 \end{bmatrix} + c_{2}\begin{bmatrix} -6 \\ -4 \end{bmatrix} + c_{3}\begin{bmatrix} 3 \\ 2 \end{bmatrix}$ for some scalars $c_1, c_2, c_3$.
This gives us the linear system $\left[\begin{array}{ccc|c} -3 & -6 & 3 & 2 \\ -2 & -4 & 2 & -3 \end{array}\right]$, which row reduces to $\left[\begin{array}{ccc|c} 1 & 2 & -1 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right]$.
Row 2 has the form [0 ... 0 | 1], which represents the equation 0 = 1, indicating no solution exists for the system.

Q: Is $\begin{bmatrix} 3 \\ -3 \end{bmatrix}$ a linear combination of $\begin{bmatrix} -3 \\ -3 \end{bmatrix}$, $\begin{bmatrix} -6 \\ -6 \end{bmatrix}$, and $\begin{bmatrix} 6 \\ 6 \end{bmatrix}$?
A: No. $\begin{bmatrix} 3 \\ -3 \end{bmatrix}$ is NOT in the span of $\begin{bmatrix} -3 \\ -3 \end{bmatrix}$, $\begin{bmatrix} -6 \\ -6 \end{bmatrix}$, and $\begin{bmatrix} 6 \\ 6 \end{bmatrix}$.
For $\begin{bmatrix} 3 \\ -3 \end{bmatrix}$ to be in the span, we need $\begin{bmatrix} 3 \\ -3 \end{bmatrix} = c_{1}\begin{bmatrix} -3 \\ -3 \end{bmatrix} + c_{2}\begin{bmatrix} -6 \\ -6 \end{bmatrix} + c_{3}\begin{bmatrix} 6 \\ 6 \end{bmatrix}$ for some scalars $c_1, c_2, c_3$.
This gives us the linear system $\left[\begin{array}{ccc|c} -3 & -6 & 6 & 3 \\ -3 & -6 & 6 & -3 \end{array}\right]$, which row reduces to $\left[\begin{array}{ccc|c} 1 & 2 & -2 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right]$.
Row 2 has the form [0 ... 0 | 1], which represents the equation 0 = 1, indicating no solution exists for the system.

Q: Is the vector $\begin{bmatrix} -2 \\ 2 \\ 0 \end{bmatrix}$ in the span of $\left\{ \begin{bmatrix} 2 \\ 2 \\ 1 \end{bmatrix}, \begin{bmatrix} 2 \\ 2 \\ -1 \end{bmatrix} \right\}$?
A: No. $\begin{bmatrix} -2 \\ 2 \\ 0 \end{bmatrix}$ is NOT in the span of $\begin{bmatrix} 2 \\ 2 \\ 1 \end{bmatrix}$ and $\begin{bmatrix} 2 \\ 2 \\ -1 \end{bmatrix}$.
For $\begin{bmatrix} -2 \\ 2 \\ 0 \end{bmatrix}$ to be in the span, we need $\begin{bmatrix} -2 \\ 2 \\ 0 \end{bmatrix} = c_{1}\begin{bmatrix} 2 \\ 2 \\ 1 \end{bmatrix} + c_{2}\begin{bmatrix} 2 \\ 2 \\ -1 \end{bmatrix}$ for some scalars $c_1, c_2$.
This gives us the linear system $\left[\begin{array}{cc|c} 2 & 2 & -2 \\ 2 & 2 & 2 \\ 1 & -1 & 0 \end{array}\right]$, which row reduces to $\left[\begin{array}{cc|c} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]$.
Row 3 has the form [0 ... 0 | 1], which represents the equation 0 = 1, indicating no solution exists for the system.

Q: Is the vector $\begin{bmatrix} -2 \\ 2 \\ 2 \end{bmatrix}$ in the span of $\left\{ \begin{bmatrix} 3 \\ -1 \\ 3 \end{bmatrix}, \begin{bmatrix} 2 \\ 3 \\ -1 \end{bmatrix} \right\}$?
A: No. $\begin{bmatrix} -2 \\ 2 \\ 2 \end{bmatrix}$ is NOT in the span of $\begin{bmatrix} 3 \\ -1 \\ 3 \end{bmatrix}$ and $\begin{bmatrix} 2 \\ 3 \\ -1 \end{bmatrix}$.
For $\begin{bmatrix} -2 \\ 2 \\ 2 \end{bmatrix}$ to be in the span, we need $\begin{bmatrix} -2 \\ 2 \\ 2 \end{bmatrix} = c_{1}\begin{bmatrix} 3 \\ -1 \\ 3 \end{bmatrix} + c_{2}\begin{bmatrix} 2 \\ 3 \\ -1 \end{bmatrix}$ for some scalars $c_1, c_2$.
This gives us the linear system $\left[\begin{array}{cc|c} 3 & 2 & -2 \\ -1 & 3 & 2 \\ 3 & -1 & 2 \end{array}\right]$, which row reduces to $\left[\begin{array}{cc|c} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]$.
Row 3 has the form [0 ... 0 | 1], which represents the equation 0 = 1, indicating no solution exists for the system.

Q: Is $\begin{bmatrix} 3 \\ 0 \\ 3 \end{bmatrix}$ a linear combination of $\begin{bmatrix} 3 \\ -3 \\ -3 \end{bmatrix}$ and $\begin{bmatrix} 2 \\ 2 \\ -3 \end{bmatrix}$?
A: No. $\begin{bmatrix} 3 \\ 0 \\ 3 \end{bmatrix}$ is NOT in the span of $\begin{bmatrix} 3 \\ -3 \\ -3 \end{bmatrix}$ and $\begin{bmatrix} 2 \\ 2 \\ -3 \end{bmatrix}$.
For $\begin{bmatrix} 3 \\ 0 \\ 3 \end{bmatrix}$ to be in the span, we need $\begin{bmatrix} 3 \\ 0 \\ 3 \end{bmatrix} = c_{1}\begin{bmatrix} 3 \\ -3 \\ -3 \end{bmatrix} + c_{2}\begin{bmatrix} 2 \\ 2 \\ -3 \end{bmatrix}$ for some scalars $c_1, c_2$.
This gives us the linear system $\left[\begin{array}{cc|c} 3 & 2 & 3 \\ -3 & 2 & 0 \\ -3 & -3 & 3 \end{array}\right]$, which row reduces to $\left[\begin{array}{cc|c} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]$.
Row 3 has the form [0 ... 0 | 1], which represents the equation 0 = 1, indicating no solution exists for the system.

Q: Is the vector $\begin{bmatrix} 3 \\ -2 \\ 1 \end{bmatrix}$ in the span of $\left\{ \begin{bmatrix} 1 \\ 2 \\ 0 \end{bmatrix}, \begin{bmatrix} 0 \\ 3 \\ 3 \end{bmatrix} \right\}$?
A: No. $\begin{bmatrix} 3 \\ -2 \\ 1 \end{bmatrix}$ is NOT in the span of $\begin{bmatrix} 1 \\ 2 \\ 0 \end{bmatrix}$ and $\begin{bmatrix} 0 \\ 3 \\ 3 \end{bmatrix}$.
For $\begin{bmatrix} 3 \\ -2 \\ 1 \end{bmatrix}$ to be in the span, we need $\begin{bmatrix} 3 \\ -2 \\ 1 \end{bmatrix} = c_{1}\begin{bmatrix} 1 \\ 2 \\ 0 \end{bmatrix} + c_{2}\begin{bmatrix} 0 \\ 3 \\ 3 \end{bmatrix}$ for some scalars $c_1, c_2$.
This gives us the linear system $\left[\begin{array}{cc|c} 1 & 0 & 3 \\ 2 & 3 & -2 \\ 0 & 3 & 1 \end{array}\right]$, which row reduces to $\left[\begin{array}{cc|c} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]$.
Row 3 has the form [0 ... 0 | 1], which represents the equation 0 = 1, indicating no solution exists for the system.

Q: Is $\begin{bmatrix} -3 \\ 3 \\ 0 \end{bmatrix}$ a linear combination of $\begin{bmatrix} -3 \\ -2 \\ -3 \end{bmatrix}$ and $\begin{bmatrix} -1 \\ -1 \\ 3 \end{bmatrix}$?
A: No. $\begin{bmatrix} -3 \\ 3 \\ 0 \end{bmatrix}$ is NOT in the span of $\begin{bmatrix} -3 \\ -2 \\ -3 \end{bmatrix}$ and $\begin{bmatrix} -1 \\ -1 \\ 3 \end{bmatrix}$.
For $\begin{bmatrix} -3 \\ 3 \\ 0 \end{bmatrix}$ to be in the span, we need $\begin{bmatrix} -3 \\ 3 \\ 0 \end{bmatrix} = c_{1}\begin{bmatrix} -3 \\ -2 \\ -3 \end{bmatrix} + c_{2}\begin{bmatrix} -1 \\ -1 \\ 3 \end{bmatrix}$ for some scalars $c_1, c_2$.
This gives us the linear system $\left[\begin{array}{cc|c} -3 & -1 & -3 \\ -2 & -1 & 3 \\ -3 & 3 & 0 \end{array}\right]$, which row reduces to $\left[\begin{array}{cc|c} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]$.
Row 3 has the form [0 ... 0 | 1], which represents the equation 0 = 1, indicating no solution exists for the system.

Q: Is the vector $\begin{bmatrix} -1 \\ -2 \\ 2 \end{bmatrix}$ in the span of $\left\{ \begin{bmatrix} 1 \\ 2 \\ 2 \end{bmatrix}, \begin{bmatrix} -1 \\ 2 \\ 1 \end{bmatrix} \right\}$?
A: No. $\begin{bmatrix} -1 \\ -2 \\ 2 \end{bmatrix}$ is NOT in the span of $\begin{bmatrix} 1 \\ 2 \\ 2 \end{bmatrix}$ and $\begin{bmatrix} -1 \\ 2 \\ 1 \end{bmatrix}$.
For $\begin{bmatrix} -1 \\ -2 \\ 2 \end{bmatrix}$ to be in the span, we need $\begin{bmatrix} -1 \\ -2 \\ 2 \end{bmatrix} = c_{1}\begin{bmatrix} 1 \\ 2 \\ 2 \end{bmatrix} + c_{2}\begin{bmatrix} -1 \\ 2 \\ 1 \end{bmatrix}$ for some scalars $c_1, c_2$.
This gives us the linear system $\left[\begin{array}{cc|c} 1 & -1 & -1 \\ 2 & 2 & -2 \\ 2 & 1 & 2 \end{array}\right]$, which row reduces to $\left[\begin{array}{cc|c} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]$.
Row 3 has the form [0 ... 0 | 1], which represents the equation 0 = 1, indicating no solution exists for the system.

Q: Is $\begin{bmatrix} -2 \\ -2 \\ 0 \end{bmatrix}$ a linear combination of $\begin{bmatrix} -2 \\ 2 \\ 2 \end{bmatrix}$ and $\begin{bmatrix} -1 \\ 1 \\ -3 \end{bmatrix}$?
A: No. $\begin{bmatrix} -2 \\ -2 \\ 0 \end{bmatrix}$ is NOT in the span of $\begin{bmatrix} -2 \\ 2 \\ 2 \end{bmatrix}$ and $\begin{bmatrix} -1 \\ 1 \\ -3 \end{bmatrix}$.
For $\begin{bmatrix} -2 \\ -2 \\ 0 \end{bmatrix}$ to be in the span, we need $\begin{bmatrix} -2 \\ -2 \\ 0 \end{bmatrix} = c_{1}\begin{bmatrix} -2 \\ 2 \\ 2 \end{bmatrix} + c_{2}\begin{bmatrix} -1 \\ 1 \\ -3 \end{bmatrix}$ for some scalars $c_1, c_2$.
This gives us the linear system $\left[\begin{array}{cc|c} -2 & -1 & -2 \\ 2 & 1 & -2 \\ 2 & -3 & 0 \end{array}\right]$, which row reduces to $\left[\begin{array}{cc|c} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]$.
Row 3 has the form [0 ... 0 | 1], which represents the equation 0 = 1, indicating no solution exists for the system.

Q: Is the vector $\begin{bmatrix} -1 \\ -3 \\ 2 \end{bmatrix}$ in the span of $\left\{ \begin{bmatrix} 3 \\ 1 \\ 3 \end{bmatrix}, \begin{bmatrix} -2 \\ 0 \\ -1 \end{bmatrix} \right\}$?
A: No. $\begin{bmatrix} -1 \\ -3 \\ 2 \end{bmatrix}$ is NOT in the span of $\begin{bmatrix} 3 \\ 1 \\ 3 \end{bmatrix}$ and $\begin{bmatrix} -2 \\ 0 \\ -1 \end{bmatrix}$.
For $\begin{bmatrix} -1 \\ -3 \\ 2 \end{bmatrix}$ to be in the span, we need $\begin{bmatrix} -1 \\ -3 \\ 2 \end{bmatrix} = c_{1}\begin{bmatrix} 3 \\ 1 \\ 3 \end{bmatrix} + c_{2}\begin{bmatrix} -2 \\ 0 \\ -1 \end{bmatrix}$ for some scalars $c_1, c_2$.
This gives us the linear system $\left[\begin{array}{cc|c} 3 & -2 & -1 \\ 1 & 0 & -3 \\ 3 & -1 & 2 \end{array}\right]$, which row reduces to $\left[\begin{array}{cc|c} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]$.
Row 3 has the form [0 ... 0 | 1], which represents the equation 0 = 1, indicating no solution exists for the system.

Q: Is the vector $\begin{bmatrix} -1 \\ -1 \\ -3 \end{bmatrix}$ in the span of $\left\{ \begin{bmatrix} -3 \\ -3 \\ 2 \end{bmatrix}, \begin{bmatrix} -2 \\ 2 \\ 0 \end{bmatrix} \right\}$?
A: No. $\begin{bmatrix} -1 \\ -1 \\ -3 \end{bmatrix}$ is NOT in the span of $\begin{bmatrix} -3 \\ -3 \\ 2 \end{bmatrix}$ and $\begin{bmatrix} -2 \\ 2 \\ 0 \end{bmatrix}$.
For $\begin{bmatrix} -1 \\ -1 \\ -3 \end{bmatrix}$ to be in the span, we need $\begin{bmatrix} -1 \\ -1 \\ -3 \end{bmatrix} = c_{1}\begin{bmatrix} -3 \\ -3 \\ 2 \end{bmatrix} + c_{2}\begin{bmatrix} -2 \\ 2 \\ 0 \end{bmatrix}$ for some scalars $c_1, c_2$.
This gives us the linear system $\left[\begin{array}{cc|c} -3 & -2 & -1 \\ -3 & 2 & -1 \\ 2 & 0 & -3 \end{array}\right]$, which row reduces to $\left[\begin{array}{cc|c} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]$.
Row 3 has the form [0 ... 0 | 1], which represents the equation 0 = 1, indicating no solution exists for the system.

Q: Is $\begin{bmatrix} 2 \\ 1 \\ -2 \end{bmatrix}$ a linear combination of $\begin{bmatrix} -1 \\ 2 \\ 0 \end{bmatrix}$ and $\begin{bmatrix} 1 \\ 2 \\ 2 \end{bmatrix}$?
A: No. $\begin{bmatrix} 2 \\ 1 \\ -2 \end{bmatrix}$ is NOT in the span of $\begin{bmatrix} -1 \\ 2 \\ 0 \end{bmatrix}$ and $\begin{bmatrix} 1 \\ 2 \\ 2 \end{bmatrix}$.
For $\begin{bmatrix} 2 \\ 1 \\ -2 \end{bmatrix}$ to be in the span, we need $\begin{bmatrix} 2 \\ 1 \\ -2 \end{bmatrix} = c_{1}\begin{bmatrix} -1 \\ 2 \\ 0 \end{bmatrix} + c_{2}\begin{bmatrix} 1 \\ 2 \\ 2 \end{bmatrix}$ for some scalars $c_1, c_2$.
This gives us the linear system $\left[\begin{array}{cc|c} -1 & 1 & 2 \\ 2 & 2 & 1 \\ 0 & 2 & -2 \end{array}\right]$, which row reduces to $\left[\begin{array}{cc|c} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]$.
Row 3 has the form [0 ... 0 | 1], which represents the equation 0 = 1, indicating no solution exists for the system.

Q: Is the vector $\begin{bmatrix} -2 \\ 0 \\ 1 \end{bmatrix}$ in the span of $\left\{ \begin{bmatrix} -2 \\ 3 \\ -2 \end{bmatrix}, \begin{bmatrix} -3 \\ -3 \\ -3 \end{bmatrix} \right\}$?
A: No. $\begin{bmatrix} -2 \\ 0 \\ 1 \end{bmatrix}$ is NOT in the span of $\begin{bmatrix} -2 \\ 3 \\ -2 \end{bmatrix}$ and $\begin{bmatrix} -3 \\ -3 \\ -3 \end{bmatrix}$.
For $\begin{bmatrix} -2 \\ 0 \\ 1 \end{bmatrix}$ to be in the span, we need $\begin{bmatrix} -2 \\ 0 \\ 1 \end{bmatrix} = c_{1}\begin{bmatrix} -2 \\ 3 \\ -2 \end{bmatrix} + c_{2}\begin{bmatrix} -3 \\ -3 \\ -3 \end{bmatrix}$ for some scalars $c_1, c_2$.
This gives us the linear system $\left[\begin{array}{cc|c} -2 & -3 & -2 \\ 3 & -3 & 0 \\ -2 & -3 & 1 \end{array}\right]$, which row reduces to $\left[\begin{array}{cc|c} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]$.
Row 3 has the form [0 ... 0 | 1], which represents the equation 0 = 1, indicating no solution exists for the system.

Q: Is the vector $\begin{bmatrix} -1 \\ -2 \\ -2 \end{bmatrix}$ in the span of $\left\{ \begin{bmatrix} 3 \\ -3 \\ 2 \end{bmatrix}, \begin{bmatrix} -3 \\ -1 \\ 3 \end{bmatrix} \right\}$?
A: No. $\begin{bmatrix} -1 \\ -2 \\ -2 \end{bmatrix}$ is NOT in the span of $\begin{bmatrix} 3 \\ -3 \\ 2 \end{bmatrix}$ and $\begin{bmatrix} -3 \\ -1 \\ 3 \end{bmatrix}$.
For $\begin{bmatrix} -1 \\ -2 \\ -2 \end{bmatrix}$ to be in the span, we need $\begin{bmatrix} -1 \\ -2 \\ -2 \end{bmatrix} = c_{1}\begin{bmatrix} 3 \\ -3 \\ 2 \end{bmatrix} + c_{2}\begin{bmatrix} -3 \\ -1 \\ 3 \end{bmatrix}$ for some scalars $c_1, c_2$.
This gives us the linear system $\left[\begin{array}{cc|c} 3 & -3 & -1 \\ -3 & -1 & -2 \\ 2 & 3 & -2 \end{array}\right]$, which row reduces to $\left[\begin{array}{cc|c} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]$.
Row 3 has the form [0 ... 0 | 1], which represents the equation 0 = 1, indicating no solution exists for the system.

Q: Is the vector $\begin{bmatrix} 3 \\ -1 \\ 2 \end{bmatrix}$ in the span of $\left\{ \begin{bmatrix} 2 \\ -3 \\ -3 \end{bmatrix}, \begin{bmatrix} 2 \\ 2 \\ -2 \end{bmatrix} \right\}$?
A: No. $\begin{bmatrix} 3 \\ -1 \\ 2 \end{bmatrix}$ is NOT in the span of $\begin{bmatrix} 2 \\ -3 \\ -3 \end{bmatrix}$ and $\begin{bmatrix} 2 \\ 2 \\ -2 \end{bmatrix}$.
For $\begin{bmatrix} 3 \\ -1 \\ 2 \end{bmatrix}$ to be in the span, we need $\begin{bmatrix} 3 \\ -1 \\ 2 \end{bmatrix} = c_{1}\begin{bmatrix} 2 \\ -3 \\ -3 \end{bmatrix} + c_{2}\begin{bmatrix} 2 \\ 2 \\ -2 \end{bmatrix}$ for some scalars $c_1, c_2$.
This gives us the linear system $\left[\begin{array}{cc|c} 2 & 2 & 3 \\ -3 & 2 & -1 \\ -3 & -2 & 2 \end{array}\right]$, which row reduces to $\left[\begin{array}{cc|c} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]$.
Row 3 has the form [0 ... 0 | 1], which represents the equation 0 = 1, indicating no solution exists for the system.

Q: Is the vector $\begin{bmatrix} -1 \\ -2 \\ 1 \end{bmatrix}$ in the span of $\left\{ \begin{bmatrix} 2 \\ -1 \\ 0 \end{bmatrix}, \begin{bmatrix} -1 \\ 1 \\ 1 \end{bmatrix} \right\}$?
A: No. $\begin{bmatrix} -1 \\ -2 \\ 1 \end{bmatrix}$ is NOT in the span of $\begin{bmatrix} 2 \\ -1 \\ 0 \end{bmatrix}$ and $\begin{bmatrix} -1 \\ 1 \\ 1 \end{bmatrix}$.
For $\begin{bmatrix} -1 \\ -2 \\ 1 \end{bmatrix}$ to be in the span, we need $\begin{bmatrix} -1 \\ -2 \\ 1 \end{bmatrix} = c_{1}\begin{bmatrix} 2 \\ -1 \\ 0 \end{bmatrix} + c_{2}\begin{bmatrix} -1 \\ 1 \\ 1 \end{bmatrix}$ for some scalars $c_1, c_2$.
This gives us the linear system $\left[\begin{array}{cc|c} 2 & -1 & -1 \\ -1 & 1 & -2 \\ 0 & 1 & 1 \end{array}\right]$, which row reduces to $\left[\begin{array}{cc|c} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]$.
Row 3 has the form [0 ... 0 | 1], which represents the equation 0 = 1, indicating no solution exists for the system.

Q: Is $\begin{bmatrix} 2 \\ -1 \\ -2 \end{bmatrix}$ a linear combination of $\begin{bmatrix} 0 \\ 3 \\ -1 \end{bmatrix}$ and $\begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix}$?
A: No. $\begin{bmatrix} 2 \\ -1 \\ -2 \end{bmatrix}$ is NOT in the span of $\begin{bmatrix} 0 \\ 3 \\ -1 \end{bmatrix}$ and $\begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix}$.
For $\begin{bmatrix} 2 \\ -1 \\ -2 \end{bmatrix}$ to be in the span, we need $\begin{bmatrix} 2 \\ -1 \\ -2 \end{bmatrix} = c_{1}\begin{bmatrix} 0 \\ 3 \\ -1 \end{bmatrix} + c_{2}\begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix}$ for some scalars $c_1, c_2$.
This gives us the linear system $\left[\begin{array}{cc|c} 0 & 1 & 2 \\ 3 & 1 & -1 \\ -1 & 1 & -2 \end{array}\right]$, which row reduces to $\left[\begin{array}{cc|c} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]$.
Row 3 has the form [0 ... 0 | 1], which represents the equation 0 = 1, indicating no solution exists for the system.

Q: Is $\begin{bmatrix} 1 \\ -1 \\ -3 \end{bmatrix}$ a linear combination of $\begin{bmatrix} 3 \\ -1 \\ 2 \end{bmatrix}$ and $\begin{bmatrix} 0 \\ -3 \\ 1 \end{bmatrix}$?
A: No. $\begin{bmatrix} 1 \\ -1 \\ -3 \end{bmatrix}$ is NOT in the span of $\begin{bmatrix} 3 \\ -1 \\ 2 \end{bmatrix}$ and $\begin{bmatrix} 0 \\ -3 \\ 1 \end{bmatrix}$.
For $\begin{bmatrix} 1 \\ -1 \\ -3 \end{bmatrix}$ to be in the span, we need $\begin{bmatrix} 1 \\ -1 \\ -3 \end{bmatrix} = c_{1}\begin{bmatrix} 3 \\ -1 \\ 2 \end{bmatrix} + c_{2}\begin{bmatrix} 0 \\ -3 \\ 1 \end{bmatrix}$ for some scalars $c_1, c_2$.
This gives us the linear system $\left[\begin{array}{cc|c} 3 & 0 & 1 \\ -1 & -3 & -1 \\ 2 & 1 & -3 \end{array}\right]$, which row reduces to $\left[\begin{array}{cc|c} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]$.
Row 3 has the form [0 ... 0 | 1], which represents the equation 0 = 1, indicating no solution exists for the system.

Q: Is the vector $\begin{bmatrix} 0 \\ -1 \\ 0 \end{bmatrix}$ in the span of $\left\{ \begin{bmatrix} -1 \\ 0 \\ -1 \end{bmatrix}, \begin{bmatrix} 0 \\ 0 \\ -1 \end{bmatrix} \right\}$?
A: No. $\begin{bmatrix} 0 \\ -1 \\ 0 \end{bmatrix}$ is NOT in the span of $\begin{bmatrix} -1 \\ 0 \\ -1 \end{bmatrix}$ and $\begin{bmatrix} 0 \\ 0 \\ -1 \end{bmatrix}$.
For $\begin{bmatrix} 0 \\ -1 \\ 0 \end{bmatrix}$ to be in the span, we need $\begin{bmatrix} 0 \\ -1 \\ 0 \end{bmatrix} = c_{1}\begin{bmatrix} -1 \\ 0 \\ -1 \end{bmatrix} + c_{2}\begin{bmatrix} 0 \\ 0 \\ -1 \end{bmatrix}$ for some scalars $c_1, c_2$.
This gives us the linear system $\left[\begin{array}{cc|c} -1 & 0 & 0 \\ 0 & 0 & -1 \\ -1 & -1 & 0 \end{array}\right]$, which row reduces to $\left[\begin{array}{cc|c} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]$.
Row 3 has the form [0 ... 0 | 1], which represents the equation 0 = 1, indicating no solution exists for the system.

Q: Is the vector $\begin{bmatrix} 2 \\ -3 \\ -1 \end{bmatrix}$ in the span of $\left\{ \begin{bmatrix} -2 \\ -1 \\ -3 \end{bmatrix}, \begin{bmatrix} 2 \\ 2 \\ -2 \end{bmatrix} \right\}$?
A: No. $\begin{bmatrix} 2 \\ -3 \\ -1 \end{bmatrix}$ is NOT in the span of $\begin{bmatrix} -2 \\ -1 \\ -3 \end{bmatrix}$ and $\begin{bmatrix} 2 \\ 2 \\ -2 \end{bmatrix}$.
For $\begin{bmatrix} 2 \\ -3 \\ -1 \end{bmatrix}$ to be in the span, we need $\begin{bmatrix} 2 \\ -3 \\ -1 \end{bmatrix} = c_{1}\begin{bmatrix} -2 \\ -1 \\ -3 \end{bmatrix} + c_{2}\begin{bmatrix} 2 \\ 2 \\ -2 \end{bmatrix}$ for some scalars $c_1, c_2$.
This gives us the linear system $\left[\begin{array}{cc|c} -2 & 2 & 2 \\ -1 & 2 & -3 \\ -3 & -2 & -1 \end{array}\right]$, which row reduces to $\left[\begin{array}{cc|c} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]$.
Row 3 has the form [0 ... 0 | 1], which represents the equation 0 = 1, indicating no solution exists for the system.

Q: Is the vector $\begin{bmatrix} -1 \\ 0 \\ 2 \end{bmatrix}$ in the span of $\left\{ \begin{bmatrix} 3 \\ 3 \\ 0 \end{bmatrix}, \begin{bmatrix} -2 \\ 1 \\ -1 \end{bmatrix} \right\}$?
A: No. $\begin{bmatrix} -1 \\ 0 \\ 2 \end{bmatrix}$ is NOT in the span of $\begin{bmatrix} 3 \\ 3 \\ 0 \end{bmatrix}$ and $\begin{bmatrix} -2 \\ 1 \\ -1 \end{bmatrix}$.
For $\begin{bmatrix} -1 \\ 0 \\ 2 \end{bmatrix}$ to be in the span, we need $\begin{bmatrix} -1 \\ 0 \\ 2 \end{bmatrix} = c_{1}\begin{bmatrix} 3 \\ 3 \\ 0 \end{bmatrix} + c_{2}\begin{bmatrix} -2 \\ 1 \\ -1 \end{bmatrix}$ for some scalars $c_1, c_2$.
This gives us the linear system $\left[\begin{array}{cc|c} 3 & -2 & -1 \\ 3 & 1 & 0 \\ 0 & -1 & 2 \end{array}\right]$, which row reduces to $\left[\begin{array}{cc|c} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]$.
Row 3 has the form [0 ... 0 | 1], which represents the equation 0 = 1, indicating no solution exists for the system.

Q: Is $\begin{bmatrix} -2 \\ 1 \\ -1 \end{bmatrix}$ a linear combination of $\begin{bmatrix} -2 \\ -3 \\ 2 \end{bmatrix}$ and $\begin{bmatrix} -1 \\ -1 \\ 3 \end{bmatrix}$?
A: No. $\begin{bmatrix} -2 \\ 1 \\ -1 \end{bmatrix}$ is NOT in the span of $\begin{bmatrix} -2 \\ -3 \\ 2 \end{bmatrix}$ and $\begin{bmatrix} -1 \\ -1 \\ 3 \end{bmatrix}$.
For $\begin{bmatrix} -2 \\ 1 \\ -1 \end{bmatrix}$ to be in the span, we need $\begin{bmatrix} -2 \\ 1 \\ -1 \end{bmatrix} = c_{1}\begin{bmatrix} -2 \\ -3 \\ 2 \end{bmatrix} + c_{2}\begin{bmatrix} -1 \\ -1 \\ 3 \end{bmatrix}$ for some scalars $c_1, c_2$.
This gives us the linear system $\left[\begin{array}{cc|c} -2 & -1 & -2 \\ -3 & -1 & 1 \\ 2 & 3 & -1 \end{array}\right]$, which row reduces to $\left[\begin{array}{cc|c} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]$.
Row 3 has the form [0 ... 0 | 1], which represents the equation 0 = 1, indicating no solution exists for the system.

Q: Is $\begin{bmatrix} 2 \\ -3 \\ -3 \end{bmatrix}$ a linear combination of $\begin{bmatrix} 0 \\ 2 \\ -2 \end{bmatrix}$, $\begin{bmatrix} 3 \\ 3 \\ -1 \end{bmatrix}$, and $\begin{bmatrix} 6 \\ 8 \\ -4 \end{bmatrix}$?
A: No. $\begin{bmatrix} 2 \\ -3 \\ -3 \end{bmatrix}$ is NOT in the span of $\begin{bmatrix} 0 \\ 2 \\ -2 \end{bmatrix}$, $\begin{bmatrix} 3 \\ 3 \\ -1 \end{bmatrix}$, and $\begin{bmatrix} 6 \\ 8 \\ -4 \end{bmatrix}$.
For $\begin{bmatrix} 2 \\ -3 \\ -3 \end{bmatrix}$ to be in the span, we need $\begin{bmatrix} 2 \\ -3 \\ -3 \end{bmatrix} = c_{1}\begin{bmatrix} 0 \\ 2 \\ -2 \end{bmatrix} + c_{2}\begin{bmatrix} 3 \\ 3 \\ -1 \end{bmatrix} + c_{3}\begin{bmatrix} 6 \\ 8 \\ -4 \end{bmatrix}$ for some scalars $c_1, c_2, c_3$.
This gives us the linear system $\left[\begin{array}{ccc|c} 0 & 3 & 6 & 2 \\ 2 & 3 & 8 & -3 \\ -2 & -1 & -4 & -3 \end{array}\right]$, which row reduces to $\left[\begin{array}{ccc|c} 1 & 0 & 1 & 0 \\ 0 & 1 & 2 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right]$.
Row 3 has the form [0 ... 0 | 1], which represents the equation 0 = 1, indicating no solution exists for the system.

Q: Is $\begin{bmatrix} 3 \\ 1 \\ -2 \end{bmatrix}$ a linear combination of $\begin{bmatrix} -2 \\ 0 \\ -3 \end{bmatrix}$, $\begin{bmatrix} 0 \\ 2 \\ 1 \end{bmatrix}$, and $\begin{bmatrix} -4 \\ 4 \\ -4 \end{bmatrix}$?
A: No. $\begin{bmatrix} 3 \\ 1 \\ -2 \end{bmatrix}$ is NOT in the span of $\begin{bmatrix} -2 \\ 0 \\ -3 \end{bmatrix}$, $\begin{bmatrix} 0 \\ 2 \\ 1 \end{bmatrix}$, and $\begin{bmatrix} -4 \\ 4 \\ -4 \end{bmatrix}$.
For $\begin{bmatrix} 3 \\ 1 \\ -2 \end{bmatrix}$ to be in the span, we need $\begin{bmatrix} 3 \\ 1 \\ -2 \end{bmatrix} = c_{1}\begin{bmatrix} -2 \\ 0 \\ -3 \end{bmatrix} + c_{2}\begin{bmatrix} 0 \\ 2 \\ 1 \end{bmatrix} + c_{3}\begin{bmatrix} -4 \\ 4 \\ -4 \end{bmatrix}$ for some scalars $c_1, c_2, c_3$.
This gives us the linear system $\left[\begin{array}{ccc|c} -2 & 0 & -4 & 3 \\ 0 & 2 & 4 & 1 \\ -3 & 1 & -4 & -2 \end{array}\right]$, which row reduces to $\left[\begin{array}{ccc|c} 1 & 0 & 2 & 0 \\ 0 & 1 & 2 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right]$.
Row 3 has the form [0 ... 0 | 1], which represents the equation 0 = 1, indicating no solution exists for the system.

Q: Is the vector $\begin{bmatrix} 1 \\ 2 \\ 0 \end{bmatrix}$ in the span of $\left\{ \begin{bmatrix} 1 \\ 0 \\ -3 \end{bmatrix}, \begin{bmatrix} -3 \\ 2 \\ -1 \end{bmatrix}, \begin{bmatrix} -2 \\ 2 \\ -4 \end{bmatrix} \right\}$?
A: No. $\begin{bmatrix} 1 \\ 2 \\ 0 \end{bmatrix}$ is NOT in the span of $\begin{bmatrix} 1 \\ 0 \\ -3 \end{bmatrix}$, $\begin{bmatrix} -3 \\ 2 \\ -1 \end{bmatrix}$, and $\begin{bmatrix} -2 \\ 2 \\ -4 \end{bmatrix}$.
For $\begin{bmatrix} 1 \\ 2 \\ 0 \end{bmatrix}$ to be in the span, we need $\begin{bmatrix} 1 \\ 2 \\ 0 \end{bmatrix} = c_{1}\begin{bmatrix} 1 \\ 0 \\ -3 \end{bmatrix} + c_{2}\begin{bmatrix} -3 \\ 2 \\ -1 \end{bmatrix} + c_{3}\begin{bmatrix} -2 \\ 2 \\ -4 \end{bmatrix}$ for some scalars $c_1, c_2, c_3$.
This gives us the linear system $\left[\begin{array}{ccc|c} 1 & -3 & -2 & 1 \\ 0 & 2 & 2 & 2 \\ -3 & -1 & -4 & 0 \end{array}\right]$, which row reduces to $\left[\begin{array}{ccc|c} 1 & 0 & 1 & 0 \\ 0 & 1 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right]$.
Row 3 has the form [0 ... 0 | 1], which represents the equation 0 = 1, indicating no solution exists for the system.

Q: Is the vector $\begin{bmatrix} -3 \\ 1 \\ -3 \end{bmatrix}$ in the span of $\left\{ \begin{bmatrix} 2 \\ 0 \\ -2 \end{bmatrix}, \begin{bmatrix} -3 \\ -3 \\ 2 \end{bmatrix}, \begin{bmatrix} -4 \\ -6 \\ 2 \end{bmatrix} \right\}$?
A: No. $\begin{bmatrix} -3 \\ 1 \\ -3 \end{bmatrix}$ is NOT in the span of $\begin{bmatrix} 2 \\ 0 \\ -2 \end{bmatrix}$, $\begin{bmatrix} -3 \\ -3 \\ 2 \end{bmatrix}$, and $\begin{bmatrix} -4 \\ -6 \\ 2 \end{bmatrix}$.
For $\begin{bmatrix} -3 \\ 1 \\ -3 \end{bmatrix}$ to be in the span, we need $\begin{bmatrix} -3 \\ 1 \\ -3 \end{bmatrix} = c_{1}\begin{bmatrix} 2 \\ 0 \\ -2 \end{bmatrix} + c_{2}\begin{bmatrix} -3 \\ -3 \\ 2 \end{bmatrix} + c_{3}\begin{bmatrix} -4 \\ -6 \\ 2 \end{bmatrix}$ for some scalars $c_1, c_2, c_3$.
This gives us the linear system $\left[\begin{array}{ccc|c} 2 & -3 & -4 & -3 \\ 0 & -3 & -6 & 1 \\ -2 & 2 & 2 & -3 \end{array}\right]$, which row reduces to $\left[\begin{array}{ccc|c} 1 & 0 & 1 & 0 \\ 0 & 1 & 2 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right]$.
Row 3 has the form [0 ... 0 | 1], which represents the equation 0 = 1, indicating no solution exists for the system.

Q: Is the vector $\begin{bmatrix} -2 \\ 2 \\ -2 \end{bmatrix}$ in the span of $\left\{ \begin{bmatrix} 1 \\ -1 \\ -2 \end{bmatrix}, \begin{bmatrix} -2 \\ 0 \\ 2 \end{bmatrix}, \begin{bmatrix} -2 \\ -2 \\ 0 \end{bmatrix} \right\}$?
A: No. $\begin{bmatrix} -2 \\ 2 \\ -2 \end{bmatrix}$ is NOT in the span of $\begin{bmatrix} 1 \\ -1 \\ -2 \end{bmatrix}$, $\begin{bmatrix} -2 \\ 0 \\ 2 \end{bmatrix}$, and $\begin{bmatrix} -2 \\ -2 \\ 0 \end{bmatrix}$.
For $\begin{bmatrix} -2 \\ 2 \\ -2 \end{bmatrix}$ to be in the span, we need $\begin{bmatrix} -2 \\ 2 \\ -2 \end{bmatrix} = c_{1}\begin{bmatrix} 1 \\ -1 \\ -2 \end{bmatrix} + c_{2}\begin{bmatrix} -2 \\ 0 \\ 2 \end{bmatrix} + c_{3}\begin{bmatrix} -2 \\ -2 \\ 0 \end{bmatrix}$ for some scalars $c_1, c_2, c_3$.
This gives us the linear system $\left[\begin{array}{ccc|c} 1 & -2 & -2 & -2 \\ -1 & 0 & -2 & 2 \\ -2 & 2 & 0 & -2 \end{array}\right]$, which row reduces to $\left[\begin{array}{ccc|c} 1 & 0 & 2 & 0 \\ 0 & 1 & 2 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right]$.
Row 3 has the form [0 ... 0 | 1], which represents the equation 0 = 1, indicating no solution exists for the system.

Q: Is the vector $\begin{bmatrix} -1 \\ -2 \\ 0 \end{bmatrix}$ in the span of $\left\{ \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix}, \begin{bmatrix} -2 \\ 1 \\ 3 \end{bmatrix}, \begin{bmatrix} -2 \\ 1 \\ 5 \end{bmatrix} \right\}$?
A: No. $\begin{bmatrix} -1 \\ -2 \\ 0 \end{bmatrix}$ is NOT in the span of $\begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix}$, $\begin{bmatrix} -2 \\ 1 \\ 3 \end{bmatrix}$, and $\begin{bmatrix} -2 \\ 1 \\ 5 \end{bmatrix}$.
For $\begin{bmatrix} -1 \\ -2 \\ 0 \end{bmatrix}$ to be in the span, we need $\begin{bmatrix} -1 \\ -2 \\ 0 \end{bmatrix} = c_{1}\begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix} + c_{2}\begin{bmatrix} -2 \\ 1 \\ 3 \end{bmatrix} + c_{3}\begin{bmatrix} -2 \\ 1 \\ 5 \end{bmatrix}$ for some scalars $c_1, c_2, c_3$.
This gives us the linear system $\left[\begin{array}{ccc|c} 0 & -2 & -2 & -1 \\ 0 & 1 & 1 & -2 \\ 1 & 3 & 5 & 0 \end{array}\right]$, which row reduces to $\left[\begin{array}{ccc|c} 1 & 0 & 2 & 0 \\ 0 & 1 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right]$.
Row 3 has the form [0 ... 0 | 1], which represents the equation 0 = 1, indicating no solution exists for the system.

Q: Is the vector $\begin{bmatrix} 1 \\ 0 \\ -2 \end{bmatrix}$ in the span of $\left\{ \begin{bmatrix} 2 \\ -2 \\ 1 \end{bmatrix}, \begin{bmatrix} 2 \\ -3 \\ 1 \end{bmatrix}, \begin{bmatrix} 8 \\ -10 \\ 4 \end{bmatrix} \right\}$?
A: No. $\begin{bmatrix} 1 \\ 0 \\ -2 \end{bmatrix}$ is NOT in the span of $\begin{bmatrix} 2 \\ -2 \\ 1 \end{bmatrix}$, $\begin{bmatrix} 2 \\ -3 \\ 1 \end{bmatrix}$, and $\begin{bmatrix} 8 \\ -10 \\ 4 \end{bmatrix}$.
For $\begin{bmatrix} 1 \\ 0 \\ -2 \end{bmatrix}$ to be in the span, we need $\begin{bmatrix} 1 \\ 0 \\ -2 \end{bmatrix} = c_{1}\begin{bmatrix} 2 \\ -2 \\ 1 \end{bmatrix} + c_{2}\begin{bmatrix} 2 \\ -3 \\ 1 \end{bmatrix} + c_{3}\begin{bmatrix} 8 \\ -10 \\ 4 \end{bmatrix}$ for some scalars $c_1, c_2, c_3$.
This gives us the linear system $\left[\begin{array}{ccc|c} 2 & 2 & 8 & 1 \\ -2 & -3 & -10 & 0 \\ 1 & 1 & 4 & -2 \end{array}\right]$, which row reduces to $\left[\begin{array}{ccc|c} 1 & 0 & 2 & 0 \\ 0 & 1 & 2 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right]$.
Row 3 has the form [0 ... 0 | 1], which represents the equation 0 = 1, indicating no solution exists for the system.

Q: Is $\begin{bmatrix} -2 \\ -1 \\ -1 \end{bmatrix}$ a linear combination of $\begin{bmatrix} -1 \\ 3 \\ -1 \end{bmatrix}$, $\begin{bmatrix} 0 \\ 2 \\ -3 \end{bmatrix}$, and $\begin{bmatrix} -1 \\ 7 \\ -7 \end{bmatrix}$?
A: No. $\begin{bmatrix} -2 \\ -1 \\ -1 \end{bmatrix}$ is NOT in the span of $\begin{bmatrix} -1 \\ 3 \\ -1 \end{bmatrix}$, $\begin{bmatrix} 0 \\ 2 \\ -3 \end{bmatrix}$, and $\begin{bmatrix} -1 \\ 7 \\ -7 \end{bmatrix}$.
For $\begin{bmatrix} -2 \\ -1 \\ -1 \end{bmatrix}$ to be in the span, we need $\begin{bmatrix} -2 \\ -1 \\ -1 \end{bmatrix} = c_{1}\begin{bmatrix} -1 \\ 3 \\ -1 \end{bmatrix} + c_{2}\begin{bmatrix} 0 \\ 2 \\ -3 \end{bmatrix} + c_{3}\begin{bmatrix} -1 \\ 7 \\ -7 \end{bmatrix}$ for some scalars $c_1, c_2, c_3$.
This gives us the linear system $\left[\begin{array}{ccc|c} -1 & 0 & -1 & -2 \\ 3 & 2 & 7 & -1 \\ -1 & -3 & -7 & -1 \end{array}\right]$, which row reduces to $\left[\begin{array}{ccc|c} 1 & 0 & 1 & 0 \\ 0 & 1 & 2 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right]$.
Row 3 has the form [0 ... 0 | 1], which represents the equation 0 = 1, indicating no solution exists for the system.

Q: Is the vector $\begin{bmatrix} -2 \\ 0 \\ 1 \end{bmatrix}$ in the span of $\left\{ \begin{bmatrix} -1 \\ 2 \\ -2 \end{bmatrix}, \begin{bmatrix} 1 \\ -3 \\ 2 \end{bmatrix}, \begin{bmatrix} 0 \\ -2 \\ 0 \end{bmatrix} \right\}$?
A: No. $\begin{bmatrix} -2 \\ 0 \\ 1 \end{bmatrix}$ is NOT in the span of $\begin{bmatrix} -1 \\ 2 \\ -2 \end{bmatrix}$, $\begin{bmatrix} 1 \\ -3 \\ 2 \end{bmatrix}$, and $\begin{bmatrix} 0 \\ -2 \\ 0 \end{bmatrix}$.
For $\begin{bmatrix} -2 \\ 0 \\ 1 \end{bmatrix}$ to be in the span, we need $\begin{bmatrix} -2 \\ 0 \\ 1 \end{bmatrix} = c_{1}\begin{bmatrix} -1 \\ 2 \\ -2 \end{bmatrix} + c_{2}\begin{bmatrix} 1 \\ -3 \\ 2 \end{bmatrix} + c_{3}\begin{bmatrix} 0 \\ -2 \\ 0 \end{bmatrix}$ for some scalars $c_1, c_2, c_3$.
This gives us the linear system $\left[\begin{array}{ccc|c} -1 & 1 & 0 & -2 \\ 2 & -3 & -2 & 0 \\ -2 & 2 & 0 & 1 \end{array}\right]$, which row reduces to $\left[\begin{array}{ccc|c} 1 & 0 & 2 & 0 \\ 0 & 1 & 2 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right]$.
Row 3 has the form [0 ... 0 | 1], which represents the equation 0 = 1, indicating no solution exists for the system.

Q: Is $\begin{bmatrix} -3 \\ 1 \\ 0 \end{bmatrix}$ a linear combination of $\begin{bmatrix} -1 \\ -3 \\ 0 \end{bmatrix}$, $\begin{bmatrix} 1 \\ 3 \\ 1 \end{bmatrix}$, and $\begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix}$?
A: No. $\begin{bmatrix} -3 \\ 1 \\ 0 \end{bmatrix}$ is NOT in the span of $\begin{bmatrix} -1 \\ -3 \\ 0 \end{bmatrix}$, $\begin{bmatrix} 1 \\ 3 \\ 1 \end{bmatrix}$, and $\begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix}$.
For $\begin{bmatrix} -3 \\ 1 \\ 0 \end{bmatrix}$ to be in the span, we need $\begin{bmatrix} -3 \\ 1 \\ 0 \end{bmatrix} = c_{1}\begin{bmatrix} -1 \\ -3 \\ 0 \end{bmatrix} + c_{2}\begin{bmatrix} 1 \\ 3 \\ 1 \end{bmatrix} + c_{3}\begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix}$ for some scalars $c_1, c_2, c_3$.
This gives us the linear system $\left[\begin{array}{ccc|c} -1 & 1 & 0 & -3 \\ -3 & 3 & 0 & 1 \\ 0 & 1 & 1 & 0 \end{array}\right]$, which row reduces to $\left[\begin{array}{ccc|c} 1 & 0 & 1 & 0 \\ 0 & 1 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right]$.
Row 3 has the form [0 ... 0 | 1], which represents the equation 0 = 1, indicating no solution exists for the system.

Q: Is $\begin{bmatrix} 2 \\ -1 \\ -3 \end{bmatrix}$ a linear combination of $\begin{bmatrix} 2 \\ -2 \\ 3 \end{bmatrix}$, $\begin{bmatrix} -2 \\ -3 \\ -1 \end{bmatrix}$, and $\begin{bmatrix} 0 \\ -5 \\ 2 \end{bmatrix}$?
A: No. $\begin{bmatrix} 2 \\ -1 \\ -3 \end{bmatrix}$ is NOT in the span of $\begin{bmatrix} 2 \\ -2 \\ 3 \end{bmatrix}$, $\begin{bmatrix} -2 \\ -3 \\ -1 \end{bmatrix}$, and $\begin{bmatrix} 0 \\ -5 \\ 2 \end{bmatrix}$.
For $\begin{bmatrix} 2 \\ -1 \\ -3 \end{bmatrix}$ to be in the span, we need $\begin{bmatrix} 2 \\ -1 \\ -3 \end{bmatrix} = c_{1}\begin{bmatrix} 2 \\ -2 \\ 3 \end{bmatrix} + c_{2}\begin{bmatrix} -2 \\ -3 \\ -1 \end{bmatrix} + c_{3}\begin{bmatrix} 0 \\ -5 \\ 2 \end{bmatrix}$ for some scalars $c_1, c_2, c_3$.
This gives us the linear system $\left[\begin{array}{ccc|c} 2 & -2 & 0 & 2 \\ -2 & -3 & -5 & -1 \\ 3 & -1 & 2 & -3 \end{array}\right]$, which row reduces to $\left[\begin{array}{ccc|c} 1 & 0 & 1 & 0 \\ 0 & 1 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right]$.
Row 3 has the form [0 ... 0 | 1], which represents the equation 0 = 1, indicating no solution exists for the system.

Q: Is $\begin{bmatrix} 2 \\ -1 \\ -1 \end{bmatrix}$ a linear combination of $\begin{bmatrix} -1 \\ 1 \\ -3 \end{bmatrix}$, $\begin{bmatrix} 1 \\ 3 \\ 2 \end{bmatrix}$, and $\begin{bmatrix} -1 \\ 5 \\ -4 \end{bmatrix}$?
A: No. $\begin{bmatrix} 2 \\ -1 \\ -1 \end{bmatrix}$ is NOT in the span of $\begin{bmatrix} -1 \\ 1 \\ -3 \end{bmatrix}$, $\begin{bmatrix} 1 \\ 3 \\ 2 \end{bmatrix}$, and $\begin{bmatrix} -1 \\ 5 \\ -4 \end{bmatrix}$.
For $\begin{bmatrix} 2 \\ -1 \\ -1 \end{bmatrix}$ to be in the span, we need $\begin{bmatrix} 2 \\ -1 \\ -1 \end{bmatrix} = c_{1}\begin{bmatrix} -1 \\ 1 \\ -3 \end{bmatrix} + c_{2}\begin{bmatrix} 1 \\ 3 \\ 2 \end{bmatrix} + c_{3}\begin{bmatrix} -1 \\ 5 \\ -4 \end{bmatrix}$ for some scalars $c_1, c_2, c_3$.
This gives us the linear system $\left[\begin{array}{ccc|c} -1 & 1 & -1 & 2 \\ 1 & 3 & 5 & -1 \\ -3 & 2 & -4 & -1 \end{array}\right]$, which row reduces to $\left[\begin{array}{ccc|c} 1 & 0 & 2 & 0 \\ 0 & 1 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right]$.
Row 3 has the form [0 ... 0 | 1], which represents the equation 0 = 1, indicating no solution exists for the system.

Q: Is the vector $\begin{bmatrix} 3 \\ 0 \\ 2 \end{bmatrix}$ in the span of $\left\{ \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix}, \begin{bmatrix} -2 \\ 2 \\ 3 \end{bmatrix}, \begin{bmatrix} -2 \\ 4 \\ 3 \end{bmatrix} \right\}$?
A: No. $\begin{bmatrix} 3 \\ 0 \\ 2 \end{bmatrix}$ is NOT in the span of $\begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix}$, $\begin{bmatrix} -2 \\ 2 \\ 3 \end{bmatrix}$, and $\begin{bmatrix} -2 \\ 4 \\ 3 \end{bmatrix}$.
For $\begin{bmatrix} 3 \\ 0 \\ 2 \end{bmatrix}$ to be in the span, we need $\begin{bmatrix} 3 \\ 0 \\ 2 \end{bmatrix} = c_{1}\begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix} + c_{2}\begin{bmatrix} -2 \\ 2 \\ 3 \end{bmatrix} + c_{3}\begin{bmatrix} -2 \\ 4 \\ 3 \end{bmatrix}$ for some scalars $c_1, c_2, c_3$.
This gives us the linear system $\left[\begin{array}{ccc|c} 0 & -2 & -2 & 3 \\ 1 & 2 & 4 & 0 \\ 0 & 3 & 3 & 2 \end{array}\right]$, which row reduces to $\left[\begin{array}{ccc|c} 1 & 0 & 2 & 0 \\ 0 & 1 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right]$.
Row 3 has the form [0 ... 0 | 1], which represents the equation 0 = 1, indicating no solution exists for the system.

Q: Is $\begin{bmatrix} 0 \\ 1 \\ 2 \end{bmatrix}$ a linear combination of $\begin{bmatrix} -2 \\ 0 \\ 0 \end{bmatrix}$, $\begin{bmatrix} -1 \\ -2 \\ 1 \end{bmatrix}$, and $\begin{bmatrix} -3 \\ -2 \\ 1 \end{bmatrix}$?
A: No. $\begin{bmatrix} 0 \\ 1 \\ 2 \end{bmatrix}$ is NOT in the span of $\begin{bmatrix} -2 \\ 0 \\ 0 \end{bmatrix}$, $\begin{bmatrix} -1 \\ -2 \\ 1 \end{bmatrix}$, and $\begin{bmatrix} -3 \\ -2 \\ 1 \end{bmatrix}$.
For $\begin{bmatrix} 0 \\ 1 \\ 2 \end{bmatrix}$ to be in the span, we need $\begin{bmatrix} 0 \\ 1 \\ 2 \end{bmatrix} = c_{1}\begin{bmatrix} -2 \\ 0 \\ 0 \end{bmatrix} + c_{2}\begin{bmatrix} -1 \\ -2 \\ 1 \end{bmatrix} + c_{3}\begin{bmatrix} -3 \\ -2 \\ 1 \end{bmatrix}$ for some scalars $c_1, c_2, c_3$.
This gives us the linear system $\left[\begin{array}{ccc|c} -2 & -1 & -3 & 0 \\ 0 & -2 & -2 & 1 \\ 0 & 1 & 1 & 2 \end{array}\right]$, which row reduces to $\left[\begin{array}{ccc|c} 1 & 0 & 1 & 0 \\ 0 & 1 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right]$.
Row 3 has the form [0 ... 0 | 1], which represents the equation 0 = 1, indicating no solution exists for the system.

Q: Is the vector $\begin{bmatrix} 3 \\ 2 \\ -1 \end{bmatrix}$ in the span of $\left\{ \begin{bmatrix} 1 \\ -1 \\ 2 \end{bmatrix}, \begin{bmatrix} 2 \\ -3 \\ -1 \end{bmatrix}, \begin{bmatrix} 6 \\ -8 \\ 2 \end{bmatrix} \right\}$?
A: No. $\begin{bmatrix} 3 \\ 2 \\ -1 \end{bmatrix}$ is NOT in the span of $\begin{bmatrix} 1 \\ -1 \\ 2 \end{bmatrix}$, $\begin{bmatrix} 2 \\ -3 \\ -1 \end{bmatrix}$, and $\begin{bmatrix} 6 \\ -8 \\ 2 \end{bmatrix}$.
For $\begin{bmatrix} 3 \\ 2 \\ -1 \end{bmatrix}$ to be in the span, we need $\begin{bmatrix} 3 \\ 2 \\ -1 \end{bmatrix} = c_{1}\begin{bmatrix} 1 \\ -1 \\ 2 \end{bmatrix} + c_{2}\begin{bmatrix} 2 \\ -3 \\ -1 \end{bmatrix} + c_{3}\begin{bmatrix} 6 \\ -8 \\ 2 \end{bmatrix}$ for some scalars $c_1, c_2, c_3$.
This gives us the linear system $\left[\begin{array}{ccc|c} 1 & 2 & 6 & 3 \\ -1 & -3 & -8 & 2 \\ 2 & -1 & 2 & -1 \end{array}\right]$, which row reduces to $\left[\begin{array}{ccc|c} 1 & 0 & 2 & 0 \\ 0 & 1 & 2 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right]$.
Row 3 has the form [0 ... 0 | 1], which represents the equation 0 = 1, indicating no solution exists for the system.

Q: Is $\begin{bmatrix} -2 \\ -2 \\ -2 \end{bmatrix}$ a linear combination of $\begin{bmatrix} -2 \\ 0 \\ 2 \end{bmatrix}$, $\begin{bmatrix} -2 \\ 2 \\ 0 \end{bmatrix}$, and $\begin{bmatrix} -4 \\ 2 \\ 2 \end{bmatrix}$?
A: No. $\begin{bmatrix} -2 \\ -2 \\ -2 \end{bmatrix}$ is NOT in the span of $\begin{bmatrix} -2 \\ 0 \\ 2 \end{bmatrix}$, $\begin{bmatrix} -2 \\ 2 \\ 0 \end{bmatrix}$, and $\begin{bmatrix} -4 \\ 2 \\ 2 \end{bmatrix}$.
For $\begin{bmatrix} -2 \\ -2 \\ -2 \end{bmatrix}$ to be in the span, we need $\begin{bmatrix} -2 \\ -2 \\ -2 \end{bmatrix} = c_{1}\begin{bmatrix} -2 \\ 0 \\ 2 \end{bmatrix} + c_{2}\begin{bmatrix} -2 \\ 2 \\ 0 \end{bmatrix} + c_{3}\begin{bmatrix} -4 \\ 2 \\ 2 \end{bmatrix}$ for some scalars $c_1, c_2, c_3$.
This gives us the linear system $\left[\begin{array}{ccc|c} -2 & -2 & -4 & -2 \\ 0 & 2 & 2 & -2 \\ 2 & 0 & 2 & -2 \end{array}\right]$, which row reduces to $\left[\begin{array}{ccc|c} 1 & 0 & 1 & 0 \\ 0 & 1 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right]$.
Row 3 has the form [0 ... 0 | 1], which represents the equation 0 = 1, indicating no solution exists for the system.

Q: Is $\begin{bmatrix} -2 \\ -1 \\ 1 \end{bmatrix}$ a linear combination of $\begin{bmatrix} 2 \\ -1 \\ 3 \end{bmatrix}$, $\begin{bmatrix} 0 \\ 1 \\ 1 \end{bmatrix}$, and $\begin{bmatrix} 4 \\ -1 \\ 7 \end{bmatrix}$?
A: No. $\begin{bmatrix} -2 \\ -1 \\ 1 \end{bmatrix}$ is NOT in the span of $\begin{bmatrix} 2 \\ -1 \\ 3 \end{bmatrix}$, $\begin{bmatrix} 0 \\ 1 \\ 1 \end{bmatrix}$, and $\begin{bmatrix} 4 \\ -1 \\ 7 \end{bmatrix}$.
For $\begin{bmatrix} -2 \\ -1 \\ 1 \end{bmatrix}$ to be in the span, we need $\begin{bmatrix} -2 \\ -1 \\ 1 \end{bmatrix} = c_{1}\begin{bmatrix} 2 \\ -1 \\ 3 \end{bmatrix} + c_{2}\begin{bmatrix} 0 \\ 1 \\ 1 \end{bmatrix} + c_{3}\begin{bmatrix} 4 \\ -1 \\ 7 \end{bmatrix}$ for some scalars $c_1, c_2, c_3$.
This gives us the linear system $\left[\begin{array}{ccc|c} 2 & 0 & 4 & -2 \\ -1 & 1 & -1 & -1 \\ 3 & 1 & 7 & 1 \end{array}\right]$, which row reduces to $\left[\begin{array}{ccc|c} 1 & 0 & 2 & 0 \\ 0 & 1 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right]$.
Row 3 has the form [0 ... 0 | 1], which represents the equation 0 = 1, indicating no solution exists for the system.

Q: Is the vector $\begin{bmatrix} -3 \\ 2 \\ 1 \end{bmatrix}$ in the span of $\left\{ \begin{bmatrix} 1 \\ -1 \\ 3 \end{bmatrix}, \begin{bmatrix} 1 \\ 2 \\ -1 \end{bmatrix}, \begin{bmatrix} 3 \\ 3 \\ 1 \end{bmatrix} \right\}$?
A: No. $\begin{bmatrix} -3 \\ 2 \\ 1 \end{bmatrix}$ is NOT in the span of $\begin{bmatrix} 1 \\ -1 \\ 3 \end{bmatrix}$, $\begin{bmatrix} 1 \\ 2 \\ -1 \end{bmatrix}$, and $\begin{bmatrix} 3 \\ 3 \\ 1 \end{bmatrix}$.
For $\begin{bmatrix} -3 \\ 2 \\ 1 \end{bmatrix}$ to be in the span, we need $\begin{bmatrix} -3 \\ 2 \\ 1 \end{bmatrix} = c_{1}\begin{bmatrix} 1 \\ -1 \\ 3 \end{bmatrix} + c_{2}\begin{bmatrix} 1 \\ 2 \\ -1 \end{bmatrix} + c_{3}\begin{bmatrix} 3 \\ 3 \\ 1 \end{bmatrix}$ for some scalars $c_1, c_2, c_3$.
This gives us the linear system $\left[\begin{array}{ccc|c} 1 & 1 & 3 & -3 \\ -1 & 2 & 3 & 2 \\ 3 & -1 & 1 & 1 \end{array}\right]$, which row reduces to $\left[\begin{array}{ccc|c} 1 & 0 & 1 & 0 \\ 0 & 1 & 2 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right]$.
Row 3 has the form [0 ... 0 | 1], which represents the equation 0 = 1, indicating no solution exists for the system.

Q: Is the vector $\begin{bmatrix} 0 \\ -3 \\ 3 \end{bmatrix}$ in the span of $\left\{ \begin{bmatrix} -3 \\ -2 \\ -1 \end{bmatrix}, \begin{bmatrix} 3 \\ -3 \\ -3 \end{bmatrix}, \begin{bmatrix} 0 \\ -10 \\ -8 \end{bmatrix} \right\}$?
A: No. $\begin{bmatrix} 0 \\ -3 \\ 3 \end{bmatrix}$ is NOT in the span of $\begin{bmatrix} -3 \\ -2 \\ -1 \end{bmatrix}$, $\begin{bmatrix} 3 \\ -3 \\ -3 \end{bmatrix}$, and $\begin{bmatrix} 0 \\ -10 \\ -8 \end{bmatrix}$.
For $\begin{bmatrix} 0 \\ -3 \\ 3 \end{bmatrix}$ to be in the span, we need $\begin{bmatrix} 0 \\ -3 \\ 3 \end{bmatrix} = c_{1}\begin{bmatrix} -3 \\ -2 \\ -1 \end{bmatrix} + c_{2}\begin{bmatrix} 3 \\ -3 \\ -3 \end{bmatrix} + c_{3}\begin{bmatrix} 0 \\ -10 \\ -8 \end{bmatrix}$ for some scalars $c_1, c_2, c_3$.
This gives us the linear system $\left[\begin{array}{ccc|c} -3 & 3 & 0 & 0 \\ -2 & -3 & -10 & -3 \\ -1 & -3 & -8 & 3 \end{array}\right]$, which row reduces to $\left[\begin{array}{ccc|c} 1 & 0 & 2 & 0 \\ 0 & 1 & 2 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right]$.
Row 3 has the form [0 ... 0 | 1], which represents the equation 0 = 1, indicating no solution exists for the system.

Q: Is the vector $\begin{bmatrix} -2 \\ 1 \\ 1 \end{bmatrix}$ in the span of $\left\{ \begin{bmatrix} 1 \\ 3 \\ -2 \end{bmatrix}, \begin{bmatrix} -1 \\ 1 \\ -3 \end{bmatrix}, \begin{bmatrix} 0 \\ 8 \\ -10 \end{bmatrix} \right\}$?
A: No. $\begin{bmatrix} -2 \\ 1 \\ 1 \end{bmatrix}$ is NOT in the span of $\begin{bmatrix} 1 \\ 3 \\ -2 \end{bmatrix}$, $\begin{bmatrix} -1 \\ 1 \\ -3 \end{bmatrix}$, and $\begin{bmatrix} 0 \\ 8 \\ -10 \end{bmatrix}$.
For $\begin{bmatrix} -2 \\ 1 \\ 1 \end{bmatrix}$ to be in the span, we need $\begin{bmatrix} -2 \\ 1 \\ 1 \end{bmatrix} = c_{1}\begin{bmatrix} 1 \\ 3 \\ -2 \end{bmatrix} + c_{2}\begin{bmatrix} -1 \\ 1 \\ -3 \end{bmatrix} + c_{3}\begin{bmatrix} 0 \\ 8 \\ -10 \end{bmatrix}$ for some scalars $c_1, c_2, c_3$.
This gives us the linear system $\left[\begin{array}{ccc|c} 1 & -1 & 0 & -2 \\ 3 & 1 & 8 & 1 \\ -2 & -3 & -10 & 1 \end{array}\right]$, which row reduces to $\left[\begin{array}{ccc|c} 1 & 0 & 2 & 0 \\ 0 & 1 & 2 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right]$.
Row 3 has the form [0 ... 0 | 1], which represents the equation 0 = 1, indicating no solution exists for the system.