Consider the following matrix: $$ A=\left[\begin{array}{ccc}1 & -2 & -3 \ -1 & 2 & 8 \ 2 & -4 & -4\end{array} ight] $$ For each of the following vectors, determine whether it is in the image and/or null space of $$ A $$. If the vector is in the image of $$ A $$ find a vector $$ \mathbf{x} $$ so that $$ A \mathbf{x}-\mathbf{v}_{i} $$. $$ \mathbf{v}_{1}=\left[\begin{array}{c}2 \ 1 \ 0\end{array} ight] $$ \mathbf{v}_{2}=\left[\begin{array}{c}-1 \ 1 \ -2\end{array} ight] $$

Question

Consider the following matrix: $$ A=\left[\begin{array}{ccc}1 & -2 & -3 \ -1 & 2 & 8 \ 2 & -4 & -4\end{array}ight] $$ For each of the following vectors, determine whether it is in the image and/or null space of $$ A $$. If the vector is in the image of $$ A $$ find a vector $$ \mathbf{x} $$ so that $$ A \mathbf{x}-\mathbf{v}_{i} $$. $$ \mathbf{v}_{1}=\left[\begin{array}{c}2 \ 1 \ 0\end{array}ight]< $$ Select an answer > $$ \mathbf{v}_{2}=\left[\begin{array}{c}-1 \ 1 \ -2\end{array}ight] $$

UpStudy AI · Accepted Answer

To determine whether a vector is in the image or null space of a matrix $$ A $$, we need to check the following:

1. Image of $$ A $$: A vector $$ \mathbf{v} $$ is in the image of $$ A $$ if there exists a vector $$ \mathbf{x} $$ such that $$ A\mathbf{x} = \mathbf{v} $$.
2. Null space of $$ A $$: A vector $$ \mathbf{v} $$ is in the null space of $$ A $$ if $$ A\mathbf{v} = \mathbf{0} $$.

Let's check each vector:

### Vector $$ \mathbf{v}_{1} = \left[\begin{array}{c}2 \ 1 \ 0\end{array}\right] $$

To check if $$ \mathbf{v}_{1} $$ is in the image of $$ A $$, we need to find a vector $$ \mathbf{x} $$ such that $$ A\mathbf{x} = \mathbf{v}_{1} $$.

Let $$ \mathbf{x} = \left[\begin{array}{c}x_1 \ x_2 \ x_3\end{array}\right] $$. Then we have the following system of equations:

$$
\begin{align*}
x_1 - 2x_2 - 3x_3 &= 2 \
-x_1 + 2x_2 + 8x_3 &= 1 \
2x_1 - 4x_2 - 4x_3 &= 0
\end{align*}
$$

We can solve this system using row reduction or any other method. Let's solve it using row reduction:

$$
\begin{bmatrix}
1 & -2 & -3 \
-1 & 2 & 8 \
2 & -4 & -4
\end{bmatrix}
\rightarrow
\begin{bmatrix}
1 & -2 & -3 \
0 & 4 & 11 \
0 & 0 & 2
\end{bmatrix}
\rightarrow
\begin{bmatrix}
1 & -2 & -3 \
0 & 1 & \frac{11}{4} \
0 & 0 & 1
\end{bmatrix}
$$

From the reduced row echelon form, we can see that the system is consistent and has a unique solution. Therefore, $$ \mathbf{v}_{1} $$ is in the image of $$ A $$.

To find $$ \mathbf{x} $$, we can back-substitute:

$$
\begin{align*}
x_3 &= 0 \
x_2 + \frac{11}{4}x_3 &= 1 \Rightarrow x_2 = 1 \
x_1 - 2x_2 - 3x_3 &= 2 \Rightarrow x_1 = 2 + 2(1) + 3(0) = 4
\end{align*}
$$

So, $$ \mathbf{x} = \left[\begin{array}{c}4 \ 1 \ 0\end{array}\right] $$.

### Vector $$ \mathbf{v}_{2} = \left[\begin{array}{c}-1 \ 1 \ -2\end{array}\right] $$

To check if $$ \mathbf{v}_{2} $$ is in the image of $$ A $$, we need to find a vector $$ \mathbf{x} $$ such that $$ A\mathbf{x} = \mathbf{v}_{2} $$.

Let $$ \mathbf{x} = \left[\begin{array}{c}x_1 \ x_2 \ x_3\end{array}\right] $$. Then we have the following system of equations:

$$
\begin{align*}
x_1 - 2x_2 - 3x_3 &= -1 \
-x_1 + 2x_2 + 8x_3 &= 1 \
2x_1 - 4x_2 - 4x_3 &= -2
\end{align*}
$$

We can solve this system using row reduction or any other method. Let's solve it using row reduction:

\[
\begin{bmatrix}
1 & -2 & -3 \
-1 & 2 & 8 \
2 & -4 & -4
\end{bmatrix}
\rightarrow
\begin{bmatrix}
1 & -2 & -3 \
0 & 4 & 11 \
0 & 0 & 2
\end{bmatrix}
\rightarrow
\begin{bmatrix}
1 & -2 & -3 \
0 & 1 & \frac{11}{4
Vector $$ \mathbf{v}_{1} $$ is in the image of $$ A $$ with $$ \mathbf{x} = \left[\begin{array}{c}4 \ 1 \ 0\end{array}\right] $$. Vector $$ \mathbf{v}_{2} $$ is not in the image of $$ A $$.