If A is a matrix such that left[begin{array}{ | vedclass

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(D) Let $A = \begin{bmatrix} x_1 \ x_2 \end{bmatrix}$.Given the equation $\begin{bmatrix} 2 & 1 \ 3 & 2 \end{bmatrix} \begin{bmatrix} x_1 \ x_2 \en

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$\left[\begin{array}{r} 2 \ -3 \end{array}ight]$

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(D) Let $A = \begin{bmatrix} x_1 \ x_2 \end{bmatrix}$.Given the equation $\begin{bmatrix} 2 & 1 \ 3 & 2 \end{bmatrix} \begin{bmatrix} x_1 \ x_2 \end{bmatrix} \begin{bmatrix} 1 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 1 \ 0 & 0 \end{bmatrix}$.First, compute the product $\begin{bmatrix} 2 & 1 \ 3 & 2 \end{bmatrix} \begin{bmatrix} x_1 \ x_2 \end{bmatrix} = \begin{bmatrix} 2x_1 + x_2 \ 3x_1 + 2x_2 \end{bmatrix}$.Now, multiply by $\begin{bmatrix} 1 & 1 \end{bmatrix}$:$\begin{bmatrix} 2x_1 + x_2 \ 3x_1 + 2x_2 \end{bmatrix} \begin{bmatrix} 1 & 1 \end{bmatrix} = \begin{bmatrix} 2x_1 + x_2 & 2x_1 + x_2 \ 3x_1 + 2x_2 & 3x_1 + 2x_2 \end{bmatrix}$.Equating this to the given matrix $\begin{bmatrix} 1 & 1 \ 0 & 0 \end{bmatrix}$, we get:$2x_1 + x_2 = 1$ and $3x_1 + 2x_2 = 0$.From the second equation, $x_2 = -\frac{3}{2}x_1$.Substituting into the first: $2x_1 - \frac{3}{2}x_1 = 1 \Rightarrow \frac{1}{2}x_1 = 1 \Rightarrow x_1 = 2$.Then $x_2 = -\frac{3}{2}(2) = -3$.Thus, $A = \begin{bmatrix} 2 \ -3 \end{bmatrix}$.

If $A$ is a matrix such that $\left[\begin{array}{ll} 2 & 1 \\ 3 & 2 \end{array}\right] A \left[\begin{array}{ll} 1 & 1 \end{array}\right] = \left[\begin{array}{ll} 1 & 1 \\ 0 & 0 \end{array}\right]$, then $A$ is equal to

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