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(D) Given that $A \cdot \begin{bmatrix} 1 & 2 \ 0 & 3 \end{bmatrix}$ is a scalar matrix. Let this scalar matrix be $K = \begin{bmatrix} k & 0 \ 0 &

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$\begin{bmatrix} 36 & -32 \ 0 & 4 \end{bmatrix}$

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(D) Given that $A \cdot \begin{bmatrix} 1 & 2 \ 0 & 3 \end{bmatrix}$ is a scalar matrix. Let this scalar matrix be $K = \begin{bmatrix} k & 0 \ 0 & k \end{bmatrix}$.Then $A = \begin{bmatrix} k & 0 \ 0 & k \end{bmatrix} \cdot \begin{bmatrix} 1 & 2 \ 0 & 3 \end{bmatrix}^{-1}$.The inverse of $\begin{bmatrix} 1 & 2 \ 0 & 3 \end{bmatrix}$ is $\frac{1}{3} \begin{bmatrix} 3 & -2 \ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & -2/3 \ 0 & 1/3 \end{bmatrix}$.Thus,$A = \begin{bmatrix} k & 0 \ 0 & k \end{bmatrix} \begin{bmatrix} 1 & -2/3 \ 0 & 1/3 \end{bmatrix} = \begin{bmatrix} k & -2k/3 \ 0 & k/3 \end{bmatrix}$.Given $|3A| = 108$. Since $A$ is a $2 	imes 2$ matrix,$|3A| = 3^2 |A| = 9|A|$.So,$9|A| = 108 \Rightarrow |A| = 12$.Calculating $|A|$ from our expression: $|A| = (k)(k/3) - 0 = k^2/3$.Equating the two: $k^2/3 = 12 \Rightarrow k^2 = 36 \Rightarrow k = \pm 6$.For $k = 6$,$A = \begin{bmatrix} 6 & -4 \ 0 & 2 \end{bmatrix}$,so $A^2 = \begin{bmatrix} 6 & -4 \ 0 & 2 \end{bmatrix} \begin{bmatrix} 6 & -4 \ 0 & 2 \end{bmatrix} = \begin{bmatrix} 36 & -32 \ 0 & 4 \end{bmatrix}$.For $k = -6$,$A = \begin{bmatrix} -6 & 4 \ 0 & -2 \end{bmatrix}$,so $A^2 = \begin{bmatrix} -6 & 4 \ 0 & -2 \end{bmatrix} \begin{bmatrix} -6 & 4 \ 0 & -2 \end{bmatrix} = \begin{bmatrix} 36 & -32 \ 0 & 4 \end{bmatrix}$.In both cases,$A^2 = \begin{bmatrix} 36 & -32 \ 0 & 4 \end{bmatrix}$.

Let $A$ be a matrix such that $A \cdot \begin{bmatrix} 1 & 2 \\ 0 & 3 \end{bmatrix}$ is a scalar matrix and $|3A| = 108$. Then $A^2$ equals

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