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(D) Let $A = \begin{bmatrix} a & b \ c & d \end{bmatrix}$.Given $AB = \begin{bmatrix} a & b \ c & d \end{bmatrix} \begin{bmatrix} 1 & 2 \ 0 & 3 \en

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$\begin{bmatrix} 10 & -6 \ 0 & 4 \end{bmatrix}$

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(D) Let $A = \begin{bmatrix} a & b \ c & d \end{bmatrix}$.Given $AB = \begin{bmatrix} a & b \ c & d \end{bmatrix} \begin{bmatrix} 1 & 2 \ 0 & 3 \end{bmatrix} = \begin{bmatrix} a & 2a+3b \ c & 2c+3d \end{bmatrix}$.Since $AB$ is a scalar matrix,$c = 0$ and $2a+3b = 0$,and $a = 2c+3d = 3d$.Thus,$a = 3d$ and $b = -\frac{2}{3}a = -2d$.So,$A = \begin{bmatrix} 3d & -2d \ 0 & d \end{bmatrix}$.Given $\det(3A) = 27$,we have $3^2 \det(A) = 27$,so $\det(A) = 3$.$\det(A) = (3d)(d) - 0 = 3d^2 = 3$,which implies $d^2 = 1$. Assuming $d=1$,we get $A = \begin{bmatrix} 3 & -2 \ 0 & 1 \end{bmatrix}$.Then $A^2 = \begin{bmatrix} 3 & -2 \ 0 & 1 \end{bmatrix} \begin{bmatrix} 3 & -2 \ 0 & 1 \end{bmatrix} = \begin{bmatrix} 9 & -8 \ 0 & 1 \end{bmatrix}$.$A^{-1} = \frac{1}{\det(A)} 	ext{adj}(A) = \frac{1}{3} \begin{bmatrix} 1 & 2 \ 0 & 3 \end{bmatrix} = \begin{bmatrix} 1/3 & 2/3 \ 0 & 1 \end{bmatrix}$.Therefore,$3A^{-1} + A^2 = 3 \begin{bmatrix} 1/3 & 2/3 \ 0 & 1 \end{bmatrix} + \begin{bmatrix} 9 & -8 \ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 2 \ 0 & 3 \end{bmatrix} + \begin{bmatrix} 9 & -8 \ 0 & 1 \end{bmatrix} = \begin{bmatrix} 10 & -6 \ 0 & 4 \end{bmatrix}$.

Market	$x, y, z$
$I$	$10,000, 2,000, 18,000$
$II$	$6,000, 20,000, 8,000$

Let $A$ be a matrix such that $AB$ is a scalar matrix where $B = \begin{bmatrix} 1 & 2 \\ 0 & 3 \end{bmatrix}$ and $\det(3A) = 27$. Then $3A^{-1} + A^2 =$

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