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(A) Let $P = \begin{bmatrix} 2 & 1 \ 3 & 2 \end{bmatrix}$ and $Q = \begin{bmatrix} -3 & 2 \ 5 & -3 \end{bmatrix}$. The given equation is $P \cdot A

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$\begin{bmatrix} 1 & 1 \ 1 & 0 \end{bmatrix}$

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(A) Let $P = \begin{bmatrix} 2 & 1 \ 3 & 2 \end{bmatrix}$ and $Q = \begin{bmatrix} -3 & 2 \ 5 & -3 \end{bmatrix}$. The given equation is $P \cdot A \cdot Q = I$, where $I$ is the identity matrix. Multiplying by $P^{-1}$ on the left and $Q^{-1}$ on the right, we get $A = P^{-1} \cdot I \cdot Q^{-1} = P^{-1} \cdot Q^{-1}$. First, find $P^{-1}$: $\det(P) = (2)(2) - (1)(3) = 4 - 3 = 1$. Thus, $P^{-1} = \frac{1}{1} \begin{bmatrix} 2 & -1 \ -3 & 2 \end{bmatrix} = \begin{bmatrix} 2 & -1 \ -3 & 2 \end{bmatrix}$. Next, find $Q^{-1}$: $\det(Q) = (-3)(-3) - (2)(5) = 9 - 10 = -1$. Thus, $Q^{-1} = \frac{1}{-1} \begin{bmatrix} -3 & -2 \ -5 & -3 \end{bmatrix} = \begin{bmatrix} 3 & 2 \ 5 & 3 \end{bmatrix}$. Now, calculate $A = P^{-1} \cdot Q^{-1} = \begin{bmatrix} 2 & -1 \ -3 & 2 \end{bmatrix} \begin{bmatrix} 3 & 2 \ 5 & 3 \end{bmatrix} = \begin{bmatrix} (2)(3) + (-1)(5) & (2)(2) + (-1)(3) \ (-3)(3) + (2)(5) & (-3)(2) + (2)(3) \end{bmatrix} = \begin{bmatrix} 6-5 & 4-3 \ -9+10 & -6+6 \end{bmatrix} = \begin{bmatrix} 1 & 1 \ 1 & 0 \end{bmatrix}$.

If $\begin{bmatrix} 2 & 1 \\ 3 & 2 \end{bmatrix} \cdot A \cdot \begin{bmatrix} -3 & 2 \\ 5 & -3 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$, then $A =$

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