Let A be a square matrix of order 2 such that $\begin{bmatrix} 2 & 1 \\ 3 & 2 \end{bmatrix} A \begin{bmatrix} -3 & 2 \\ 5 & -3 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$, then A is:

Given the equation:

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

Let $P = \begin{bmatrix} 2 & 1 \\ 3 & 2 \end{bmatrix}$, $Q = \begin{bmatrix} -3 & 2 \\ 5 & -3 \end{bmatrix}$, and $I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$

The equation becomes: $PAQ = I$

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To isolate $A$, multiply both sides on the left by $P^{-1}$ and on the right by $Q^{-1}$:

$P^{-1}(PAQ)Q^{-1} = P^{-1}IQ^{-1}$

$A = P^{-1}Q^{-1}$

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For matrix $P = \begin{bmatrix} 2 & 1 \\ 3 & 2 \end{bmatrix}$:

$\text{det}(P) = (2)(2) - (1)(3)$

$= 4 - 3$

$= 1$

Using the inverse formula for a 2×2 matrix:

$P^{-1} = \frac{1}{1}\begin{bmatrix} 2 & -1 \\ -3 & 2 \end{bmatrix}$

$= \begin{bmatrix} 2 & -1 \\ -3 & 2 \end{bmatrix}$

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For matrix $Q = \begin{bmatrix} -3 & 2 \\ 5 & -3 \end{bmatrix}$:

$\text{det}(Q) = (-3)(-3) - (2)(5)$

$= 9 - 10$

$= -1$

$Q^{-1} = \frac{1}{-1}\begin{bmatrix} -3 & -2 \\ -5 & -3 \end{bmatrix}$

$= \begin{bmatrix} 3 & 2 \\ 5 & 3 \end{bmatrix}$

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$A = P^{-1}Q^{-1}$

$= \begin{bmatrix} 2 & -1 \\ -3 & 2 \end{bmatrix}\begin{bmatrix} 3 & 2 \\ 5 & 3 \end{bmatrix}$

First row, first column: $(2)(3) + (-1)(5) = 6 - 5 = 1$

First row, second column: $(2)(2) + (-1)(3) = 4 - 3 = 1$

Second row, first column: $(-3)(3) + (2)(5) = -9 + 10 = 1$

Second row, second column: $(-3)(2) + (2)(3) = -6 + 6 = 0$

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