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# If $\begin{bmatrix}2&1\\3&2\end{bmatrix}A\begin{bmatrix}-3&2\\5&-3\end{bmatrix}=\begin{bmatrix}1&0\\0&1\end{bmatrix}$ then the matrix A is equal to

$\begin{array}{1 1}(a)\;\begin{bmatrix}1&0\\1&1\end{bmatrix}&(b)\;\begin{bmatrix}0&1\\1&1\end{bmatrix}\\(c)\;\begin{bmatrix}1&1\\1&0\end{bmatrix}&(d)\;\begin{bmatrix}1&1\\0&1\end{bmatrix}\end{array}$

We have $\begin{bmatrix}2&1\\3&2\end{bmatrix}A\begin{bmatrix}-3&2\\5&-3\end{bmatrix}=\begin{bmatrix}1 &0\\0&1\end{bmatrix}$
$\Rightarrow \begin{bmatrix}2&1\\3&2\end{bmatrix}A=\begin{bmatrix}1&0\\0&1\end{bmatrix}\begin{bmatrix}-3&2\\5&-3\end{bmatrix}^{-1}$
$A=\begin{bmatrix}2&1\\3&2\end{bmatrix}^{-1}\begin{bmatrix}1&0\\0&1\end{bmatrix}\begin{bmatrix}-3&2\\5&-3\end{bmatrix}^{-1}$
$\;\;=\large\frac{1}{1}$$\begin{bmatrix}2&-1\\-3&2\end{bmatrix}\large\frac{-1}{1}$$\begin{bmatrix}-3&-2\\-5&-3\end{bmatrix}$
$\;\;=\begin{bmatrix}2 &-1\\-3&2\end{bmatrix}\begin{bmatrix}3&2\\5&3\end{bmatrix}$
$\;\;=\begin{bmatrix}6-5&4-3\\-9+10&-6+6\end{bmatrix}$
$\;\;=\begin{bmatrix}1&1\\1&0\end{bmatrix}$
Hence (c) is the correct answer.