Find the matrix A satisfying the matrix equation
$\left[ {\begin{array}{llllllllllllllllllll}2&1\\3&2\end{array}} \right]A\left[ {\begin{array}{cccccccccccccccccccc}{ - 3}&2\\5&{ - 3}\end{array}} \right] = \left[ {\begin{array}{llllllllllllllllllll}1&0\\0&1\end{array}} \right]$.

We have, ${\left[ {\begin{array}{llllllllllllllllllll}2&1\\3&2\end{array}} \right]_{2 \times 2}}A \cdot {\left[ {\begin{array}{cccccccccccccccccccc}{ - 3}&2\\5&{ - 3}\end{array}} \right]_{2 \times 2}} = {\left[ {\begin{array}{cccccccccccccccccccc}1&0\\0&1\end{array}} \right]_{2 \times 2}}$

Let $A = {\left[ {\begin{array}{llllllllllllllllllll}a&b\\c&d\end{array}} \right]_{2 \times 2}}$

$\therefore$ $\left[ {\begin{array}{llllllllllllllllllll}2&1\\3&2\end{array}} \right]\left[ {\begin{array}{llllllllllllllllllll}a&b\\c&d\end{array}} \right]\left[ {\begin{array}{cccccccccccccccccccc}{ - 3}&2\\5&{ - 3}\end{array}} \right]$ $ = \left[ {\begin{array}{cccccccccccccccccccc}1&0\\0&1\end{array}} \right]$

$\Rightarrow$ $\left[ {\begin{array}{llllllllllllllllllll}{2a + c}&{2b + d}\\{3a + 2c}&{3b + 2d}\end{array}} \right]\left[ {\begin{array}{cccccccccccccccccccc}{ - 3}&2\\5&{ - 3}\end{array}} \right]$ $ = \left[ {\begin{array}{llllllllllllllllllll}1&0\\0&1\end{array}} \right]$

$\Rightarrow$ $\left[ {\begin{array}{llllllllllllllllllll}{ - 6a - 3c + 10b + 5d}&{4a + 2c - 6b - 3d}\\{ - 9a - 6c + 15b + 10d}&{6a + 4c - 9b - 6d}\end{array}} \right]$ $ = \left[ {\begin{array}{llllllllllllllllllll}1&0\\0&1\end{array}} \right]$
$\Rightarrow$ $- 6a - 3c + 10b + 5d = 1$

….(i)
$\Rightarrow$ $4a + 2c - 6b - 3d = 0$

…..(ii)
$\Rightarrow$ $- 9a - 6c + 15b + 10d = 0$

…….(iii)
$\Rightarrow$ $6a + 4c - 9b - 6d = 1$

…….(iv)
On adding Eqs. (i) and (iv),

we get
$c + b - d = 2 \Rightarrow d = c + b - 2$

…….(v)
On adding Eqs. (ii) and (iii),

we get
$- 5a - 4c + 9b + 7d = 0$

…….(vi)
On adding Eqs. (vi) and (iv),

we get
$a + 0 + 0 + d = 1 \Rightarrow d = 1 - a$

…….(vii)
From Eqs. (v) and (vii),
$c + b - 2 = 1 - a \Rightarrow a + b + c = 3$

…..(viii)
$\Rightarrow$ $a = 3 - b - c$

Now, using the values of A and $d$ in Eq. (iii),

we get
$- 9(3 - b - c) - 6c + 15b + 10( - 2 + b + c) = 0$

$\Rightarrow$ $- 27 + 9b + 9c - 6c + 15b - 20 + 10b + 10c = 0$

$\Rightarrow$ $34b + 13c = 47$
…….(ix)
Now, using the values of $a$ and $d$ in Eq. (ii),

we get
$4(3 - b - c) + 2c - 6b - 3(b + c - 2) = 0$

$\Rightarrow$ $12 - 4b - 4c + 2c - 6b - 3b - 3c + 6 $ $= 0$
$\Rightarrow$ $- 13b - 5c = - 18$

…..(x)
On multiplying Eq. (ix) by 5 and Eq. (x) by 13, then adding,

we get
$\frac{\begin{array}{l} - 169b - 65c = - 234\\\,\,\,\,170b + 65c = 235\end{array}}{{b = 1}}$

$\Rightarrow$ $- 13 \times 1 - 5c = - 18$

[from Eq. $({\rm{x}})]$
$\Rightarrow$ $- 5c = - 18 + 13 = - 5 \Rightarrow c = 1$

$\therefore$ $a = 3 - 1 - 1 = 1$ and $d = 1 - 1 = 0$
$\therefore$ $A = \left[ {\begin{array}{cccccccccccccccccccc}1&1\\1&0\end{array}} \right]$