Find the matrix A satisfying the matrix equation
 $ \left[ {\begin{array}{*{20}{c}}
  2&1 \\
  3&2
\end{array}} \right]A\left[ {\begin{array}{*{20}{c}}
  { - 3}&2 \\
  5&{ - 3}
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
  1&0 \\
  0&1
\end{array}} \right] $

Hint: Here we will entitle the given matrix as B and C matrix and then will use the properties of the matrix, identity matrix and the inverse of the matrix. Then accordingly satisfying the conditions will find matrix A.

Complete step-by-step answer:
 $ \left[ {\begin{array}{*{20}{c}}
  2&1 \\
  3&2
\end{array}} \right]A\left[ {\begin{array}{*{20}{c}}
  { - 3}&2 \\
  5&{ - 3}
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
  1&0 \\
  0&1
\end{array}} \right] $
Let us assume that –
 $ B = \left[ {\begin{array}{*{20}{c}}
  2&1 \\
  3&2
\end{array}} \right] $ and
 $ C = \left[ {\begin{array}{*{20}{c}}
  { - 3}&2 \\
  5&{ - 3}
\end{array}} \right] $
Also, Identity matrix $ I = \left[ {\begin{array}{*{20}{c}}
  1&0 \\
  0&1
\end{array}} \right] $
So, the given expression can be written as –
 $ BAC = I $
Multiply the above expression with the matrix inverse of B.
 $ \Rightarrow {B^{ - 1}}BAC = I.{B^{ - 1}} $
We know that product of any matrix with its inverse gives identity matrix and product of identity matrix with inverse gives inverse. Since the identity matrix multiplied with any matrix gives the matrix itself.
 $ \Rightarrow IAC = {B^{ - 1}} $
Therefore,
 $ \Rightarrow AC = {B^{ - 1}} $
Similarly, multiply the above expression with the matrix inverse of C.
 $ \Rightarrow AC{C^{ - 1}} = {B^{ - 1}}.{C^{ - 1}} $
Apply the same concept, inverse of the matrix and the matrix gives identity matrix.
 $ \Rightarrow AI = {B^{ - 1}}.{C^{ - 1}} $
By property-
 $ \Rightarrow A = {B^{ - 1}}{C^{ - 1}} $ ..... (I)
Now, find the inverse of matrix B- first find the adjoint of the matrix and the determinant of the product of matrix and at last will place the values in the standard formula. $ {(B)^{ - 1}} = \dfrac{{adj(B)}}{{\left| B \right|}} $
 $ \left| B \right| = \left| {\begin{array}{*{20}{c}}
  2&1 \\
  3&2
\end{array}} \right| $
Expand the determinant-
 $ \left| B \right| = 4 - 3 = 1 $
Adjoint of B $ = {\left[ {\begin{array}{*{20}{c}}
  2&{ - 3} \\
  { - 1}&2
\end{array}} \right]^1} $
 $ adjB = \left[ {\begin{array}{*{20}{c}}
  2&{ - 1} \\
  { - 3}&2
\end{array}} \right] $
Now, inverse
 $ {B^{ - 1}} = \dfrac{{adjB}}{{\left| B \right|}} $
Place values-
 $ {B^{ - 1}} = \dfrac{{\left[ {\begin{array}{*{20}{c}}
  2&{ - 1} \\
  { - 3}&2
\end{array}} \right]}}{{(1)}} $
Take negative sign common from the matrix
 $ {B^{ - 1}} = \left[ {\begin{array}{*{20}{c}}
  2&{ - 1} \\
  { - 3}&2
\end{array}} \right] $ ..... (II)
Similarly for the matrix C
 $ \left| C \right| = \left| {\begin{array}{*{20}{c}}
  { - 3}&2 \\
  5&{ - 3}
\end{array}} \right| $
Expand the determinant-
 $ \left| C \right| = 9 - 10 = - 1 $
Adjoint of C $ = {\left[ {\begin{array}{*{20}{c}}
  { - 3}&{ - 5} \\
  { - 2}&{ - 3}
\end{array}} \right]^1} $
 $ adjC = \left[ {\begin{array}{*{20}{c}}
  { - 3}&{ - 2} \\
  { - 5}&{ - 3}
\end{array}} \right] $
Now, inverse
 $ {C^{ - 1}} = \dfrac{{adjC}}{{\left| C \right|}} $
Place values-
 $ {C^{ - 1}} = \dfrac{{\left[ {\begin{array}{*{20}{c}}
  { - 3}&{ - 2} \\
  { - 5}&{ - 3}
\end{array}} \right]}}{{( - 1)}} $
Take negative sign common from the matrix
 $ {C^{ - 1}} = \left[ {\begin{array}{*{20}{c}}
  3&2 \\
  5&3
\end{array}} \right] $ .... (III)
Place the values of equation (II) and (III) in (I)
 $ \Rightarrow A = \left[ {\begin{array}{*{20}{c}}
  2&{ - 1} \\
  { - 3}&2
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
  3&2 \\
  5&3
\end{array}} \right] $
Find the product of two matrices-
 $ \Rightarrow A = \left[ {\begin{array}{*{20}{c}}
  {2(3) + ( - 1)(5)}&{2(2) + ( - 1)(3)} \\
  { - 3(3) + 2(5)}&{( - 3)2 + 3(2)}
\end{array}} \right] $
Simplify
 $ \Rightarrow A = \left[ {\begin{array}{*{20}{c}}
  {6 - 5}&{4 - 3} \\
  { - 9 + 10}&{ - 6 + 6}
\end{array}} \right] $
 $ \Rightarrow A = \left[ {\begin{array}{*{20}{c}}
  1&1 \\
  1&0
\end{array}} \right] $
So, the correct answer is “$A = \left[ {\begin{array}{*{20}{c}}
  1&1 \\
  1&0
\end{array}} \right] $ ”.

Note: In mathematics, particularly in linear algebra, matrix multiplication is a binary operation that produces a matrix from two matrices. For matrix multiplication, the number of columns in the first matrix must be equal to the number of rows in the second matrix.