Find the inverse of the matrix$\left( {\begin{array}{*{20}{c}}
  { - 3}&2 \\
  5&{ - 3}
\end{array}} \right)$. Hence find the matrix P satisfying the matrix equation:
$P\left( {\begin{array}{*{20}{c}}
  { - 3}&2 \\
  5&{ - 3}
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
  1&2 \\
  2&{ - 1}
\end{array}} \right)$

Hint: 1. When a matrix A is multiplied with its inverse A-1, it gives out identity matrix as a result.
i.e. $A{A^{ - 1}} = I$
2. Inverse of a matrix is defined as the adjoint of the matrix divided by its determinant.
i.e. ${A^{ - 1}} = \dfrac{{adjA}}{{\left| A \right|}}$
3. If we are given that$A = \left( {\begin{array}{*{20}{c}}
  a&b \\
  d&c
\end{array}} \right)$
Then, adjoint of a matrix is defined as $adjA = \left( {\begin{array}{*{20}{c}}
  c&{ - b} \\
  d&a
\end{array}} \right)$

Complete step-by-step answer:
We are given that $A = \left( {\begin{array}{*{20}{c}}
  { - 3}&2 \\
  5&{ - 3}
\end{array}} \right)$
1. Then by using the hint $adjA = \left( {\begin{array}{*{20}{c}}
  { - 3}&{ - 2} \\
  { - 5}&{ - 3}
\end{array}} \right)$
2. Determinant of matrix A can be found out as,
$\begin{gathered}
  \left| A \right| = \left( {( - 3) \times ( - 3)} \right) - \left( {5 \times 2} \right) \\
  \left| A \right| = 9 - 10 \\
  \left| A \right| = - 1 \\
\end{gathered} $
3. Now for finding the inverse of the matrix, then
${A^{ - 1}} = \dfrac{{adjA}}{{\left| A \right|}}$
Putting the values of $adjA$and $\left| A \right|$ in the above equation, we get
$\begin{gathered}
  {A^{ - 1}} = \dfrac{{\left( {\begin{array}{*{20}{c}}
  { - 3}&{ - 2} \\
  { - 5}&{ - 3}
\end{array}} \right)}}{{ - 1}} \\
  {A^{ - 1}} = \left( {\begin{array}{*{20}{c}}
  3&2 \\
  5&3
\end{array}} \right) \\
\end{gathered} $
The inverse of the matrix A is${A^{ - 1}} = \left( {\begin{array}{*{20}{c}}
  3&2 \\
  5&3
\end{array}} \right)$.
4. Now for finding out the value of P such that
$P\left( {\begin{array}{*{20}{c}}
  { - 3}&2 \\
  5&{ - 3}
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
  1&2 \\
  2&{ - 1}
\end{array}} \right)$ ...... (1)
Since we know that
$A = \left( {\begin{array}{*{20}{c}}
  { - 3}&2 \\
  5&{ - 3}
\end{array}} \right)$
Let, $B = \left( {\begin{array}{*{20}{c}}
  1&2 \\
  2&{ - 1}
\end{array}} \right)$
So Eqn (1) can also be written as
$PA = B$
Multiplying by ${A^{ - 1}}$both sides we get
$PA{A^{ - 1}} = B{A^{ - 1}}$
Since we know that when a matrix is multiplied it with its inverse than it results in identity matrix
i.e. $A{A^{ - 1}} = I$
$PI = B{A^{ - 1}}$
$P = B{A^{ - 1}}$
Since, now we know the values of ${A^{ - 1}}$and$B$. Putting in the above equation, we get
$\begin{gathered}
  P = \left( {\begin{array}{*{20}{c}}
  1&2 \\
  2&{ - 1}
\end{array}} \right) \times \left( {\begin{array}{*{20}{c}}
  3&2 \\
  5&3
\end{array}} \right) \\
  P = \left( {\begin{array}{*{20}{c}}
  {(1 \times 3) + (2 \times 5)}&{(1 \times 2) + (2 \times 3)} \\
  {(2 \times 3) + ( - 1 \times 5)}&{(2 \times 2) + ( - 1 \times 3)}
\end{array}} \right) \\
  P = \left( {\begin{array}{*{20}{c}}
  {3 + 10}&{2 + 6} \\
  {6 + ( - 5)}&{4 + ( - 3)}
\end{array}} \right) \\
  P = \left( {\begin{array}{*{20}{c}}
  {13}&8 \\
  {6 - 5}&{4 - 3}
\end{array}} \right) \\
  P = \left( {\begin{array}{*{20}{c}}
  {13}&8 \\
  1&1
\end{array}} \right) \\
\end{gathered} $
Therefore, the value of P is:
$P = \left( {\begin{array}{*{20}{c}}
  {13}&8 \\
  1&1
\end{array}} \right)$

Note: When an identity matrix $I$is multiplied to any matrix then the resultant matrix is same as the multiplied matrix
i.e. $A{I^{ - 1}} = A$