If \[A = \left[ {\begin{array}{*{20}{c}}2&2\\{ - 3}&2\end{array}} \right]\] and \[B = \left[ {\begin{array}{*{20}{c}}0&{ - 1}\\1&0\end{array}} \right]\] , then find \[{\left( {{B^{ - 1}}{A^{ - 1}}} \right)^{ - 1}}\].
A. \[\left[ {\begin{array}{*{20}{c}}2&{ - 2}\\2&3\end{array}} \right]\]
B. \[\left[ {\begin{array}{*{20}{c}}3&{ - 2}\\2&2\end{array}} \right]\]
C. \[\dfrac{1}{{10}}\left[ {\begin{array}{*{20}{c}}2&2\\{ - 2}&3\end{array}} \right]\]
D. \[\dfrac{1}{{10}}\left[ {\begin{array}{*{20}{c}}3&2\\{ - 2}&2\end{array}} \right]\]

Hint: First we will check whether the given matrices are invertible or not. To solve the question, we will apply the formula of the inverse of the matrix product. Again, apply the formula inverse of the inverse matrix. Then substitute the matrix A and B and solve it.

Formula Used:
A matrix A is invertible, if \[\left| A \right| \ne 0\].
Inverse of matrix product formula: \[{\left( {AB} \right)^{ - 1}} = {B^{ - 1}}{A^{ - 1}}\], where A and B invertible matrix.
Inverse of inverse matrix formula is \[{\left( {{A^{ - 1}}} \right)^{ - 1}} = A\]

Complete step by step solution:
Given matrices are \[A = \left[ {\begin{array}{*{20}{c}}2&2\\{ - 3}&2\end{array}} \right]\] and \[B = \left[ {\begin{array}{*{20}{c}}0&{ - 1}\\1&0\end{array}} \right]\].
Now calculating the determinate of both matrix
\[\left| A \right| = \left| {\begin{array}{*{20}{c}}2&2\\{ - 3}&2\end{array}} \right|\]
\[ = 2 \cdot 2 - 2 \cdot \left( { - 3} \right)\]
\[ = 10 \ne 0\]
\[\left| A \right| = \left| {\begin{array}{*{20}{c}}0&{ - 1}\\1&0\end{array}} \right|\]
\[ = 0 \cdot 0 - 1 \cdot \left( { - 1} \right)\]
\[ = 1 \ne 0\]
Sine, determinant of both the matrices \[ne 0\]
Thus both matrices are invertible.
Apply the formula of the inverse of matrix product on \[{\left( {{B^{ - 1}}{A^{ - 1}}} \right)^{ - 1}}\]
\[{\left( {{B^{ - 1}}{A^{ - 1}}} \right)^{ - 1}}\]
\[ = {\left( {{A^{ - 1}}} \right)^{ - 1}}{\left( {{B^{ - 1}}} \right)^{ - 1}}\]
Now applying the inverse of the inverse matrix
\[ = AB\]
Substitute \[A = \left[ {\begin{array}{*{20}{c}}2&2\\{ - 3}&2\end{array}} \right]\] and \[B = \left[ {\begin{array}{*{20}{c}}0&{ - 1}\\1&0\end{array}} \right]\]
By matrix multiplication,
\[ = \left[ {\begin{array}{*{20}{c}}2&2\\{ - 3}&2\end{array}} \right]\left[ {\begin{array}{*{20}{c}}0&{ - 1}\\1&0\end{array}} \right]\]
\[ = \left[ {\begin{array}{*{20}{c}}{2 \cdot 0 + 2 \cdot 1}&{2 \cdot \left( { - 1} \right) + 2 \cdot 0}\\{ - 3 \cdot 0 + 2 \cdot 1}&{ - 3 \cdot \left( { - 1} \right) + 2 \cdot 0}\end{array}} \right]\]
\[ = \left[ {\begin{array}{*{20}{c}}2&{ - 2}\\2&3\end{array}} \right]\]
Hence option A is the correct option.

Note: Students often do a common mistake to solve the question. They do not apply the inverse of matrix multiplication. The correct way is: we apply the inverse of matrix multiplication, then the formula inverse of the inverse matrix. Remember the matrix must be invertible.