Let $A = \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}3&7\\2&5\end{array}} \right]$ and $B = \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}6&8\\7&9\end{array}} \right]$ . Verify that ${(AB)^{ - 1}} = {B^{ - 1}}{A^{ - 1}}$ .

We have, $AB = \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}3&7\\2&5\end{array}} \right]\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}6&8\\7&9\end{array}} \right] = \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{67}&{87}\\{47}&{61}\end{array}} \right]$

Since, $|AB| = 67 \times 61 - 47 \times 87 = - 2 \ne 0$

So, AB is non-singular matrix and therefore,

${(AB)^{ - 1}}$ exists and is given by ${(AB)^{ - 1}} = \cfrac{1}{{|AB|}}adj(AB)$

$= - \cfrac{1}{2}\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{61}&{ - 87}\\{ - 47}&{67}\end{array}} \right] = \cfrac{1}{2}\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{ - 61}&{87}\\{47}&{ - 67}\end{array}} \right]$

Further, $|A| = 15 - 14 = 1 \ne 0$ and $|B| = 54 - 56 = - 2 \ne 0.$

So, A and B are both non-singular matrices and

therefore, ${A^{ - 1}}$and ${B^{ - 1}}$

both exist and are given by
${A^{ - 1}} = 1\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}5&{ - 7}\\{ - 2}&3\end{array}} \right],{B^{ - 1}} = - \cfrac{1}{2}\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}9&{ - 8}\\{ - 7}&6\end{array}} \right]$

$\therefore$ ${B^{ - 1}}{A^{ - 1}} = - \cfrac{1}{2}\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}9&{ - 8}\\{ - 7}&6\end{array}} \right]\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}5&{ - 7}\\{ - 2}&3\end{array}} \right]$

$= - \cfrac{1}{2}\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{61}&{ - 87}\\{ - 47}&{67}\end{array}} \right] = \cfrac{1}{2}\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{ - 61}&{87}\\{47}&{ - 67}\end{array}} \right]$

Hence, ${(AB)^{ - 1}} = {B^{ - 1}}{A^{ - 1}}.$