If ${A^{ - 1}} = \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}3&{ - 1}&1\\{ - 15}&6&{ - 5}\\5&{ - 2}&2\end{array}} \right]$ and $B = \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}1&2&{ - 2}\\{ - 1}&3&0\\0&{ - 2}&1\end{array}} \right]$ find ${(AB)^{ - 1}}.$

Given, ${A^{ - 1}} = \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}3&{ - 1}&1\\{ - 15}&6&{ - 5}\\5&{ - 2}&2\end{array}} \right]$, $B = \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}1&2&{ - 2}\\{ - 1}&3&0\\0&{ - 2}&1\end{array}} \right]$

Now,$|B| = 1(3) - 2( - 1) - 2(2) = 3 + 2 - 4 = 1 \ne 0$
$\therefore$ ${B^{ - 1}}$ exists.

Since ${(AB)^{ - 1}} = {B^{ - 1}}{A^{ - 1}},$ so,

we need to calculate ${B^{ - 1}}.$

Cofactors of elements of B are :

${B_{11}} = {( - 1)^{1 + 1}}\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}3&0\\{ - 2}&1\end{array}} \right| = 3,$

${B_{12}} = {( - 1)^{1 + 2}}\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{ - 1}&0\\0&1\end{array}} \right| = 1,$

${B_{13}} = {( - 1)^{1 + 3}}\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{ - 1}&3\\0&{ - 2}\end{array}} \right| = 2,$

${B_{21}} = {( - 1)^{2 + 1}}\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}2&{ - 2}\\{ - 2}&1\end{array}} \right| = 2,$

${B_{22}} = {( - 1)^{2 + 2}}\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}1&{ - 2}\\0&1\end{array}} \right| = 1,$

${B_{23}} = {( - 1)^{2 + 3}}\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}1&2\\0&{ - 2}\end{array}} \right| = 2,$,

${B_{31}} = {( - 1)^{3 + 1}}\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}2&{ - 2}\\3&0\end{array}} \right| = 6,$

${B_{32}} = {( - 1)^{3 + 2}}\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}1&{ - 2}\\{ - 1}&0\end{array}} \right| = 2,$

${B_{33}} = {( - 1)^{3 + 3}}\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}1&2\\{ - 1}&3\end{array}} \right| = 5$

%$\therefore$ $adj\;B = {\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}3&1&2\\2&1&2\\6&2&5\end{array}} \right]^{\prime}} = \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}3&2&6\\1&1&2\\2&2&5\end{array}} \right]$

Hence, ${B^{ - 1}} = \cfrac{1}{{|B|}}(adj\;B) = \cfrac{1}{1}\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}3&2&6\\1&1&2\\2&2&5\end{array}} \right]$

Now, ${B^{ - 1}}{A^{ - 1}} = \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}3&2&6\\1&1&2\\2&2&5\end{array}} \right]\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}3&{ - 1}&1\\{ - 15}&6&{ - 5}\\5&{ - 2}&2\end{array}} \right]$

$= \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{9 - 30 + 30}&{ - 3 + 12 - 12}&{3 - 10 + 12}\\{3 - 15 + 10}&{ - 1 + 6 - 4}&{1 - 5 + 4}\\{6 - 30 + 25}&{ - 2 + 12 - 10}&{2 - 10 + 10}\end{array}} \right] = \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}9&{ - 3}&5\\{ - 2}&1&0\\1&0&2\end{array}} \right]$