If $A = \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}2&{ - 3}&5\\3&2&{ - 4}\\1&1&{ - 2}\end{array}} \right]$ , find ${A^{ - 1}}.$Using ${A^{ - 1}},$ solve the system of equations
$2x - 3y + 5z = 11,3x + 2y - 4z = - 5,x + y - 2z = - 3.$

We have, $A = \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}2&{ - 3}&5\\3&2&{ - 4}\\1&1&{ - 2}\end{array}} \right]$
$\therefore$ $|A| = 2( - 4 + 4) + 3( - 6 + 4) + 5(3 - 2) = 0 - 6 + 5 = - 1 \ne 0.$

So, A is invertible.

Cofactors of elements of A are :

${A_{11}} = {( - 1)^{1 + 1}}\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}2&{ - 4}\\1&{ - 2}\end{array}} \right| = ( - 4 + 4) = 0,$

${A_{12}} = {( - 1)^{1 + 2}}\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}3&{ - 4}\\1&{ - 2}\end{array}} \right| = - ( - 6 + 4) = 2,$

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

${A_{21}} = {( - 1)^{2 + 1}}\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{ - 3}&5\\1&{ - 2}\end{array}} \right| = - (6 - 5) = - 1,$

${A_{22}} = {( - 1)^{2 + 2}}\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}2&5\\1&{ - 2}\end{array}} \right| = ( - 4 - 5) = - 9,$

${A_{23}} = {( - 1)^{2 + 3}}\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}2&{ - 3}\\1&1\end{array}} \right| = - (2 + 3) = - 5,$,

${A_{31}} = {( - 1)^{3 + 1}}\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{ - 3}&5\\2&{ - 4}\end{array}} \right| = (12 - 10) = 2,$

${A_{32}} = {( - 1)^{3 + 2}}\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}2&5\\3&{ - 4}\end{array}} \right| = - ( - 8 - 15) = 23,$

${A_{33}} = {( - 1)^{3 + 3}}\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}2&{ - 3}\\3&2\end{array}} \right| = (4 + 9) = 13,$

%$\therefore$ $adj\;A = {\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}0&2&1\\{ - 1}&{ - 9}&{ - 5}\\2&{23}&{13}\end{array}} \right]^{\prime}} = \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}0&{ - 1}&2\\2&{ - 9}&{23}\\1&{ - 5}&{13}\end{array}} \right]$

Hence, ${A^{ - 1}} = \cfrac{1}{{|A|}}(adj\;\;A) = \cfrac{1}{{ - 1}}\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}0&{ - 1}&2\\2&{ - 9}&{23}\\1&{ - 5}&{13}\end{array}} \right]$

$= \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}0&1&{ - 2}\\{ - 2}&9&{ - 23}\\{ - 1}&5&{ - 13}\end{array}} \right]$

Now, the given system of equations can be written in form $AX = B,$

where
$A = \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}2&{ - 3}&5\\3&2&{ - 4}\\1&1&{ - 2}\end{array}} \right],X = \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}x\\y\\z\end{array}} \right]$

and $B = \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{11}\\{ - 5}\\{ - 3}\end{array}} \right]$

As, $|A| = - 1 \ne 0,$

so given system of equations has a unique solution given by $X = {A^{ - 1}}B.$

$\Rightarrow$ $\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}x\\y\\z\end{array}} \right] = \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}0&1&{ - 2}\\{ - 2}&9&{ - 23}\\{ - 1}&5&{ - 13}\end{array}} \right]\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{11}\\{ - 5}\\{ - 3}\end{array}} \right]$

$= \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{0 - 5 + 6}\\{ - 22 - 45 + 69}\\{ - 11 - 25 + 39}\end{array}} \right] = \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}1\\2\\3\end{array}} \right]$
Hence, $x = 1,y = 2$ and $z = 3.$