If $A = \left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}1&2&0\\{ - 2}&{ - 1}&{ - 2}\\0&{ - 1}&1\end{array}} \right|$, then find the value of ${A^{ - 1}}$.
Using ${A^{ - 1}}$, solve the system of linear equations $x - 2y = 10$, $2x - y - z = 8$ and $- 2y + z = 7$.

We have, $A = \left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}1&2&0\\{ - 2}&{ - 1}&{ - 2}\\0&{ - 1}&1\end{array}} \right|$

……(i)
$\therefore$ $|A| = 1( - 3) - 2( - 2) + 0 = 1 \ne 0$

Now, ${A_{11}} = - 3,{A_{12}} = 2,{A_{13}} = 2,{A_{21}} = - 2,{A_{22}} = 1,{A_{23}} = 1,{A_{31}} = - 4,{A_{32}} = 2$ and ${A_{33}} = 3$

${\mathop{\rm adj}\nolimits} (A) = {\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{ - 3}&2&2\\{ - 2}&1&1\\{ - 4}&2&3\end{array}} \right|^T} = \left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{ - 3}&{ - 2}&{ - 4}\\2&1&2\\2&1&3\end{array}} \right|$

$\therefore$ ${A^{ - 1}} = \frac{{{\mathop{\rm adj}\nolimits} A}}{{|A|}}$

$= \frac{1}{1}\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{ - 3}&{ - 2}&{ - 4}\\2&1&2\\2&1&3\end{array}} \right|$

$\Rightarrow$ ${A^{ - 1}} = \left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{ - 3}&{ - 2}&{ - 4}\\2&1&2\\2&1&3\end{array}} \right|$

……(ii)
Also. we have the system of linear equations as
$x - 2y = 10$,

$2x - y - z = 8$

and $- 2y + z = 7$

In the form of $CX = D$

$\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}1&2&0\\2&{ - 1}&{ - 1}\\0&{ - 2}&1\end{array}} \right]\left[ {\begin{array}{llllllllllllllllllll}x\\y\\z\end{array}} \right] = \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{10}\\8\\7\end{array}} \right]$

where$C = \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}1&{ - 2}&0\\2&{ - 1}&{ - 1}\\0&{ - 2}&1\end{array}} \right],X = \left[ {\begin{array}{llllllllllllllllllll}x\\y\\z\end{array}} \right]$

and $D = \left[ {\begin{array}{cccccccccccccccccccc}{10}\\8\\7\end{array}} \right]$
We know that, ${\left. {{A^T}} \right)^{ - 1}} = {\left( {{A^{ - 1}}} \right)^T}$

$\therefore$ ${C^T} = \left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}1&2&0\\{ - 2}&{ - 1}&{ - 2}\\0&{ - 1}&1\end{array}} \right| = A$

[using Eq. (i)]
$\therefore$ $X = {C^{ - 1}}D$

$\Rightarrow$ $\left[ {\begin{array}{llllllllllllllllllll}x\\y\\z\end{array}} \right]$

$= \left[ {\begin{array}{llllllllllllllllllll}{ - 3}&2&2\\{ - 2}&1&1\\{ - 4}&2&3\end{array}} \right]\left[ {\begin{array}{cccccccccccccccccccc}{10}\\8\\7\end{array}} \right]$

$= \left[ {\begin{array}{cccccccccccccccccccc}{ - 30 + 16 + 14}\\{ - 20 + 8 + 7}\\{ - 40 + 16 + 21}\end{array}} \right]$

$= \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}0\\{ - 5}\\{ - 3}\end{array}} \right]$

$\therefore$ $x = 0,y = - 5$ and $z = - 3$