Find ${A^{ - 1}}$, if $A = \left| {\begin{array}{llllllllllllllllllll}0&1&1\\1&0&1\\1&1&0\end{array}} \right|$ and show that ${A^{ - 1}} = \frac{{{A^2} - 3I}}{2}$.

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

$\therefore$ ${A_{11}} = - 1,{A_{12}} = 1,{A_{13}} = 1,{A_{21}} = 1,{A_{22}} = - 1,{A_{23}} = 1,{A_{31}} = 1,{A_{32}} = 1$ and ${A_{33}} = - 1$

adj $A = {\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{ - 1}&1&1\\1&{ - 1}&1\\1&1&{ - 1}\end{array}} \right|^T} = \left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{ - 1}&1&1\\1&{ - 1}&1\\1&1&{ - 1}\end{array}} \right|$

and $|A| = - 1( - 1) + 1 \cdot 1 = 2$

$\therefore$ ${A^{ - 1}} = \frac{{{\mathop{\rm adj}\nolimits} A}}{{|A|}} = \frac{1}{2}\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{ - 1}&1&1\\1&{ - 1}&1\\1&1&{ - 1}\end{array}} \right]$

…….(i)
and ${A^2} = \left[ {\begin{array}{llllllllllllllllllll}0&1&1\\1&0&1\\1&1&0\end{array}} \right] \cdot \left[ {\begin{array}{llllllllllllllllllll}0&1&1\\1&0&1\\1&1&0\end{array}} \right] = \left[ {\begin{array}{llllllllllllllllllll}2&1&1\\1&2&1\\1&1&2\end{array}} \right]$

……..(ii)
$\therefore$ $\frac{{{A^2} - 3I}}{2} = \frac{1}{2}\left\{ {\left| {\begin{array}{llllllllllllllllllll}2&1&1\\1&2&1\\1&1&2\end{array}} \right| - \left| {\begin{array}{llllllllllllllllllll}3&0&0\\0&3&0\\0&0&3\end{array}} \right|} \right\}$

$= \frac{1}{2}\left| {\begin{array}{cccccccccccccccccccc}{ - 1}&1&1\\1&{ - 1}&1\\1&1&{ - 1}\end{array}} \right|$
$= {A^{ - 1}}$

[using Eq. (i)]