Find A–1 if and show that A-1 = (A2 – 3I)/ 2.

Solution:

Given,

$A=\left[\begin{array}{lll}0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0\end{array}\right]$

Co-factors are:

$A_{11}=-1, A_{12}=1, A_{13}=1$

$A_{21}=1, A_{22}=-1, A_{23}=1$

$A_{31}=1, A_{31}=1, A_{32}=1 A_{33}=-1$

Now, $\operatorname{adj} A=\left[\begin{array}{ccc}-1 & 1 & 1 \\ 1 & -1 & 1 \\ 1 & 1 & -1\end{array}\right]^{T}=\left[\begin{array}{ccc}-1 & 1 & 1 \\ 1 & -1 & 1 \\ 1 & 1 & -1\end{array}\right]$

$|A|=0-1(-1)+1.1=2$

Thus, $A^{-1}=\frac{\operatorname{adj} A}{|A|}=\frac{1}{2}\left[\begin{array}{ccc}-1 & 1 & 1 \\ 1 & -1 & 1 \\ 1 & 1 & -1\end{array}\right]$

Now, 

$A^{2}=\left[\begin{array}{lll}0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0\end{array}\right] \cdot\left[\begin{array}{lll}0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0\end{array}\right]=\left[\begin{array}{lll}2 & 1 & 1 \\ 1 & 2 & 1 \\ 1 & 1 & 2\end{array}\right]$

Hence,

$\frac{A^{2}-3 I}{2}=\frac{1}{2}\left\{\left[\begin{array}{lll}2 & 1 & 1^{\circ} \\ 1 & 2 & 1 \\ 1 & 1 & 2\end{array}\right]-\left[\begin{array}{lll}3 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 3\end{array}\right]\right\}$

$=\frac{1}{2}\left[\begin{array}{ccc}-1 & 1 & 1 \\ 1 & -1 & 1 \\ 1 & 1 & -1\end{array}\right]=A^{-1}$

Hence proved.