# Solve this

Question:

If $A=\left[\begin{array}{lll}0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0\end{array}\right]$, find $A^{-1}$ and show that $A^{-1}=\frac{1}{2}\left(A^{2}-3 I\right)$

Solution:

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

$A^{-1}=\frac{1}{|A|}$ Adj. $A$

Now,

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

$=-1(-1)+1(1)$

$=2$

Now, to find Adj. $A$

$A_{11}=(-1)^{1+1}(-1)=-1 \quad A_{21}=(-1)^{2+1}(-1)=1 \quad A_{31}=(-1)^{3+1}(1)=1$

$A_{12}=(-1)^{1+2}(-1)=1 \quad A_{22}=(-1)^{2+2}(-1)=-1 \quad A_{32}=(-1)^{3+2}(-1)=1$

$A_{13}=(-1)^{1+3}(1)=1 \quad A_{23}=(-1)^{2+3}(-1)=1 \quad A_{33}=(-1)^{3+3}(-1)=-1$

Therefore,

Adj. $A=\left[\begin{array}{ccc}-1 & 1 & 1 \\ 1 & -1 & 1 \\ 1 & 1 & -1\end{array}\right]$

Thus,

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

Now,

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

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

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

Now, to show $A^{-1}=\frac{1}{2}\left(A^{2}-3 I\right)$

RHS

$=\frac{1}{2}\left(A^{2}-3 I\right)$

$=\frac{1}{2}\left(\left[\begin{array}{lll}2 & 1 & 1 \\ 1 & 2 & 1 \\ 1 & 1 & 2\end{array}\right]-3\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]\right)$

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

$=A^{-1}$

$=\mathrm{LHS}$

Hence, $A^{-1}=\frac{1}{2}\left(A^{2}-3 I\right)$.