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### Answers (1)

Answer:

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

Hint:

Here, we use basic concept of determinant and inverse of matrix

$A^{-1}= \frac{1}{\left | A \right |}\times Adj\left ( A\right )$

Given:

$A= \begin{bmatrix} 0 &1 &1 \\ 1 & 0 & 1\\ 1& 1 & 0 \end{bmatrix}$

Solution:

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

Cofactor of A

\begin{aligned} &C_{11}=-1, C_{21}=1, C_{31}=1 \\ &C_{12}=1, C_{22}=-1, C_{32}=1 \\ &C_{13}=1, C_{23}=1, C_{33}=-1 \end{aligned}

\begin{aligned} &\operatorname{Adj}(A)=\left[\begin{array}{ccc} -1 & 1 & 1 \\ 1 & -1 & 1 \\ 1 & 1 & -1 \end{array}\right] \\ & \end{aligned}

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

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

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

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

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