Browse Questions

# Find $A^{-1}$ if $A=\begin{bmatrix}0 & 1 & 1\\1 & 0 & 1\\1 & 1 & 0\end{bmatrix}$ and show that $A^{-1}=\Large \frac{A^2-3I}{2}$

Toolbox:
• If |A|$\neq 0$,then it is a non-singular matrix.
• $A^{-1}=\frac{1}{|A|}adj\;A$
• Identity matrix is $\begin{bmatrix}1 & 0 & 0\\0 & 1 & 0\\0 & 0 & 1\end{bmatrix}$
Let A=$\begin{bmatrix}0 & 1 & 1\\1 & 0 & 1\\1 & 1 &0\end{bmatrix}$
To find $A^{-1}$,let us first find if the given matrix is singular or non-singular.
For this |A|$\neq 0.$
To find the value of |A|,let us expand along $R_1$.
|A|=0(0-1)-1(0-1)+1(1-0)
$\;\;=1+1=2\neq 0$
Hence it is a non-singular matrix.
Next let us find the adjoint of A
To find adj A let us find the cofactors.
$A_{11}=(-1)^{1+1}\begin{vmatrix}0 & 1\\1 & 0\end{vmatrix}=-1.$
$A_{12}=(-1)^{1+2}\begin{vmatrix}1 & 1\\1 & 0\end{vmatrix}=-(-1)=1.$
$A_{13}=(-1)^{1+3}\begin{vmatrix}1 & 0\\1 & 1\end{vmatrix}=1.$
$A_{21}=(-1)^{2+1}\begin{vmatrix}1 & 1\\1 & 0\end{vmatrix}=-(0-1)=1.$
$A_{22}=(-1)^{2+2}\begin{vmatrix}0 & 1\\1 & 0\end{vmatrix}=-1.$
$A_{23}=(-1)^{2+3}\begin{vmatrix}0 & 1\\1 & 1\end{vmatrix}=0-1=-(-1)=1.$
$A_{31}=(-1)^{3+1}\begin{vmatrix}1 & 1\\0 & 1\end{vmatrix}=1-0=1.$
$A_{32}=(-1)^{3+2}\begin{vmatrix}0 & 1\\1 & 1\end{vmatrix}=-(0-1)=-(-1)=1.$
$A_{33}=(-1)^{1+1}\begin{vmatrix}0 & 1\\1 & 0\end{vmatrix}=0-1=-1.$
adj A=$\begin{bmatrix}A_{11} & A_{21} & A_{31}\\A_{12} & A_{22} & A_{32}\\A_{13} & A_{23} & A_{23}\end{bmatrix}$
$\;\;=\begin{bmatrix}-1 & 1 & 1\\1 & -1 & 1\\1 & 1 &-1\end{bmatrix}$
$A^{-1}=\frac{1}{|A|}(adj A)$
We know |A|=2.
Therefore $A^{-1}=\frac{1}{2}\begin{bmatrix}-1 & 1 & 1\\1 & -1 & 1\\1 & 1 & -1\end{bmatrix}$=LHS.
Consider the RHS=$\frac{A^2-3I}{2}=\frac{1}{2}[A^2-3I]$
$A^2=A.A$
$\;\;=\begin{bmatrix}0 & 1 & 1\\1 & 0 & 1\\1 & 1 &0\end{bmatrix}\begin{bmatrix}0 & 1 & 1\\1 & 0 & 1\\1 & 1 &0\end{bmatrix}$
Matrix multiplication can be done by multiplying the rows of matrix A with columns of matrix be as below:
$A^2=\begin{bmatrix}0+1+1 & 0+0+1 & 0+1+0\\0+0+1 & 1+0+1 & 1+0+1\\0+1+0 & 1+0+0 & 1+1+0\end{bmatrix}$
$\;\;=\begin{bmatrix}2 & 1 & 1\\1 & 2 & 1\\1 & 1 & 2\end{bmatrix}$
Now $3I=3\begin{bmatrix}1 & 0 & 0\\0 & 1 & 0\\0 & 0& 1\end{bmatrix}=\begin{bmatrix}3 & 0 & 0\\0 & 3 & 0\\0 & 0 & 3\end{bmatrix}$
Therefore $\frac{1}{2}[A^2-3I]=\frac{1}{2}\begin{bmatrix}2 & 1 &1\\1 & 2 & 1\\1 & 1 & 2\end{bmatrix}-\begin{bmatrix}3 & 0 & 0\\0 & 3 & 0\\0 & 0 &3\end{bmatrix}$
$\frac{1}{2}[A^2-3I]=\frac{1}{2}\begin{bmatrix}-1 & 1 & 1\\1 & -1 & 1\\1 & 1 & -1\end{bmatrix}$
LHS=RHS.
Hence proved.