#### Need solution for RD Sharma Maths Class 12 Chapter 6 Adjoint and Inverse of Matrices Excercise 6.1 Question 29

$A^{-1}= A^{2}$

Hint:

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

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

Given:

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

Solution:

\begin{aligned} &|A|=-1\left|\begin{array}{ll} 1 & 1 \\ 1 & 0 \end{array}\right|-2\left|\begin{array}{cc} -1 & 1 \\ 0 & 0 \end{array}\right|+0 \\ &|A|=-1(-1)-0+0= 1 \\ &|A|=1 \end{aligned}

Cofactor of A

\begin{aligned} &C_{11}=-1, C_{21}=0, C_{31}=2 \\ &C_{12}=0, C_{22}=0, C_{32}=1 \\ &C_{13}=-1, C_{23}=1, C_{33}=1 \\ &\operatorname{Adj}(A)=\left[\begin{array}{ccc} -1 & 0 & -1 \\ 0 & 0 & 1 \\ 2 & 1 & 1 \end{array}\right]^{T}=\left[\begin{array}{ccc} -1 & 0 & 2 \\ 0 & 0 & 1 \\ -1 & 1 & 1 \end{array}\right] \end{aligned}

\begin{aligned} &A^{-1}=\frac{1}{|A|} \times \operatorname{Adj}(A) \\ &=\frac{1}{1}\left[\begin{array}{ccc} -1 & 0 & 2 \\ 0 & 0 & 1 \\ -1 & 1 & 1 \end{array}\right]=\left[\begin{array}{ccc} -1 & 0 & 2 \\ 0 & 0 & 1 \\ -1 & 1 & 1 \end{array}\right] \end{aligned}

\begin{aligned} &A^{2}=A \times A \\ &=\left[\begin{array}{ccc} -1 & 2 & 0 \\ -1 & 1 & 1 \\ 0 & 1 & 0 \end{array}\right]\left[\begin{array}{ccc} -1 & 2 & 0 \\ -1 & 1 & 1 \\ 0 & 1 & 0 \end{array}\right]=\left[\begin{array}{ccc} -1-2+0 & -2+2+0 & 0+2+0 \\ 1-1+1 & -2+1+1 & -1+1-0 \\ 0-1+0 & 0+1-0 & 0+1-0 \end{array}\right] \\ &=\left[\begin{array}{ccc} -1 & 0 & 2 \\ 0 & 0 & 1 \\ -1 & 1 & 1 \end{array}\right] \end{aligned}

Hence  $A^{-1}= A^{2}$