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#### Need solution for RD Sharma Maths Class 12 Chapter 6 Adjoint and Inverse of Matrices Excercise 6.1 Question 8 Subquestion (iii)

$A^{-1}=\frac{1}{4}\left[\begin{array}{ccc} 3 & 1 & -1 \\ 1 & 3 & 1 \\ -1 & 1 & 3 \end{array}\right]$

Hint:

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

Given:

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

Solution:

\begin{aligned} &A^{-1}=\frac{1}{|A|} \times \operatorname{Adj}(A) \\ &|A|=2\left|\begin{array}{cc} 2 & -1 \\ -1 & 2 \end{array}\right|-(-1)\left|\begin{array}{cc} -1 & -1 \\ 1 & 2 \end{array}\right|+1\left|\begin{array}{cc} -1 & 2 \\ 1 & -1 \end{array}\right| \\ &=2(4-1)+1(-2+1)+1(1-2) \\ &=6-2 \\ &=4 \neq 0 \end{aligned}

Hence $A^{-1}$  exist

Cofactor of A

\begin{aligned} &C_{11}=3, C_{21}=1, C_{31}=-1 \\ &C_{12}=3, C_{22}=3, C_{32}=1 \\ &C_{13}=-1, C_{23}=1, C_{33}=3 \end{aligned}

\begin{aligned} &\operatorname{Adj}(A)=C_{i j}^{T} \\ &\operatorname{Adj}(A)=\left[\begin{array}{ccc} 3 & 1 & -1 \\ 1 & 3 & 1 \\ -1 & 1 & 3 \end{array}\right]^{T}=\left[\begin{array}{ccc} 3 & 1 & -1 \\ 1 & 3 & 1 \\ -1 & 1 & 3 \end{array}\right] \end{aligned}

\begin{aligned} &A^{-1}=\frac{1}{|A|} \times \operatorname{Adj}(A) \\ &=\frac{1}{4}\left[\begin{array}{ccc} 3 & 1 & -1 \\ 1 & 3 & 1 \\ -1 & 1 & 3 \end{array}\right] \end{aligned}