#### Explain solution RD Sharma class 12 Chapter 6 Adjoint and Inverse of a Matrix exercise 6.1 question 9 Subquestion (ii)

$A^{-1}=\frac{1}{2}\left[\begin{array}{ccc} 1 & 1 & -1 \\ -3 & 1 & 1 \\ 9 & -5 & -1 \end{array}\right]$

Hint:

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

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

Given:

$A=\left[\begin{array}{lll} 2 & 3 & 1 \\ 3 & 4 & 1 \\ 3 & 7 & 2 \end{array}\right]$

Solution:

\begin{aligned} &|A|=2\left|\begin{array}{ll} 4 & 1 \\ 7 & 2 \end{array}\right|-3\left|\begin{array}{ll} 3 & 1 \\ 3 & 2 \end{array}\right|+1\left|\begin{array}{ll} 3 & 4 \\ 3 & 7 \end{array}\right| \\ &=2(8-7)-3(6-3)+1(21-12) \\ &=2-9+9 \\ &=2 \end{aligned}

Hence $A^{-1}$  exist

Cofactor of A

\begin{aligned} &C_{11}=1, C_{21}=1, C_{31}=-1 \\ &C_{12}=-3, C_{22}=1, C_{32}=1 \\ &C_{13}=9, C_{23}=-5, C_{33}=-1 \end{aligned}

\begin{aligned} &\operatorname{Adj}(A)=C_{i j}^{T} \\ &\operatorname{Adj}(A)=\left[\begin{array}{ccc} 1 & -3 & 9 \\ 1 & 1 & -5 \\ -1 & 1 & -1 \end{array}\right]^{T}=\left[\begin{array}{ccc} 1 & 1 & -1 \\ -3 & 1 & 1 \\ 9 & -5 & -1 \end{array}\right] \end{aligned}

\begin{aligned} A^{-1} &=\frac{1}{|A|} \times \operatorname{Adj}(A) \\ A^{-1} &=\frac{1}{2}\left[\begin{array}{ccc} 1 & 1 & -1 \\ -3 & 1 & 1 \\ 9 & -5 & -2 \end{array}\right] \end{aligned}

\begin{aligned} &A^{-1} \times A=\frac{1}{2}\left[\begin{array}{ccc} 1 & 1 & -1 \\ -3 & 1 & 1 \\ 9 & -5 & -1 \end{array}\right]\left[\begin{array}{ccc} 2 & 3 & 1 \\ 3 & 4 & 1 \\ 3 & 7 & 2 \end{array}\right] \\ &=\frac{1}{2}\left[\begin{array}{ccc} 2+3-3 & 3+4-7 & 1+1-2 \\ -6+3+3 & -9+4+7 & -3+1+2 \\ 18-15-3 & 27-20-7 & 9-5-2 \end{array}\right]=\frac{1}{2}\left[\begin{array}{lll} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{array}\right]=\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] \end{aligned}

Hence $A^{^{-1 }}\times A=I$