#### Provide Solution for RD Sharma Class 12 Chapter 6 Adjoint and Inverse Matrix Exercise 6.1 Question 27

$A^{-1}=\frac{1}{9}\left[\begin{array}{ccc} -8 & 4 & 1 \\ 1 & 4 & -8 \\ 4 & 7 & 4 \end{array}\right]=A^{T}$

Hint:

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

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

Given:

$A=\frac{1}{9}\left[\begin{array}{ccc} -8 & 1 & 4 \\ 1 & 4 & 7 \\ 1 & -8 & 4 \end{array}\right]$   Find   $A^{T}=A^{-1}$

Solution:

\begin{aligned} &A^{T}=A^{-1} \\ &L H S \\ &A^{T}=\frac{1}{9}\left[\begin{array}{ccc} 8 & 1 & 4 \\ 4 & 4 & 7 \\ 1 & -8 & 4 \end{array}\right]^{T}=\frac{1}{9}\left[\begin{array}{ccc} -8 & 4 & 1 \\ 1 & 4 & -8 \\ 4 & 7 & 4 \end{array}\right] \end{aligned}

\begin{aligned} &R H S \\ &A^{-1}=\frac{1}{|A|} \times \operatorname{Adj}(A) \end{aligned}

Let’s find

\begin{aligned} &|A| \& \operatorname{Adj}(A) \\ &|A|=\frac{1}{9}[-8(16+56)-1(16-7)+4(-32-4)] \\ &|A|=-81 \end{aligned}

Cofactor of A

\begin{aligned} &C_{11}=72, C_{21}=-36, C_{31}=-9 \\ &C_{12}=-9, C_{22}=-36, C_{32}=72 \\ &C_{13}=-36, C_{23}=-63, C_{33}=-36 \\&\operatorname{Adj}(A)=\left[\begin{array}{ccc} 72 & -9 & -36 \\ -36 & -36 & -63 \\ 9 & 72 & -36 \end{array}\right]^{T}=\left[\begin{array}{ccc} 72 & -36 & -9 \\ -9 & -36 & 72 \\ -36 & -63 & -36 \end{array}\right] \end{aligned}

\begin{aligned} &A^{-1}=\frac{1}{|A|} \times A d j(A)=\frac{1}{-9}\left[\begin{array}{ccc} 72 & -36 & -9 \\ -9 & -36 & 72 \\ 36 & -63 & -36 \end{array}\right] \\ &A^{-1}=\frac{1}{9}\left[\begin{array}{ccc} -8 & 4 & 1 \\ 1 & 4 & -8 \\ 4 & 7 & 4 \end{array}\right]=A^{T} \end{aligned}