#### Please Solve RD Sharma Class 12 Chapter 6 Adjoint and Inverese of Matrices Exercise 6.1 Question 5 Maths Textbook Solution.

$\operatorname{Adj}(A)=3 A^{T}$

Hint:

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

Given:

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

Solution:

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

Let’s find cofactors  $C_{ij}=\left ( -1 \right )^{i+j}$

\begin{aligned} &C_{11}=+(1-4)=-3 \\ &C_{12}=-(2+4)=-6 \\ &C_{13}=+(-4-2)=-6 \\ &C_{21}=-(-2-4)=6 \\ &C_{22}=+(-1+4)=3 \\ &C_{23}=-(2+4)=-6 \\ &C_{31}=+(4+2)=6 \\ &C_{32}=-(2+4)=-6 \\ &C_{33}=+(-1+4)=3 \end{aligned}

$C_{i j}=\left[\begin{array}{ccc} -3 & -6 & -6 \\ 6 & 3 & -6 \\ 6 & -6 & 3 \end{array}\right]$

So,

$Adj\left ( A \right )$ = Transpose of  $C_{ij}$

$\operatorname{Adj}(A)=\left[\begin{array}{ccc} -3 & 6 & 6 \\ -6 & 3 & -6 \\ -6 & -6 & 3 \end{array}\right]$                    (1)

$3 A^{T}=3 \times\left[\begin{array}{ccc} -1 & 2 & 2 \\ -2 & 1 & -2 \\ -2 & -2 & 1 \end{array}\right]=\left[\begin{array}{ccc} -3 & 6 & 6 \\ -6 & 3 & -6 \\ -6 & -6 & 3 \end{array}\right]$                 (2)

Here, from equation (1) and (2)

Clearly see that,

$3 A^{T}=A d j(A)$

Hence, proved