#### Need solution for RD Sharma Maths Class 12 Chapter 6 Adjoint and Inverse of Matrices Excercise 6.1 Question 2 Subquestion (iv).

$\operatorname{Adj}(A)=\left[\begin{array}{ccc} 3 & -1 & 1 \\ -15 & 7 & -5 \\ 4 & -2 & 2 \end{array}\right]$

Hint:

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

Given:

$A=\left[\begin{array}{ccc} 2 & 0 & -1 \\ 5 & 1 & 0 \\ 1 & 1 & 3 \end{array}\right]$

Solution:

Let’s find   $\left | A \right |$

\begin{aligned} &|A|=\left|\begin{array}{ccc} 2 & 0 & -1 \\ 5 & 1 & 0 \\ 1 & 1 & 3 \end{array}\right| \\ &=2\left|\begin{array}{ll} 1 & 0 \\ 1 & 3 \end{array}\right|-0\left|\begin{array}{ll} 5 & 0 \\ 1 & 3 \end{array}\right|+(-1)\left|\begin{array}{ll} 5 & 1 \\ 1 & 1 \end{array}\right| \\ &=2(3)-0-1(5-1) \\ &=6-4=2 \\ &|A|=2 \end{aligned}
Let’s find cofactor

\begin{aligned} &C_{11}=+(3-0)=3 \\ &C_{12}=-(15-0)=-15 \\ &C_{13}=+(5-1)=4 \\ &C_{21}=-(0+1)=-1 \\ &C_{22}=+(6+1)=7 \\ &C_{23}=-(-2-0)=-2 \\ &C_{31}=+(2-1)=1 \\ &C_{32}=-(0+5)=-5 \\ &C_{33}=+(2-0)=2 \end{aligned}

$C_{i j}=\left[\begin{array}{ccc} 3 & -15 & 4 \\ -1 & 7 & -2 \\ 1 & -5 & 2 \end{array}\right]$

Let’s take the transpose of  $C_{ij}$

$\operatorname{Adj}(A)=\left[\begin{array}{ccc} 3 & -1 & 1 \\ -15 & 7 & -5 \\ 4 & -2 & 2 \end{array}\right]$

Let’s prove,

\begin{aligned} &\operatorname{Adj}(A) \times A=|A| I=A \times \operatorname{Adj}(A) \\\\ &\operatorname{Adj}(A) \times A=\left[\begin{array}{ccc} 3 & -1 & 1 \\ -15 & 7 & -5 \\ 4 & -2 & 2 \end{array}\right] \times\left[\begin{array}{ccc} 2 & 0 & -1 \\ 5 & 1 & 0 \\ 1 & 1 & 3 \end{array}\right]=\left[\begin{array}{ccc} 6+(-5)+1 & 0 & 0 \\ 0 & 0+7-5 & 0 \\ 0 & 0 & -4+0+6 \end{array}\right] \end{aligned}

$=\left[\begin{array}{lll} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{array}\right]$                                                   (1)

$|A| I=2 \times\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]=\left[\begin{array}{lll} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{array}\right]$                       (2)

$A \times \operatorname{Adj}(A)=\left[\begin{array}{ccc} 2 & 0 & -1 \\ 5 & 1 & 0 \\ 1 & 1 & 3 \end{array}\right] \times\left[\begin{array}{ccc} 3 & -1 & 1 \\ -15 & 7 & -5 \\ 4 & -2 & 2 \end{array}\right]=\left[\begin{array}{ccc} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{array}\right]$                     (3)

From the equation (1), (2) and (3)

$\operatorname{Adj}(A) \times A=|A| I=A \times \operatorname{Adj}(A)$