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

$\operatorname{Adj}(A)=\left[\begin{array}{ccc} -22 & 11 & -11 \\ 4 & -2 & 2 \\ 16 & -8 & 8 \end{array}\right]$

Hint:

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

Given:

$A=\left|\begin{array}{ccc} 2 & -1 & 3 \\ 4 & 2 & 5 \\ 0 & 4 & -1 \end{array}\right|$

Solution:

Let’s find  $|A|$

$|A|=\left|\begin{array}{ccc} 2 & -1 & 3 \\ 4 & 2 & 5 \\ 0 & 4 & -1 \end{array}\right|$

\begin{aligned} &|A|=2\left|\begin{array}{cc} 2 & 5 \\ 4 & -1 \end{array}\right|-(-1)\left|\begin{array}{cc} 4 & 5 \\ 0 & -1 \end{array}\right|+3\left|\begin{array}{ll} 4 & 2 \\ 0 & 4 \end{array}\right| \\ &=2(-2-20)+1(-4-0)+3(16-0) \\ &=-44-4+48 \\ &=0 \\ &|A|=0 \end{aligned}

Let’s find cofactor

\begin{aligned} &C_{11}=+(-2-20)=-22 \\ &C_{12}=-(-4+0)=4 \\ &C_{13}=+(16-0)=16 \\ &C_{21}=-(1-12)=11 \\ &C_{22}=+(-2-0)=-2 \\ &C_{23}=-(-8-0)=-8 \\ &C_{31}=+(-5-6)=-11 \\ &C_{32}=-(10-12)=2 \\ &C_{33}=+(4+4)=8 \end{aligned}

$C_{i j}=\left[\begin{array}{ccc} -22 & 4 & 16 \\ 11 & -2 & -8 \\ -11 & 2 & 8 \end{array}\right]$

Let’s take transpose of $C_{ij}$

$\operatorname{Adj}(A)=\left[\begin{array}{ccc} -22 & 11 & -11 \\ 4 & -2 & 2 \\ 16 & -8 & 8 \end{array}\right]$

Let’s prove this below

\begin{aligned} &\operatorname{Adj}(A) \times A=|A| I=A \times \operatorname{Adj}(A) \\\\ &\operatorname{Adj}(A) \times A=\left[\begin{array}{ccc} -22 & 11 & -11 \\ 4 & -2 & 2 \\ 16 & -8 & 8 \end{array}\right] \times\left[\begin{array}{ccc} 2 & -1 & 3 \\ 4 & 2 & 5 \\ 0 & 4 & -1 \end{array}\right] \end{aligned}

$=\left[\begin{array}{ccc} -44+44+0 & 0 & 0 \\ 0 & -4+(-4)+8 & 0 \\ 0 & 0 & 48-40-8 \end{array}\right]$

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

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

$A \times \operatorname{Adj}(A)=\left[\begin{array}{ccc} 2 & -1 & 3 \\ 4 & 2 & 5 \\ 0 & 4 & -1 \end{array}\right] \times\left[\begin{array}{ccc} -22 & 4 & -11 \\ 4 & -2 & 2 \\ 16 & -8 & 8 \end{array}\right]=\left[\begin{array}{lll} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right]$                (3)

So, from equation (1), (2) and (3)

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