#### Need solution for RD Sharma Maths Class 12 Chapter 6 Adjoint and Inverse of Matrices Excercise 6.1 Question 8 Subquestion (ii)

$A^{-1}=\frac{1}{27}\left[\begin{array}{ccc} 4 & 17 & 3 \\ -1 & -11 & 6 \\ 5 & 1 & -3 \end{array}\right]$

Hint:

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

Given:

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

Solution:

We know that

$A^{-1}=\frac{1}{|A|} \times A d j(A)$

So let’s find $\left | A \right |$

\begin{aligned} &|A|=1\left|\begin{array}{ll} -1 & -1 \\ 3 & -1 \end{array}\right|-2\left|\begin{array}{ll} 1 & -1 \\ 2 & -1 \end{array}\right|+5\left|\begin{array}{cc} 1 & -1 \\ 2 & 3 \end{array}\right| \\ &=1(1+3)-2(-1+2)+5(3+2) \\ &=4-2+25 \\ &|A|=27 \end{aligned}

Hence $A^{-1}$  exist

Cofactor of A are

\begin{aligned} &C_{11}=4, C_{21}=17, C_{31}=3 \\ &C_{12}=-1, C_{22}=-11, C_{32}=6 \\ &C_{13}=5, C_{23}=1, C_{33}=-3 \end{aligned}

\begin{aligned} &\text { So } \operatorname{Adj}(A)=C_{{ij}}^{T} \\ &\qquad \operatorname{Adj}(A)=\left[\begin{array}{ccc} 4 & -1 & 5 \\ 17 & -11 & 1 \\ 3 & 6 & -3 \end{array}\right]^{T}=\left[\begin{array}{ccc} 4 & 17 & 3 \\ -1 & -11 & 6 \\ 5 & 1 & -3 \end{array}\right] \end{aligned}

\begin{aligned} A^{-1} &=\frac{1}{|A|} \times \operatorname{Adj}(A) \\ A^{-1} &=\frac{1}{27}\left[\begin{array}{ccc} 4 & 17 & 3 \\ -1 & -11 & 6 \\ 5 & 1 & -3 \end{array}\right] \end{aligned}