#### Need solution for RD Sharma Maths Class 12 Chapter 6 Adjoint and Inverse of Matrices Excercise 6.1 Question 15

$\left[\begin{array}{ccc} -2 & 19 & -27 \\ -2 & 18 & -25 \\ -3 & 29 & -42 \end{array}\right]$

Hint:

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

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

Given:

$A=\left[\begin{array}{lll} 5 & 0 & 4 \\ 2 & 3 & 2 \\ 1 & 2 & 1 \end{array}\right], B^{-1}=\left[\begin{array}{lll} 1 & 3 & 3 \\ 1 & 4 & 3 \\ 1 & 3 & 4 \end{array}\right]$

Solution:

For $(A B)^{-1}$  we know that

$(A B)^{-1}=B^{-1} A^{-1}$

Here $B^{-1}$  is also given so

Let’s fin $A^{-1}$

$|A|=-5+4=-1$

Cofactor of A are

\begin{aligned} &C_{11}=-1, C_{21}=8, C_{31}=-12 \\ &C_{12}=0, C_{22}=1, C_{32}=-2 \\ &C_{13}=1, C_{23}=-10, C_{33}=15 \end{aligned}

\begin{aligned} &\operatorname{Adj}(A)=\left[\begin{array}{ccc} -1 & 0 & 1 \\ 8 & 1 & -10 \\ -12 & -2 & 15 \end{array}\right]^{T}=\left[\begin{array}{ccc} -1 & 8 & -12 \\ 0 & 1 & -2 \\ 1 & -10 & 15 \end{array}\right] \\\\ &A^{-1}=-1\left[\begin{array}{ccc} -1 & 8 & -12 \\ 0 & 1 & -2 \\ 1 & -10 & 15 \end{array}\right]=\left[\begin{array}{ccc} 1 & -8 & 12 \\ 0 & -1 & 2 \\ -1 & 10 & -15 \end{array}\right] \end{aligned}

\begin{aligned} &(A B)^{-1}=B^{-1} A^{-1} \\ &=\left[\begin{array}{ccc} 1 & 3 & 3 \\ 1 & 4 & 3 \\ 1 & 3 & 4 \end{array}\right]\left[\begin{array}{ccc} 1 & -8 & 12 \\ 0 & -1 & 2 \\ -1 & 10 & -15 \end{array}\right]=\left[\begin{array}{ccc} 1+0-3 & -8-3+30 & 12+6-45 \\ 1+0-3 & -8-4+30 & 12+8-45 \\ 1+0-4 & -8-3+40 & 12+6-60 \end{array}\right] \end{aligned}

$(A B)^{-1}=\left[\begin{array}{ccc} -2 & 19 & -27 \\ -2 & 18 & -25 \\ -3 & 29 & -42 \end{array}\right]$