#### Please Solve RD Sharma Class 12 Chapter 6 Adjoint and Inverese of Matrices Exercise 6.1 Question 8 subquestion (iv) Maths Textbook Solution.

$A^{-1}=\left[\begin{array}{ccc} 3 & -1 & 1 \\ -15 & 6 & -5 \\ 5 & -2 & 2 \end{array}\right]$                                                .

Hint:

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

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

Given:

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

Solution:

\begin{aligned} &|A|=2\left|\begin{array}{ll} 1 & 0 \\ 1 & 3 \end{array}\right|-0\left|\begin{array}{ll} 5 & 0 \\ 0 & 3 \end{array}\right|-1\left|\begin{array}{ll} 5 & 1 \\ 0 & 1 \end{array}\right| \\ &=2(3-0)-0-1(5) \\ &=6-5 \\ &|A|=1 \end{aligned}

Hence $A^{-1}$  exist

Cofactor of A are

\begin{aligned} &C_{11}=3, C_{12}=-15, C_{13}=5 \\ &C_{21}=-1, C_{22}=6, C_{23}=-2 \\ &C_{31}=1, C_{32}=-5, C_{33}=2 \end{aligned}

\begin{aligned} &\operatorname{Adj}(A)=C_{{ij}}^{T} \\ &\operatorname{Adj}(A)=\left[\begin{array}{ccc} 3 & -15 & 5 \\ -1 & 6 & -2 \\ 1 & -5 & 2 \end{array}\right]^{T}=\left[\begin{array}{ccc} 3 & -1 & 1 \\ -15 & 6 & -5 \\ 5 & -2 & 2 \end{array}\right] \end{aligned}

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