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

$A^{-1}=\left[\begin{array}{ccc} -2 & 1 & 1 \\ \frac{11}{4} & \frac{-1}{2} & \frac{-3}{4} \\ -1 & 0 & 0 \end{array}\right]$

Hint:

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

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

Given:

$A=\left[\begin{array}{ccc} 0 & 0 & -1 \\ 3 & 4 & 5 \\ -2 & -4 & -7 \end{array}\right]$

Solution:

\begin{aligned} &|A|=0\left|\begin{array}{cc} 4 & 5 \\ -4 & -7 \end{array}\right|-0\left|\begin{array}{cc} 3 & 5 \\ -2 & -7 \end{array}\right|-1\left|\begin{array}{cc} 3 & 4 \\ -2 & -4 \end{array}\right| \\ &=0-0-1(-12+8) \\ &=4 \end{aligned}

Hence $A^{-1}$  exist

Cofactor of A are

\begin{aligned} &C_{11}=-8, C_{21}=4, C_{31}=4 \\ &C_{12}=11, C_{22}=-2, C_{32}=-3 \\ &C_{13}=-4, C_{23}=0, C_{33}=0 \end{aligned}

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

\begin{aligned} A^{-1} &=\frac{1}{4} \times\left[\begin{array}{ccc} -8 & 4 & 4 \\ 11 & -2 & -3 \\ -4 & 0 & 0 \end{array}\right] \\ A^{-1} &=\left[\begin{array}{ccc} -2 & 1 & 1 \\ \frac{11}{4} & \frac{-1}{2} & \frac{-3}{4} \\ -1 & 0 & 0 \end{array}\right] \end{aligned}