#### Provide Solution for RD Sharma Class 12 Chapter 6 Adjoint and Inverse Matrix Exercise 6.1 Question 37

$\left[\begin{array}{ccc} -9 & -8 & -2 \\ 8 & 7 & 2 \\ -5 & -4 & -1 \end{array}\right]$

Hint:

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

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

Given:

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

Solution:

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

Let’s find  $\left | A^{T} \right |$

\begin{aligned} &\left|A^{T}\right|=(-1-8)-0-2(-8+3) \\ &=-9+10=1 \end{aligned}

Cofactor of $A^{T}$

\begin{aligned} &C_{11}=-9, C_{12}=8, C_{13}=-5 \\ &C_{21}=-8, C_{22}=7, C_{23}=-4 \\ &C_{31}=-2, C_{32}=2, C_{33}=1 \end{aligned}

$\operatorname{Adj}\left(A^{T}\right)=\left[\begin{array}{ccc} -9 & -8 & -2 \\ 8 & 7 & 2 \\ -5 & -4 & -1 \end{array}\right]$

$\left(A^{T}\right)^{-1}=\frac{1}{1}\left[\begin{array}{ccc} -9 & -8 & -2 \\ 8 & 7 & 2 \\ -5 & -4 & -1 \end{array}\right]$

$\left(A^{T}\right)^{-1}=\left[\begin{array}{ccc} -9 & -8 & -2 \\ 8 & 7 & 2 \\ -5 & -4 & -1 \end{array}\right]$

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