#### Provide Solution for RD Sharma Class 12 Chapter 6 Adjoint and Inverse Matrix Exercise 6.1 Question 7 Subquestion (i)

$\operatorname{Adj}(A)=\left[\begin{array}{ccc} 25 & 0 & 0 \\ 0 & 25 & 0 \\ 0 & 0 & 25 \end{array}\right]$

Hint:

Here, we use basic concept of determinant.

Given:

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

Solution:

Let’s find cofactor of A

\begin{aligned} &C_{11}=9, C_{21}=19, C_{31}=-4 \\ &C_{12}=4, C_{22}=14, C_{32}=1 \\ &C_{13}=8, C_{23}=3, C_{33}=2 \end{aligned}

$C_{i j}=\left[\begin{array}{ccc} 9 & 4 & 8 \\ 19 & 14 & 3 \\ -4 & 1 & 2 \end{array}\right]$

$Adj(A)$  = Transpose of  $C_{ij}$

\begin{aligned} &\operatorname{Adj}(A)=\left[\begin{array}{ccc} 9 & 19 & -4 \\ 4 & 14 & 1 \\ 8 & 3 & 2 \end{array}\right] \\ &A \times \operatorname{Adj}(A)=\left[\begin{array}{ccc} 1 & -2 & 3 \\ 0 & 2 & -1 \\ -4 & 5 & 2 \end{array}\right] \times\left[\begin{array}{ccc} 9 & 19 & -4 \\ 4 & 14 & 1 \\ 8 & 3 & 2 \end{array}\right] \end{aligned}

$=\left[\begin{array}{ccc} 9-8+24 & 19-28-9 & -4-2+6 \\ 0+8-8 & 0+28-3 & 0+2-2 \\ -36+20+16 & -76+70+6 & 16+5+4 \end{array}\right]=\left[\begin{array}{ccc} 25 & 0 & 0 \\ 0 & 25 & 0 \\ 0 & 0 & 25 \end{array}\right]$

Hence, $A \times \operatorname{Adj}(A)=25 I$

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$A^{-1}=\left[\begin{array}{cc} \cos \theta &- \sin \theta \\ \sin \theta & \cos \theta \end{array}\right]$

Hint:

Here, we use basic concept of inverse

Given:

$A=\left[\begin{array}{cc} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{array}\right]$

Solution:

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

Let’s find  $\left | A \right |$

$|A|=\cos ^{2} \theta+\sin ^{2} \theta=1$

So,

$A^{-1}=\left[\begin{array}{cc} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{array}\right]$