#### Explain solution RD Sharma class 12 Chapter 6 Adjoint and Inverse of a Matrix exercise 6.1 question 1 Subquestion (iv)

$Adj\left ( A \right )=\left|\begin{array}{cc} 1 & \tan \frac{\alpha}{2} \\ -\tan \frac{\alpha}{2} & 1 \end{array}\right|$

Hint:

Here, we use basic concept of determinant.

Given:

$A=\left|\begin{array}{cc} 1 & \tan \frac{\alpha}{2} \\ -\tan \frac{\alpha}{2} & 1 \end{array}\right|$

Solution:

Let’s find $|A|$
$|A|=\left|\begin{array}{cc} 1 & \tan \frac{\alpha}{2} \\ -\tan \frac{\alpha}{2} & 1 \end{array}\right|$

\begin{aligned} &|A|=1-\left[\left(-\tan \frac{\alpha}{2}\right)\left(\tan \frac{\alpha}{2}\right)\right] \\ &|A|=1+\tan ^{2} \frac{\alpha}{2} \\ &|A|=\sec ^{2} \frac{\alpha}{2} \end{aligned}

Let’s find cofactor

$C_{11}=1, C_{12}=\tan \frac{\alpha}{2}, C_{21}=-\tan \frac{\alpha}{2}, C_{22}=1$

$C_{ij}=\left[\begin{array}{cc} 1 & \tan \frac{\alpha}{2} \\ -\tan \frac{\alpha}{2} & 1 \end{array}\right]$

Take transpose of $C_{ij}$

$\operatorname{Adj}(A)=\left[\begin{array}{cc} 1 & -\tan \frac{\alpha}{2} \\ \tan \frac{\alpha}{2} & 1 \end{array}\right]$

Let’s prove below

\begin{aligned} &\operatorname{Adj}(A) \times A=|A| I=A \times \operatorname{Adj}(A) \\\\ &\operatorname{Adj}(A) \times A=\left[\begin{array}{cc} 1 & -\tan \frac{\alpha}{2} \\ \tan \frac{\alpha}{2} & 1 \end{array}\right] \times\left[\begin{array}{cc} 1 & \tan \frac{\alpha}{2} \\ -\tan \frac{\alpha}{2} & 1 \end{array}\right] \end{aligned}

\begin{aligned} &=\left[\begin{array}{cc} 1+\tan ^{2} \frac{\alpha}{2} & \tan \frac{\alpha}{2}-\tan \frac{\alpha}{2}\\ \\ \tan \frac{\alpha}{2}-\tan \frac{\alpha}{2} & 1+\tan ^{2} \frac{\alpha}{2} \end{array}\right] \\ \\&=\left[\begin{array}{cc} \sec ^{2} \frac{\alpha}{2} & 0 \\ 0 & \sec ^{2} \frac{\alpha}{2} \end{array}\right] \end{aligned}                                                    (1)

\begin{aligned} &|A| I=\sec ^{2} \frac{\alpha}{2} \times\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right] \\\\ &=\left[\begin{array}{cc} \sec ^{2} \frac{\alpha}{2} & 0 \\ 0 & \sec ^{2} \frac{\alpha}{2} \end{array}\right] \end{aligned}                                                                        (2)

$A \times A d j(A)=\left[\begin{array}{cc} 1 & \tan \frac{\alpha}{2} \\ -\tan \frac{\alpha}{2} & 1 \end{array}\right] \times\left[\begin{array}{cc} 1 & -\tan \frac{\alpha}{2} \\ \tan \frac{\alpha}{2} & 1 \end{array}\right]$

\begin{aligned} &=\left[\begin{array}{cc} 1+\tan ^{2} \frac{\alpha}{2} & 0 \\ 0 & 1+\tan ^{2} \frac{\alpha}{2} \end{array}\right] \\\\ &=\left[\begin{array}{cc} \sec ^{2} \frac{\alpha}{2} & 0 \\ 0 & \sec ^{2} \frac{\alpha}{2} \end{array}\right] \end{aligned}                                                                 (3)

From equation (1), (2) and 3

$\operatorname{Adj}(A) \times A=|A| \times I=A \times \operatorname{Adj}(A)$

Hence proved