Find the adjoint of each of the following matrices:

Solution:

Given below are the squares matrices. Here, we will interchange the diagonal elements and change the signs of
the off-diagonal elements.

s. (i) $A=\left[\begin{array}{cc}-3 & 5 \\ 2 & 4\end{array}\right]$

$\operatorname{adj} A=\left[\begin{array}{cc}4 & -5 \\ -2 & -3\end{array}\right]$

$(\operatorname{adj} A) A=\left[\begin{array}{cc}-22 & 0 \\ 0 & -22\end{array}\right]$

$|A|=-22$

$|A| I=\left[\begin{array}{cc}-22 & 0 \\ 0 & -22\end{array}\right]$

$A(\operatorname{adj} A)=\left[\begin{array}{cc}-22 & 0 \\ 0 & -22\end{array}\right]$

$\therefore(\operatorname{adj} A) A=|A| I=A(\operatorname{adj} A)$

Hence verified.

(ii) $B=\left[\begin{array}{ll}a & b \\ c & d\end{array}\right]$

$\operatorname{adj} B=\left[\begin{array}{cc}d & -b \\ -c & a\end{array}\right]$

$(\operatorname{adj} B) B=\left[\begin{array}{cc}a d-b c & 0 \\ 0 & -c b+a d\end{array}\right]$

$|B|=a d-b c$

$|B| I=\left[\begin{array}{cc}a d-b c & 0 \\ 0 & -c b+a d\end{array}\right]$

$B(\operatorname{adj} B)=\left[\begin{array}{cc}a d-b c & 0 \\ 0 & -c b+a d\end{array}\right]$

Hence verified.

(iii) $C=\left[\begin{array}{ll}\cos \alpha & \sin \alpha \\ \sin \alpha & \cos \alpha\end{array}\right]$

$\operatorname{adj} C=\left[\begin{array}{cc}\cos \alpha & -\sin \alpha \\ -\sin \alpha & \cos \alpha\end{array}\right]$

$\therefore(\operatorname{adj} B) B=|B| I=B(\operatorname{adj} B)$

$(\operatorname{adj} C) C=\left[\begin{array}{cc}\cos ^{2} \alpha-\sin ^{2} \alpha & 0 \\ 0 & \cos ^{2} \alpha-\sin ^{2} \alpha\end{array}\right]$

$|C|=\cos ^{2} \alpha-\sin ^{2} \alpha$

$|C| I=\left[\begin{array}{cc}\cos ^{2} \alpha-\sin ^{2} \alpha & 0 \\ 0 & \cos ^{2} \alpha-\sin ^{2} \alpha\end{array}\right]$

$C(\operatorname{adj} C)=\left[\begin{array}{cc}\cos ^{2} \alpha-\sin ^{2} \alpha & 0 \\ 0 & \cos ^{2} \alpha-\sin ^{2} \alpha\end{array}\right]$

$\therefore(\operatorname{adj} C) C=|C| I=C(\operatorname{adj} C)$

Hence verified.

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

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

$(\operatorname{adj} D) D=\left[\begin{array}{cc}1+\tan ^{2} \frac{\alpha}{2} & 0 \\ 0 & 1+\tan ^{2} \frac{\alpha}{2}\end{array}\right]$

$|D|=1+\tan ^{2} \frac{\alpha}{2}$

$|D| I=\left[\begin{array}{cc}1+\tan ^{2} \frac{\alpha}{2} & 0 \\ 0 & 1+\tan ^{2} \frac{\alpha}{2}\end{array}\right]$

$D(\operatorname{adj} D)=\left[\begin{array}{cc}1+\tan ^{2} \frac{\alpha}{2} & 0 \\ 0 & 1+\tan ^{2} \frac{\alpha}{2}\end{array}\right]$

$\therefore(\operatorname{adj} D) D=|D| I=D(\operatorname{adj} D)$

Hence verified.