Find the adjoint of each of the following matrices: (i) $\left[\begin{array}{ll}\cos \alpha & \sin \alpha \\ \sin \alpha & \cos \alpha\end{array}\right]$ (ii)$\left[\begin{array}{cc}1 & \tan \frac{\alpha}{2} \\ -\tan \frac{\alpha}{2} & 1\end{array}\right]$ Verify that $(\operatorname{adj} A) A=|A| I=A(\operatorname{adj} A)$ for the above matrices.

Find the adjoint of each of the following matrices:
(i) $\left[\begin{array}{ll}\cos \alpha & \sin \alpha \\ \sin \alpha & \cos \alpha\end{array}\right]$
(ii) $\left[\begin{array}{cc}1 & \tan \frac{\alpha}{2} \\ -\tan \frac{\alpha}{2} & 1\end{array}\right]$ Verify that $(\operatorname{adj} A) A=|A| I=A(\operatorname{adj} A)$ for the above matrices.

Solution:

(i) Suppose

$A=\left[\begin{array}{ll}\cos \alpha & \sin \alpha \\ \sin \alpha & \cos \alpha\end{array}\right]$

Therefore cofactors of $A$ are

$C_{11}=\cos \alpha$

$C_{12}=-\sin \alpha$

$C_{21}=-\sin \alpha$

$C_{22}=\cos \alpha$

It is known that, $\operatorname{adj} A=\left[\begin{array}{ll}C_{11} & C_{12} \\ C_{21} & C_{22}\end{array}\right]^{T}$

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

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

Now, $\left(\operatorname{adj}\right.$ A) $A=\left[\begin{array}{cc}\cos \alpha & -\sin \alpha \\ -\sin \alpha & \cos \alpha\end{array}\right]\left[\begin{array}{cc}\cos \alpha & \sin \alpha \\ \sin \alpha & \cos \alpha\end{array}\right]$

$=\left[\begin{array}{cc}-\sin ^{2} \alpha+\cos ^{2} \alpha & \cos \alpha \cdot \sin \alpha-\sin \alpha \cdot \cos \alpha \\ -\cos \alpha \sin \alpha+\sin \alpha \cos \alpha & -\sin ^{2} \alpha+\cos ^{2} \alpha\end{array}\right]$

$\left(\right.$ adj A) $A=\left[\begin{array}{cc}\cos 2 \alpha & 0 \\ 0 & \cos 2 \alpha\end{array}\right]$

And, $|\mathrm{A}| \mathrm{I}=\left|\begin{array}{ll}\cos \alpha & \sin \alpha \\ \sin \alpha & \cos \alpha\end{array}\right|\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$

$=\left(\cos ^{2} \alpha-\sin ^{2} \alpha\right)\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$

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

Adjoint and Inverse of a Matrix

$=\left[\begin{array}{cc}\cos 2 \alpha & 0 \\ 0 & \cos 2 \alpha\end{array}\right]$

Also, $A($ adj A)

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

$=\left[\begin{array}{cc}\cos 2 \alpha & 0 \\ 0 & \cos 2 \alpha\end{array}\right]$

As a result, $(\operatorname{adj} A) A=|A| l=A(\operatorname{adj} A)$

(ii) Suppose

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

Therefore cofactors of $A$ are

$\mathrm{C}_{11}=1$

$C_{12}=\tan \alpha / 2$

$C_{21}=-\tan \alpha / 2$

$C_{22}=1$

It is known that, $\operatorname{adj} \mathrm{A}=\left[\begin{array}{ll}\mathrm{C}_{11} & \mathrm{C}_{12} \\ \mathrm{C}_{21} & \mathrm{C}_{22}\end{array}\right]^{\mathrm{T}}$

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

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

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

Adjoint and Inverse of a Matrix

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

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

And, $|A| .|=| \begin{array}{cc}1 & \tan \frac{\alpha}{2} \\ -\tan \frac{\alpha}{2} & 1\end{array} \mid\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]=\left(1+\tan ^{2} \frac{\alpha}{2}\right)\left[\begin{array}{cc}1 & 0 \\ 0 & 1\end{array}\right]$

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

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

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

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

As a result, $(\operatorname{adj} A) A=|A| l=A(\operatorname{adj} A)$

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