Question

Find the adjoint of each of the following matrices:

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

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

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

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

Verify that (adj A) A = |A| I = A (adj A) for the above matrices.

Answer 1

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

s.$$\begin{gathered} \left(i\right)A=\left[\begin{array}{cc}-3& 5\\ 2& 4\end{array}\right] \\ \mathrm{adj}A=\left[\begin{array}{cc}4& -5\\ -2& -3\end{array}\right] \\ \left(\mathrm{adj}A\right)A=\left[\begin{array}{cc}-22& 0\\ 0& -22\end{array}\right] \\ \left|A\right|=-22 \\ \left|A\right|I=\left[\begin{array}{cc}-22& 0\\ 0& -22\end{array}\right] \\ A\left(\mathrm{adj}A\right)=\left[\begin{array}{cc}-22& 0\\ 0& -22\end{array}\right] \\ \therefore \left(\mathrm{adj}A\right)A=\left|A\right|I=A\left(\mathrm{adj}A\right) \\ \mathrm{Hence}\mathrm{verified}. \\ \left(ii\right)B=\left[\begin{array}{cc}a& b\\ c& d\end{array}\right] \\ \mathrm{adj}B=\left[\begin{array}{cc}d& -b\\ -c& a\end{array}\right] \\ \left(\mathrm{adj}B\right)B=\left[\begin{array}{cc}ad-bc& 0\\ 0& -cb+ad\end{array}\right] \\ \left|B\right|=ad-bc \\ \left|B\right|I=\left[\begin{array}{cc}ad-bc& 0\\ 0& -cb+ad\end{array}\right] \\ B\left(\mathrm{adj}B\right)=\left[\begin{array}{cc}ad-bc& 0\\ 0& -cb+ad\end{array}\right] \\ \therefore \left(\mathrm{adj}B\right)B=\left|B\right|I=B\left(\mathrm{adj}B\right) \\ \mathrm{Hence}\mathrm{verified}. \\ \left(iii\right)C=\left[\begin{array}{cc}\cos \alpha & \sin \alpha \\ \sin \alpha & \cos \alpha \end{array}\right] \\ \mathrm{adj}C=\left[\begin{array}{cc}\cos \alpha & -\sin \alpha \\ -\sin \alpha & \cos \alpha \end{array}\right] \\ \left(\mathrm{adj}C\right)C=\left[\begin{array}{cc}{\cos }^2\alpha -{\sin }^2\alpha & 0\\ 0& {\cos }^2\alpha -{\sin }^2\alpha \end{array}\right] \\ \left|C\right|={\cos }^2\alpha -{\sin }^2\alpha \\ \left|C\right|I=\left[\begin{array}{cc}{\cos }^2\alpha -{\sin }^2\alpha & 0\\ 0& {\cos }^2\alpha -{\sin }^2\alpha \end{array}\right] \\ C\left(\mathrm{adj}C\right)=\left[\begin{array}{cc}{\cos }^2\alpha -{\sin }^2\alpha & 0\\ 0& {\cos }^2\alpha -{\sin }^2\alpha \end{array}\right] \\ \therefore \left(\mathrm{adj}C\right)C=\left|C\right|I=C\left(\mathrm{adj}C\right) \\ \mathrm{Hence}\mathrm{verified}. \\ \left(iv\right)D=\left[\begin{array}{cc}1& \tan \frac{\alpha }{2}\\ -\tan \frac{\alpha }{2}& 1\end{array}\right] \\ \mathrm{adj}D=\left[\begin{array}{cc}1& -\tan \frac{\alpha }{2}\\ \tan \frac{\alpha }{2}& 1\end{array}\right] \\ \left(\mathrm{adj}D\right)D=\left[\begin{array}{cc}1+{\tan }^2\frac{\alpha }{2}& 0\\ 0& 1+{\tan }^2\frac{\alpha }{2}\end{array}\right] \\ \left|D\right|=1+{\tan }^2\frac{\alpha }{2} \\ \left|D\right|I=\left[\begin{array}{cc}1+{\tan }^2\frac{\alpha }{2}& 0\\ 0& 1+{\tan }^2\frac{\alpha }{2}\end{array}\right] \\ D\left(\mathrm{adj}D\right)=\left[\begin{array}{cc}1+{\tan }^2\frac{\alpha }{2}& 0\\ 0& 1+{\tan }^2\frac{\alpha }{2}\end{array}\right] \\ \therefore \left(\mathrm{adj}D\right)D=\left|D\right|I=D\left(\mathrm{adj}D\right) \\ \mathrm{Hence}\mathrm{verified}. \end{gathered}$$