Solution:
(i) Suppose
![A=\left[\begin{array}{ll}\cos \alpha & \sin \alpha \\ \sin \alpha & \cos \alpha\end{array}\right]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-1334032adbdffac69fff3d7c56cbc7d6_l3.png "Rendered by QuickLaTeX.com")
Therefore cofactors of 
It is known that, ![\operatorname{adj} A=\left[\begin{array}{ll}C_{11} & C_{12} \\ C_{21} & C_{22}\end{array}\right]^{T}](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-47ba298b5730b20f42f54ab045b9183b_l3.png "Rendered by QuickLaTeX.com")
![(\operatorname{adj} A)=\left[\begin{array}{cc}\cos \alpha & -\sin \alpha \\ -\sin \alpha & \cos \alpha\end{array}\right]^{T}](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-e63a6b769c611a394bb47c283dd394a4_l3.png "Rendered by QuickLaTeX.com")
![=\left[\begin{array}{cc}\cos \alpha & -\sin \alpha \\ -\sin \alpha & \cos \alpha\end{array}\right]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-76c81860404a3173b96f7a61c8228e92_l3.png "Rendered by QuickLaTeX.com")
Now, 
![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]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-7a804076fbc119eef7ee2f41bdb4a606_l3.png "Rendered by QuickLaTeX.com")
![=\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]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-0067786852d7f27ed7d5af5fbeda143e_l3.png "Rendered by QuickLaTeX.com")
![A=\left[\begin{array}{cc}\cos 2 \alpha & 0 \\ 0 & \cos 2 \alpha\end{array}\right]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-5bc054861b0088506719ec02ef5d0bad_l3.png "Rendered by QuickLaTeX.com")
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]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-00eb0c2c90d0d29250a565e82c99c434_l3.png "Rendered by QuickLaTeX.com")
![=\left(\cos ^{2} \alpha-\sin ^{2} \alpha\right)\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-7f1d7eb7eb6def919fd7e7a1868c7b28_l3.png "Rendered by QuickLaTeX.com")
![=\left[\begin{array}{cc}\cos ^{2} \alpha-\sin ^{2} \alpha & 0 \\ 0 & \cos ^{2} \alpha-\sin ^{2} \alpha\end{array}\right]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-8d2b17431563e216c69532a089799397_l3.png "Rendered by QuickLaTeX.com")
Adjoint and Inverse of a Matrix
![=\left[\begin{array}{cc}\cos 2 \alpha & 0 \\ 0 & \cos 2 \alpha\end{array}\right]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-cb6c131c8614b9b96e8535910090842e_l3.png "Rendered by QuickLaTeX.com")
Also, 
![=\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]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-0779d1cb1bc4f6bdfa0ec6c1c212227d_l3.png "Rendered by QuickLaTeX.com")
As a result, 
(ii) Suppose
![A=\left[\begin{array}{cc}1 & \tan \frac{\alpha}{2} \\ -\tan \frac{\alpha}{2} & 1\end{array}\right]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-e466a12df6a524178d5d51dfd8edbfbf_l3.png "Rendered by QuickLaTeX.com")
Therefore cofactors of
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}}](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-aae8762ab65b2dd9da01ec09e5a4b13c_l3.png "Rendered by QuickLaTeX.com")
![(\operatorname{adj} A)=\left[\begin{array}{cc}1 & \tan \frac{\alpha}{2} \\ -\tan \frac{\alpha}{2} & 1\end{array}\right]^{\mathrm{T}}](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-37d3a0d859cb81240b627ea8dd297049_l3.png "Rendered by QuickLaTeX.com")
![=\left[\begin{array}{cc}1 & -\tan \frac{\alpha}{2} \\ \tan \frac{\alpha}{2} & 1\end{array}\right]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-328cea312e9e8ce219eb5fdef71f6e1d_l3.png "Rendered by QuickLaTeX.com")
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]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-296b7d3cccd0fd297d7a0d494c669cec_l3.png "Rendered by QuickLaTeX.com")
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]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-f6ad538eafb07ccfab787382163cdd3c_l3.png "Rendered by QuickLaTeX.com")
![(\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]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-59a503ed6f55946e0f00101b34350a41_l3.png "Rendered by QuickLaTeX.com")
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]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-454d9750108a1ec117f5a3734c49f9a2_l3.png "Rendered by QuickLaTeX.com")
![=\left[\begin{array}{cc}1+\tan ^{2} \frac{\alpha}{2} & 0 \\ 0 & 1+\tan ^{2} \frac{\alpha}{2}\end{array}\right]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-3663570e6642c9b338b36de585c1fe69_l3.png "Rendered by QuickLaTeX.com")
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]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-6bbe1450251f21d58b84d96f78988b75_l3.png "Rendered by QuickLaTeX.com")
As a result,
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]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-c318bc4003f878b1e2dcbf877a78843d_l3.png "Rendered by QuickLaTeX.com")
(ii)![\left[\begin{array}{cc}1 & \tan \frac{\alpha}{2} \\ -\tan \frac{\alpha}{2} & 1\end{array}\right]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-47b4368d78f6211c67a17a527eeb3eda_l3.png "Rendered by QuickLaTeX.com")
