Question

If $A =\begin{bmatrix}0 & -\tan \frac{\alpha}{2} \\ \tan \frac{\alpha}{2} & 0\end{bmatrix}$ and $I$ is the identity matrix of order 2 , then $\begin{bmatrix}\cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha\end{bmatrix}$ is equal to

• A. $I + A$
• B. $I - A$
• C. $A - I$
• D. $A$

Answer 1

Answer: (A)

Here, $A=\begin{bmatrix}0 & -t \\ t & 0\end{bmatrix}$, where $t=\tan \left(\frac{\alpha}{2}\right)$
Now, $\cos \alpha=\frac{1-\tan ^{2}\left(\frac{\alpha}{2}\right)}{1+\tan ^{2}\left(\frac{\alpha}{2}\right)}=\frac{1-t^{2}}{1+t^{2}}$
and $\sin \alpha=\frac{2 \tan \left(\frac{\alpha}{2}\right)}{1+\tan ^{2}\left(\frac{\alpha}{2}\right)}=\frac{2 t }{1+ t ^{2}}$
$=( I - A )\begin{bmatrix}\cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha\end{bmatrix}$
$=\left(\begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}-\begin{bmatrix}0 & - t \\ + t & 0\end{bmatrix}\right)\begin{bmatrix}\frac{1- t ^{2}}{1+ t ^{2}} & \frac{-2 t }{1+ t ^{2}} \\ \frac{2 t }{1+ t ^{2}} & \frac{1- t ^{2}}{1+ t ^{2}}\end{bmatrix}$
$=\begin{bmatrix}1 & t \\ -t & 1\end{bmatrix}\begin{bmatrix}\frac{1-t^{2}}{1+t^{2}} & \frac{-2 t}{1+t^{2}} \\ \frac{2 t}{1+t^{2}} & \frac{1-t^{2}}{1+t^{2}}\end{bmatrix}$
$=\begin{bmatrix}\frac{1-t^{2}+2 t^{2}}{1+t^{2}} & \frac{-2 t+t\left(1-t^{2}\right)}{1+t^{2}} \\ \frac{-t\left(1-t^{2}\right)+2 t}{1+t^{2}} & \frac{2 t^{2}+1-t^{2}}{1+t^{2}}\end{bmatrix}$
$=\begin{bmatrix}\frac{1+t^{2}}{1+t^{2}} & \frac{-2 t+t-t^{3}}{1+t^{2}} \\ \frac{-t+t^{3}+2 t}{1+t^{2}} & \frac{2 t^{2}+1-t^{2}}{1+t^{2}}\end{bmatrix}$
$=\begin{bmatrix}\frac{1+t^{2}}{1+t^{2}} & \frac{-t\left(1+t^{2}\right)}{1+t^{2}} \\ \frac{t\left(1+t^{2}\right)}{1+t^{2}} & \frac{1+t^{2}}{1+t^{2}}\end{bmatrix}=\begin{bmatrix}1 & -t \\ t & 1\end{bmatrix}$
Also, $I+A=\begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}+\begin{bmatrix}0 & -t \\ t & 0\end{bmatrix}$
$=\begin{bmatrix}0+1 & - t +0 \\ t +0 & 0+1\end{bmatrix}=\begin{bmatrix}1 & - t \\ t & 1\end{bmatrix}$
$=( I - A)\begin{bmatrix}\cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha\end{bmatrix}$