Question

Let $A= \begin{bmatrix}0 & -\tan \theta / 2 \\ \tan \theta / 2 & 0\end{bmatrix}(\theta \neq n \pi)$ $B= \begin{bmatrix}\cos \theta & -\sin \theta \\ \sin \theta & \cos \theta\end{bmatrix}, I= \begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}$ Then the matrix $I+A$ is equal to

• A. $( I - A) B$
• B. $(I-A)^{2} B$
• C. $(I+A)^{2} B$
• D. $(I-A)^{2}$

Answer 1

Answer: (A)

Put $\tan (\theta / 2)=a$ so that
$B= \begin{bmatrix}\frac{1-a^{2}}{1+a^{2}} & \frac{-2 a}{1+a^{2}} \\ \frac{2 a}{1+a^{2}} & \frac{1-a^{2}}{1+a^{2}}\end{bmatrix}$
$\therefore (I-A) B= \begin{bmatrix}1 & a \\ -a & 1\end{bmatrix} \begin{bmatrix}\frac{1-a^{2}}{1+a^{2}} & \frac{-2 a}{1+a^{2}} \\ \frac{2 a}{1+a^{2}} & \frac{1-a^{2}}{1+a^{2}}\end{bmatrix}$
$= \begin{bmatrix}\frac{1-a^{2}}{1+a^{2}}+\frac{-2 a^{2}}{1+a^{2}} & \frac{-2 a}{1+a^{2}}+\frac{a\left(1-a^{2)}\right.}{1+a^{2}} \\ \frac{-a\left(1-a^{2}\right)}{1+a^{2}}+\frac{2 a}{1+a^{2}} & \frac{2 a^{2}}{1+a^{2}}+\frac{1-a^{2}}{1+a^{2}}\end{bmatrix}$
$= \begin{bmatrix}1 & -a \\ a & 1\end{bmatrix}=I+A$