Question

Let $A=\left[\begin{array}{cc}0& -\tan \theta /2\\ \tan \theta /2& 0\end{array}\right](\theta \ne n\pi )$ $B=\left[\begin{array}{cc}\cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{array}\right],I=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$ 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
$B=\left[\begin{array}{cc}\frac{1-a^2}{1+a^2}& \frac{-2a}{1+a^2}\\ \frac{2a}{1+a^2}& \frac{1-a^2}{1+a^2}\end{array}\right]$
$\therefore (I-A)B=\left[\begin{array}{cc}1& a\\ -a& 1\end{array}\right]\left[\begin{array}{cc}\frac{1-a^2}{1+a^2}& \frac{-2a}{1+a^2}\\ \frac{2a}{1+a^2}& \frac{1-a^2}{1+a^2}\end{array}\right]$
$=\left[\begin{array}{cc}\frac{1-a^2}{1+a^2}+\frac{-2a^2}{1+a^2}& \frac{-2a}{1+a^2}+\frac{a(1-a^{2)}}{1+a^2}\\ \frac{-a\left(1-a^2\right)}{1+a^2}+\frac{2a}{1+a^2}& \frac{2a^2}{1+a^2}+\frac{1-a^2}{1+a^2}\end{array}\right]$
$=\left[\begin{array}{cc}1& -a\\ a& 1\end{array}\right]=I+A$