Question

If $A=\left[\begin{array}{cc}\cos \theta & i\sin \theta \\ i\sin \theta & \cos \theta \end{array}\right],(\theta =\frac{\pi }{24})$ and $A^5=\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]$, Where $i=\sqrt{-1}$, then which one of the following is not true?

• A. $a^2-d^2=0$
• B. $a^2-c^2=1$
• C. $0\le a^2+b^2\le 1$
• D. $a^2-b^2=\frac{1}{2}$

Answer 1

The correct option is D $a^2-b^2=\frac{1}{2}$

$A=$$\left[\begin{array}{cc}\cos \theta & i\sin \theta \\ i\sin \theta & \cos \theta \end{array}\right]$

$A^2=\left[\begin{array}{cc}\cos \theta & i\sin \theta \\ i\sin \theta & \cos \theta \end{array}\right]\left[\begin{array}{cc}\cos \theta & i\sin \theta \\ i\sin \theta & \cos \theta \end{array}\right]$
$\Rightarrow A^2=\left[\begin{array}{cc}{\cos }^2\theta -{\sin }^2\theta & 2i\sin \theta \cos \theta \\ 2i\cos \theta \sin \theta \theta & {\cos }^2\theta -{\sin }^2\theta \end{array}\right]$
$\Rightarrow A^2=\left[\begin{array}{cc}\cos 2\theta & i\sin 2\theta \\ i\sin 2\theta & \cos 2\theta \end{array}\right]$

$A^3=\left[\begin{array}{cc}\cos 2\theta & i\sin 2\theta \\ i\sin 2\theta & \cos 2\theta \end{array}\right]\left[\begin{array}{cc}\cos \theta & i\sin \theta \\ i\sin \theta & \cos \theta \end{array}\right]$
$\Rightarrow A^3=\left[\begin{array}{cc}\cos 3\theta & i\sin 3\theta \\ i\sin 3\theta & \cos 3\theta \end{array}\right]$

Similarly,
$A^5=\left[\begin{array}{cc}\cos 5\theta & i\sin 5\theta \\ i\sin 5\theta & \cos 5\theta \end{array}\right]$
$\therefore a=d=\cos 5\theta$ and $b=c=i\sin 5\theta$

$\therefore a^2-b^2={\cos }^2\left(5\theta \right)+{\sin }^2\left(5\theta \right)=1$