Question

If $A=\left[\begin{array}{cc}\cos \theta & i\sin \theta \\ i\sin \theta & \cos \theta \end{array}\right]$, $\left(\theta =\frac{\pi }{24}\right)$ and $A^5=\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]$, where $i=\surd -1$, 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

$a^2-b^2=\frac{1}{2}$

Explanation for the correct option:

Step 1. Find $A^2$ matrix :

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

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

Step 2. Find $A^3$ matrix:

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

Similarly,

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

Step 2. Compare the corresponding elements of $A^5$ matrix:

$\begin{array}{rcl}a& =& d\\ & =& \cos 5\theta \end{array}$

$\begin{array}{rcl}b& =& c\\ & =& i\sin 5\theta \end{array}$

$\begin{array}{rcl}\therefore {\mathit{a}}^{\mathbf{2}}\mathbf{}\mathbf{-}\mathbf{}{\mathit{c}}^{\mathbf{2}}& =& {\cos }^{2}5\theta +{\sin }^{2}5\theta \\ & =& \mathbf{1}\end{array}$ $\left[\mathbf{\because }\mathit{s}\mathit{i}{\mathit{n}}^{\mathbf{2}}\mathit{\theta }\mathbf{}\mathbf{+}\mathbf{}\mathit{c}\mathit{o}{\mathit{s}}^{\mathbf{2}}\mathit{\theta }\mathbf{}\mathbf{=}\mathbf{}\mathbf{1}\right]$

$\begin{array}{rcl}{\mathit{a}}^{\mathbf{2}}\mathbf{}\mathbf{-}\mathbf{}{\mathit{d}}^{\mathbf{2}}& =& {\cos }^{2}5\theta -{\cos }^{2}5\theta \\ & =& \mathbf{}\mathbf{0}\end{array}$

$\begin{array}{rcl}{a}^2+b^2& =& {\cos }^{2}5\theta -{\sin }^{2}5\theta \\ & =& \cos 10\theta \\ & =& \cos 75°\end{array}$

$\mathbf{\therefore }\mathbf{0}\mathbf{}\mathbf{\le }\mathbf{}{\mathit{a}}^{\mathbf{2}}\mathbf{}\mathbf{+}\mathbf{}{\mathit{b}}^{\mathbf{2}}\mathbf{}\mathbf{\le }\mathbf{}\mathbf{1}$

Hence, Option ‘D’ is Correct.