If $A=\begin{bmatrix} \cos \theta & i\sin \theta \\ i\sin \theta & \cos \theta \end{bmatrix}$,…

If $A=\begin{bmatrix} \cos \theta & i\sin \theta \\ i\sin \theta & \cos \theta \end{bmatrix}$, $\theta=\frac{\pi}{24}$ and $A^5=\begin{bmatrix} a & b \\ c & d \end{bmatrix}$, where $i=\sqrt{-1}$, then which one of the following is not true?
  1. 0≤a2+b2≤1
  2. a2-d2=0
  3. a2-c2=1
  4. a2-b2=12

Solution

$A^2 = \begin{bmatrix} \cos \theta & i \sin \theta \\ i \sin \theta & \cos \theta \end{bmatrix} \begin{bmatrix} \cos \theta & i \sin \theta \\ i \sin \theta & \cos \theta \end{bmatrix}$ $= \begin{bmatrix} \cos^2 \theta - \sin^2 \theta & 2i \sin \theta \cos \theta \\ 2i \sin \theta \cos \theta & \cos^2 \theta - \sin^2 \theta \end{bmatrix}$ $= \begin{bmatrix} \cos 2\theta & i \sin 2\theta \\ i \sin 2\theta & \cos 2\theta \end{bmatrix}$ $A^3 = \begin{bmatrix} \cos 2\theta & i \sin 2\theta \\ i \sin 2\theta & \cos 2\theta \end{bmatrix} \begin{bmatrix} \cos \theta & i \sin \theta \\ i \sin \theta & \cos \theta \end{bmatrix}$ $= \begin{bmatrix} \cos 3\theta & i \sin 3\theta \\ i \sin 3\theta & \cos 3\theta \end{bmatrix}$ Similarly, $A^5 = \begin{bmatrix} \cos 5\theta & i \sin 5\theta \\ i \sin 5\theta & \cos 5\theta \end{bmatrix} = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$ $a = \cos 5\theta$, $b = i \sin 5\theta$, $c = i \sin 5\theta$, $d = \cos 5\theta$ $a = d$, $b = c$ (1) $a^2 + b^2 = \cos^2 5\theta - \sin^2 5\theta = \cos 10\theta \in [0,1]$ as $\theta = \frac{\pi}{24}$ (2) $a^2 - d^2 = 0$ (3) $a^2 - b^2 = \cos^2 5\theta + \sin^2 5\theta = 1$ (4) $a^2 - c^2 = a^2 - b^2 = 1$

Asked in: JEE Main 2020 (04 Sep Shift 1)

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