Solve this
Question:

If $\mathrm{A}=\left[\begin{array}{cc}\cos \theta & \text { isin } \theta \\ i \sin \theta & \cos \theta\end{array}\right], \quad\left(\theta=\frac{\pi}{24}\right)$ and

$\mathrm{A}^{5}=\left[\begin{array}{ll}\mathrm{a} & \mathrm{b} \\ \mathrm{c} & \mathrm{d}\end{array}\right]$, where $\mathrm{i}=\sqrt{-1}$, then which one

of the following is not true?

 

  1.  $0 \leq a^{2}+b^{2} \leq 1$

  2.  $\mathrm{a}^{2}-\mathrm{d}^{2}=0$

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

  4.  $a^{2}-c^{2}=1$


Correct Option: , 3

Solution:

$A^{2}=\left(\begin{array}{cc}\cos 2 \theta & i \sin 2 \theta \\ i \sin 2 \theta & \cos 2 \theta\end{array}\right)$

Similarly, $A^{5}=\left(\begin{array}{cc}\cos 5 \theta & i \sin 5 \theta \\ i \sin 5 \theta & \cos 5 \theta\end{array}\right)=\left(\begin{array}{ll}a & b \\ c & d\end{array}\right)$

(1) $a^{2}+b^{2}=\cos ^{2} 5 \theta-\sin ^{2} 5 \theta=\cos 10 \theta=\cos 75^{\circ}$

(2) $a^{2}-d^{2}=\cos ^{2} 5 \theta-\cos ^{2} 5 \theta=0$

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

(4) $a^{2}-c^{2}=\cos ^{2} 5 \theta+\sin ^{2} 5 \theta=1$

 

Administrator

Leave a comment

Please enter comment.
Please enter your name.