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If $\quad A=\left[\begin{array}{cc}\cos \theta & \text { isin } \theta \\ \operatorname{isin} \theta & \cos \theta\end{array}\right], \left(\theta=\frac{\pi}{24}\right)$ and $A^{5}=\left[\begin{array}{ll}a & b \\ c & d\end{array}\right],$ where $i=\sqrt{-1},$ then which one of the following is not true?
$0 \leq a^{2}+b^{2} \leq 1$
$a^{2}-d^{2}=0$
$a^{2}-b^{2}=\frac{1}{2}$
$a^{2}-c^{2}=1$
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, $A ^{5}=\left(\begin{array}{cc}\cos 5 \theta & i \sin 5 \theta \\ i \sin 5 \theta & \cos 5 \theta\end{array}\right)=\left(\begin{array}{ll} a & b \\ c & d \end{array}\right)$
$a^{2}+b^{2}=\cos ^{2} 5 \theta-\sin ^{2} 5 \theta=\cos 10 \theta=\cos 75^{\circ}$
$a^{2}-d^{2}=\cos ^{2} 5 \theta-\cos ^{2} 5 \theta=0$
$a^{2}-b^{2}=\cos ^{2} 5 \theta+\sin ^{2} 5 \theta=1$
$a^{2}-c^{2}=\cos ^{2} 5 \theta+\sin ^{2} 5 \theta=1$
