If $$A = \left[ {\matrix{
{\cos \theta } & {i\sin \theta } \cr
{i\sin \theta } & {\cos \theta } \cr
} } \right]$$, $\left( {\theta = {\pi \over {24}}} \right)$
and ${A^5} = \left[ {\matrix{
a & b \cr
c & d \cr
} } \right]$, where $i = \sqrt { - 1}$ then which one of the following is
not true?
Solution
$\because$ $$A = \left[ {\matrix{
{\cos \theta } & {i\sin \theta } \cr
{i\sin \theta } & {\cos \theta } \cr
} } \right]$$<br><br>$\therefore$ $${A^n} = \left[ {\matrix{
{\cos \,n\theta } & {i\sin \,n\theta } \cr
{i\sin \,n\theta } & {\cos \,n\theta } \cr
} } \right],n \in N$$<br><br>$\therefore$$${A^5} = \left[ {\matrix{
{\cos \,5\theta } & {i\sin \,5\theta } \cr
{i\sin \,5\theta } & {\cos \,5\theta } \cr
} } \right] = \left[ {\matrix{
a & b \cr
c & d \cr
} } \right]$$<br><br>$\therefore$ $a = \cos 5\theta ,\,b = i\sin 5\theta = c,\,d = \cos 5\theta$<br><br>$\therefore$ ${a^2} - {b^2} = {\cos ^2}5\theta + {\sin ^2}5\theta = 1$<br><br>${a^2} - {c^2} = {\cos ^2}5\theta + {\sin ^2}5\theta = 1$<br><br>${a^2} - {d^2} = {\cos ^2}5\theta - {\cos ^2}5\theta = 1$<br><br>$${a^2} + {b^2} = {\cos ^2}5\theta - {\sin ^2}5\theta = \cos 10\theta = \cos {{10\pi } \over {24}}$$<br><br>and $0 < \cos {{5\pi } \over {12}} < 1$
<br><br>$\Rightarrow 0 \le {a^2} + {b^2} \le 1$
About this question
Subject: Mathematics · Chapter: Matrices and Determinants · Topic: Types of Matrices and Operations
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