Question

If $A=\left[\begin{array}{cc}\cos A& \sin A\\ \sin A& -\cos A\end{array}\right]$ then find the negative of value of $|A{|}^5$.

Answer 1

$A=\left[\begin{array}{cc}\cos A& \sin A\\ \sin A& -\cos A\end{array}\right]$

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

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

$A^2=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]=I{\sin }^{2}A+{\cos }^{2}A=1$

Now,

$A^3=A^2.A=I.A=A(\because A^2=I)$

$\Rightarrow A^3=A$

$A^5=A^3.A^2=A.I=A(\because A^2=I;A^3=A)$

$\Rightarrow A^5=A$

$\Rightarrow {|A|}^5=\left[\begin{array}{cc}\cos A& \sin A\\ \sin A& -\cos A\end{array}\right]=-(\cos A{)}^2-(\sin A{)}^2=-1$