Question

If $A=\left[\begin{array}{cc}\cos x& \sin x\\ -\sin x& \cos x\end{array}\right]$, show that $A^2=\left[\begin{array}{cc}\cos 2x& \sin 2x\\ -\sin 2x& \cos 2x\end{array}\right]and{A}^{{}^{\prime }}A=I$

Answer 1

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

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

$A^2=\left[\begin{array}{cc}co{s}^{2}x-si{n}^{2}x& cosxsinx+sinxcosx\\ -sinxcosx-sinxcosx& -si{n}^{2}x+co{s}^{2}x\end{array}\right]$

$A^2=\left[\begin{array}{cc}cos2x& sin2x\\ -sin2x& cos2x\end{array}\right]$

Now, $A^{\prime }=\left[\begin{array}{cc}cosx& -sinx\\ sinx& cosx\end{array}\right]$

Therefore,

$A{A}^{\prime }=\left[\begin{array}{cc}cosx& sinx\\ -sinx& cosx\end{array}\right]\left[\begin{array}{cc}cosx& -sinx\\ sinx& cosx\end{array}\right]$

$A{A}^{\prime }=\left[\begin{array}{cc}co{s}^{2}x+si{n}^{2}x& -cosxsinx+sinccosx\\ -sinxcosx+sinxcosx& si{n}^{2}x+co{s}^{2}x\end{array}\right]$

$A{A}^{\prime }=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]=I$