Question
If $\mathrm{A}=\left[\begin{array}{cc}\cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha\end{array}\right]$, show that $\mathrm{A}^{\top} \mathrm{A}=\mathrm{I}$, where $\mathrm{I}$ is the unit matrix of order 2 .
$\begin{aligned} & \therefore \quad A^{\mathrm{T}} \mathrm{A}=\left[\begin{array}{cc}\cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha\end{array}\right]\left[\begin{array}{cc}\cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha\end{array}\right] \\ & =\left[\begin{array}{cc}\cos ^2 \alpha+\sin ^2 \alpha & \cos \alpha \sin \alpha-\sin \alpha \cos \alpha \\ \sin \alpha \cos \alpha-\cos \alpha \sin \alpha & \sin ^2 \alpha+\cos ^2 \alpha\end{array}\right]\end{aligned}$
$=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$
$\therefore A^{\top} A=I$, where $I$ is the unit matrix of order 2 .
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$(5+4 x)^{-1 / 2}$