Question
$\text{A}=\begin{bmatrix}\sin\alpha&\cos\alpha\\-\cos\alpha&\sin\alpha\end{bmatrix}$, then verify that A'A = I
✓
Answer
$ \text{A}=\begin{bmatrix}\sin\alpha&\cos\alpha\\-\cos\alpha&\sin\alpha\end{bmatrix}$
$\therefore\ \text{A}'=\begin{bmatrix}\sin\alpha&-\cos\alpha\\\cos\alpha&\sin\alpha\end{bmatrix}$
$\text{A}'\text{A}=\begin{bmatrix}\sin\alpha&-\cos\alpha\\\cos\alpha&\sin\alpha\end{bmatrix}\begin{bmatrix}\sin\alpha&\cos\alpha\\-\cos\alpha&\sin\alpha\end{bmatrix}$
$\begin{bmatrix}\sin\alpha&-\cos\alpha\\\cos\alpha&\sin\alpha\end{bmatrix}\begin{bmatrix}\sin\alpha&\cos\alpha\\-\cos\alpha&\sin\alpha\end{bmatrix}$
$=\begin{bmatrix}(\sin\alpha)(\sin\alpha)+(-\cos\alpha)(-\cos\alpha)&(\sin\alpha)(\cos\alpha)+(-\cos\alpha)(\sin\alpha)\$\cos\alpha)(\sin\alpha)+(\sin\alpha)(-\cos\alpha)&(\cos\alpha)(\cos\alpha)+(\sin\alpha)(\sin\alpha)\end{bmatrix}$
$=\begin{bmatrix}\sin^2\alpha+\cos^2\alpha&\sin\alpha\cos\alpha-\sin\alpha\cos\alpha\\\sin\alpha\cos\alpha-\sin\alpha\cos\alpha&\cos^2\alpha+\sin^2\alpha\end{bmatrix}$
$=\begin{bmatrix}1&0\\0&1\end{bmatrix}=\text{I}$
Hence, we have verified that A'A = I
$\therefore\ \text{A}'=\begin{bmatrix}\sin\alpha&-\cos\alpha\\\cos\alpha&\sin\alpha\end{bmatrix}$
$\text{A}'\text{A}=\begin{bmatrix}\sin\alpha&-\cos\alpha\\\cos\alpha&\sin\alpha\end{bmatrix}\begin{bmatrix}\sin\alpha&\cos\alpha\\-\cos\alpha&\sin\alpha\end{bmatrix}$
$\begin{bmatrix}\sin\alpha&-\cos\alpha\\\cos\alpha&\sin\alpha\end{bmatrix}\begin{bmatrix}\sin\alpha&\cos\alpha\\-\cos\alpha&\sin\alpha\end{bmatrix}$
$=\begin{bmatrix}(\sin\alpha)(\sin\alpha)+(-\cos\alpha)(-\cos\alpha)&(\sin\alpha)(\cos\alpha)+(-\cos\alpha)(\sin\alpha)\$\cos\alpha)(\sin\alpha)+(\sin\alpha)(-\cos\alpha)&(\cos\alpha)(\cos\alpha)+(\sin\alpha)(\sin\alpha)\end{bmatrix}$
$=\begin{bmatrix}\sin^2\alpha+\cos^2\alpha&\sin\alpha\cos\alpha-\sin\alpha\cos\alpha\\\sin\alpha\cos\alpha-\sin\alpha\cos\alpha&\cos^2\alpha+\sin^2\alpha\end{bmatrix}$
$=\begin{bmatrix}1&0\\0&1\end{bmatrix}=\text{I}$
Hence, we have verified that A'A = I
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