If A= & - & , then verify that A^TA = I_2. - Vidyadip

Question

If $\text{A}=\begin{bmatrix}\cos\alpha&\sin\alpha\\-\sin\alpha&\cos\alpha\end{bmatrix},$ then verify that $A^TA = I_2$.

✓

Answer

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

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