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

Question

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

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Answer

Given: $\text{A}=\begin{bmatrix}\sin\alpha&\cos\alpha\\-\cos\alpha&\sin\alpha\end{bmatrix}$
$\text{A}^\text{T}=\begin{bmatrix}\sin\alpha&-\cos\alpha\\\cos\alpha&\sin\alpha\end{bmatrix}$
Now,
$\text{A}^\text{T}\text{A}=\begin{bmatrix}\sin\alpha&-\cos\alpha\\\cos\alpha&\sin\alpha\end{bmatrix}\begin{bmatrix}\sin\alpha&\cos\alpha\\-\cos\alpha&\sin\alpha\end{bmatrix}$
$\Rightarrow\text{A}^\text{T}\text{A}=\begin{bmatrix}(\sin\alpha)(\sin\alpha)+(-\cos\alpha)(-\cos\alpha)&(\sin\alpha)(\cos\alpha)+(-\cos\alpha)(\sin\alpha)(\cos\alpha)(\sin\alpha)+(\sin\alpha)(-\cos\alpha)&(\cos\alpha)(\cos\alpha)+(\sin\alpha)(\sin\alpha)\end{bmatrix}$
$\Rightarrow\text{A}^\text{T}\text{A}=\begin{bmatrix}\sin^2\alpha+\cos^2\alpha&\sin\alpha\cos\alpha-\sin\alpha\cos\alpha\\\sin\alpha\cos\alpha-\sin\alpha\cos\alpha&\cos^2\alpha+\sin^2\alpha\end{bmatrix}$
$\Rightarrow\text{A}^\text{T}\text{A}=\begin{bmatrix}1&0\\0&1\end{bmatrix}=\text{I}$

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