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}=[(\sin \alpha)(\sin \alpha)+(-\cos \alpha)(-\cos \alpha) \quad(\sin \alpha)(\cos \alpha)$$+(-\cos \alpha)(\sin \alpha) \ \cos \alpha)(\sin \alpha)$$+(\sin \alpha)(-\cos \alpha) \quad(\cos \alpha)(\cos \alpha)+(\sin \alpha)(\sin \alpha)]$
$\Rightarrow\text{A}^\text{T}\text{A}=\begin{bmatrix}\sin^2\alpha+\cos^2\alpha&\sin\alpha\cos\alpha-\sin\alpha\cos\alpha\\\sin\alpha\cos\alpha-\sin\alpha\cos\alpha&\cos^2\alpha+\sin^2\alpha\end{bmatrix}$
$\Rightarrow\text{A}^\text{T}\text{A}=\begin{bmatrix}1&0\\0&1\end{bmatrix}=\text{I}$

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