An orthogonal matrix is

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(B) square matrix $A$ is said to be an orthogonal matrix if $A'A = I = AA'$,where $A'$ is the transpose of $A$ and $I$ is the identity matrix.Let $A =

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$\begin{bmatrix} \cos \alpha & \sin \alpha \ -\sin \alpha & \cos \alpha \end{bmatrix}$

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(B) square matrix $A$ is said to be an orthogonal matrix if $A'A = I = AA'$,where $A'$ is the transpose of $A$ and $I$ is the identity matrix.Let $A = \begin{bmatrix} \cos \alpha & \sin \alpha \ -\sin \alpha & \cos \alpha \end{bmatrix}$.Then,the transpose $A' = \begin{bmatrix} \cos \alpha & -\sin \alpha \ \sin \alpha & \cos \alpha \end{bmatrix}$.Calculating $AA'$:$AA' = \begin{bmatrix} \cos \alpha & \sin \alpha \ -\sin \alpha & \cos \alpha \end{bmatrix} \begin{bmatrix} \cos \alpha & -\sin \alpha \ \sin \alpha & \cos \alpha \end{bmatrix} = \begin{bmatrix} \cos^2 \alpha + \sin^2 \alpha & -\cos \alpha \sin \alpha + \sin \alpha \cos \alpha \ -\sin \alpha \cos \alpha + \cos \alpha \sin \alpha & \sin^2 \alpha + \cos^2 \alpha \end{bmatrix} = \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix} = I$.Similarly,$A'A = I$.Since $AA' = A'A = I$,the matrix in option $(b)$ is an orthogonal matrix.

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