Question

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Answer 1

Given Question :-

Show that matrix A is an Orthogonal matrix, where

$\rm \: A = \begin{bmatrix} cos \alpha & sin \alpha \\ - sin \alpha & cos \alpha \end{bmatrix}$

$\large\underline{\sf{Solution-}}$

Given matrix is

$\rm \: A = \begin{bmatrix} cos \alpha & sin \alpha \\ - sin \alpha & cos \alpha \end{bmatrix}$

We know,

A square matrix A is said to be Orthogonal matrix iff AA' = I

So, Consider

$\rm \: A' = \begin{bmatrix} cos \alpha & - sin \alpha \\ sin \alpha & cos \alpha \end{bmatrix}$

Now, Consider

$\rm \: A \: A'$

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

$\rm \: = \: \begin{bmatrix} {cos}^{2} \alpha + {sin}^{2} \alpha & - sin \alpha \: cos\alpha +cos \alpha \: sin \alpha\\ - sin \alpha \: cos \alpha + cos \alpha \: sin \alpha & {cos}^{2} \alpha + {sin}^{2} \alpha \end{bmatrix}$

$\rm \: = \: \begin{bmatrix} 1 & 0\\ 0 & 1\end{bmatrix}$

$\rm \: = \: I$

$\rm\implies \:AA' = I$

$\rm\implies \:A \: is \: orthogonal \: matrix$

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ADDITIONAL INFORMATION

$\begin{gathered}\boxed{\begin{array}{c|c} \bf Square \: Matrix & \bf Condition \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf Idempotent & \sf {A}^{2} = A \\ \\ \sf Nilpotent & \sf {A}^{n} = 0 \\ \\ \sf Involutory & \sf {A}^{2} = I\\ \\ \sf Symmetric & \sf {A}'= A\\ \\ \sf Skew \: Symmetric & \sf {A}' = - A \end{array}} \\ \end{gathered}$

Answer 2

${ \pmb{ \frak {\underline \red{given \: : }}}}$

$\rm\bullet{ \: Show \: that \: matrix \begin{gathered}\rm \: A' = \begin{bmatrix} \rm cos \alpha & \rm sin \alpha \\ \rm - sin \alpha & \rm \cos \alpha \end{bmatrix} \\ \end{gathered} }\\ \rm \: is \: an \: orthogonal \: matrix.$

${ \pmb{ \frak {\underline \red{Solution \: : }}}}$

$\longmapsto\rm { Any \: matrix \: A'}$

$\longmapsto\rm \boxed{ \rm {AA'} ^{T} = I}$

$\longmapsto\rm{\begin{gathered}\rm \: A' = \begin{bmatrix} \rm cos \alpha & \rm sin \alpha \\ \rm - sin \alpha & \rm \cos \alpha \end{bmatrix} \\ \end{gathered} }$

$\longmapsto\rm{\begin{gathered}\rm \: A = \begin{bmatrix} \rm cos \alpha & \rm -sin \alpha \\ \rm sin \alpha & \rm \cos \alpha \end{bmatrix} \\ \end{gathered} }$

$\longmapsto\rm{\begin{gathered}\rm \:{A A' }^{T}= \begin{bmatrix} \rm cos \alpha & \rm sin \alpha \\ \rm -sin \alpha & \rm \cos \alpha \end{bmatrix} \\ \end{gathered} }$

$\longmapsto\rm{\begin{gathered}\rm \:{A A '}^{T}= \begin{bmatrix} \rm sin^{2}\alpha + cos^{2}\alpha & 0 \\ \rm 0 & \rm sin^{2}\alpha + cos^{2} \end{bmatrix} \\ \end{gathered} }$

$\longmapsto\rm{\begin{gathered}\rm \:{A A' }^{T}= \begin{bmatrix} \rm 1 & 0 \\ \rm 0 & \rm 1 \end{bmatrix} \\ \end{gathered} } = \: I$

$\: \: \: \: \: \bigstar\rm{ \: The \: matrix \: orthagonal }$

$\ \: \: \: \: \: \: \dag{\pmb{\frak{\underline{ \red{Proved}}}}}$

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@Shivam

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