Question

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Answer 1

Answer:

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Answer 2

Question :–

$\\ { \bold{if \: \: A = \begin{bmatrix} \cos( \alpha )& \sin( \alpha ) \\ \\ - \sin( \alpha )& \cos( \alpha ) \end{bmatrix} \: \: then \: \: Prove \: \: that \: A'A = I }}$

ANSWER :–

GIVEN :–

$\\ { \bold{ \to \: A \: \: matrix \: \: is \: \: A = \begin{bmatrix} \cos( \alpha )& \sin( \alpha ) \\ \\ - \sin( \alpha )& \cos( \alpha ) \end{bmatrix} \: \: }}$

TO PROVE :–

$\\ { \bold{ \to \: A'A = I}}$

SOLUTION :–

$\\ { \bold{ \implies \: \: A = \begin{bmatrix} \cos( \alpha )& \sin( \alpha ) \\ \\ - \sin( \alpha )& \cos( \alpha ) \end{bmatrix} \: \: }}$

• Let's find A' matrix –

$\\ { \bold{ \implies \: \: A' = \begin{bmatrix} \cos( \alpha )& - \sin( \alpha ) \\ \\ \sin( \alpha )& \cos( \alpha ) \end{bmatrix} \: \: }}$

• Now Let's take L.H.S. –

$\\ { \bold{ \: \: = \: A'A }}$

$\\ { \bold{ \: \: = \: \begin{bmatrix} \cos( \alpha )& - \sin( \alpha ) \\ \\ \sin( \alpha )& \cos( \alpha ) \end{bmatrix} \begin{bmatrix} \cos( \alpha )& \sin( \alpha ) \\ \\ - \sin( \alpha )& \cos( \alpha ) \end{bmatrix} }}$

• Now multiply –

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

• We should write this as –

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

• We know that –

$\\ { \bold{ \: \longrightarrow \: \sin {}^{2} ( \alpha ) + \cos {}^{2} ( \alpha ) = 1 }}$

• So that –

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

$\\ { \bold{ \: \: = \: I \: \: \: \: \: \: \: ( Identity \: \: \: \: \: matrix)}}$

$\\ { \bold{ \: \: = \: R.H.S. \: \: \: \: \: \: \: (Hence \: \: \: \: \: proved)}}$