An orthogonal matrix is - Vidyadip

MCQ

An orthogonal matrix is

A
$\left[ {\begin{array}{*{20}{c}}{\cos \alpha }&{2\sin \alpha }\\{ - 2\sin \alpha }&{\cos \alpha }\end{array}} \right]$
✓
$\left[ {\begin{array}{*{20}{c}}{\cos \alpha }&{\sin \alpha }\\{ - \sin \alpha }&{\cos \alpha }\end{array}} \right]$
C
$\left[ {\begin{array}{*{20}{c}}{\cos \alpha }&{\sin \alpha }\\{\sin \alpha }&{\cos \alpha }\end{array}} \right]$
D
$\left[ {\begin{array}{*{20}{c}}1&1\\1&1\end{array}} \right]$

✓

Answer

Correct option: B.

$\left[ {\begin{array}{*{20}{c}}{\cos \alpha }&{\sin \alpha }\\{ - \sin \alpha }&{\cos \alpha }\end{array}} \right]$

b
(b) A square matrix is to be orthogonal matrix if $A'A = I = AA'$

$ \Rightarrow $ $A = \left[ {\begin{array}{*{20}{c}}{\cos \alpha }&{\sin \alpha }\\{ - \sin \alpha }&{\cos \alpha }\end{array}} \right]$, $A' = \left[ {\begin{array}{*{20}{c}}{\cos \alpha }&{ - \sin \alpha }\\{\sin \alpha }&{\cos \alpha }\end{array}} \right]$

$ \Rightarrow $ $AA' = \left[ {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right],\,A'A = \left[ {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right]$

$\therefore $ $AA' = A'A = I$.

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