An orthogonal matrix is, [ cos α − 2 sin α 2 sin α cos α ], [ cos α − sin α sin α cos α ], [ cos α sin α sin α cos α ], [ 1 1 1 1 ]

A square matrix is to be orthogonal matrix if A ′ A = I = A A ′ ⇒ A = [ cos α − sin α sin α cos α ] , A ′ = [ cos α sin α − sin α cos α ] ⇒ A A ′ = [ 1 0 0 1 ] , A ′ A = [ 1 0 0 1 ] ∴ A A ′ = A ′ A = I .

An orthogonal matrix is

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An orthogonal matrix is

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Options:

A $\left[\begin{array}{cc}\cos \alpha & 2 \sin \alpha \\ -2 \sin \alpha & \cos \alpha\end{array}\right]$
B $\left[\begin{array}{cc}\cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha\end{array}\right]$
C $\left[\begin{array}{cc}\cos \alpha & \sin \alpha \\ \sin \alpha & \cos \alpha\end{array}\right]$
D $\left[\begin{array}{ll}1 & 1 \\ 1 & 1\end{array}\right]$

Solution:

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The correct answer is: $\left[\begin{array}{cc}\cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha\end{array}\right]$

A square matrix is to be orthogonal matrix if
$A^{\prime} A=I=A A^{\prime}$
$\Rightarrow A=\left[\begin{array}{cc}\cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha\end{array}\right], A^{\prime}=\left[\begin{array}{cc}\cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha\end{array}\right]$
$\Rightarrow A A^{\prime}=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right], A^{\prime} A=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$
$\therefore A A^{\prime}=A^{\prime} A=I$.

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