If A = \(\begin{bmatrix}\cos {\alpha }&\sin {\alpha }\\-\sin {\alpha }&\cos {\alpha }\end{bmatrix}\) and \(A+ { A }^{ T } \) = -2I, then the value of \(\alpha \) is
Answer & explanation
Correct answer: option 2
A = \(\begin{bmatrix}\cos {\alpha }&\sin {\alpha }\\-\sin {\alpha }&\cos {\alpha }\end{bmatrix}\)
\( { A }^{ T } \) = \(\begin{bmatrix}\cos {\alpha }&-\sin {\alpha }\\\sin {\alpha }&\cos {\alpha }\end{bmatrix}\)
\(A+ { A }^{ T } \) =\(\begin{bmatrix}\cos {\alpha }&\sin {\alpha }\\-\sin {\alpha }&\cos {\alpha }\end{bmatrix}\) + \(\begin{bmatrix}\cos {\alpha }&-\sin {\alpha }\\\sin {\alpha }&\cos {\alpha }\end{bmatrix}\)= \(\begin{bmatrix}-2&0\\0&-2\end{bmatrix}\)
\(\begin{bmatrix}2\cos {\alpha }&0\\0&2\cos {\alpha }\end{bmatrix}\) = \ \(\begin{bmatrix}-2&0\\0&-2\end{bmatrix}\)
\(2\cos {\alpha }= -2\)
\(\cos {\alpha }=-1\)
\(\cos {\alpha } = \cos {\Pi }\)
Thus , \(\alpha ={\Pi}\)