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If $A =\begin{bmatrix}\cos θ&-\sin θ\\\sin θ&\cos θ\end{bmatrix}$, then $A^T+A=I_2$, if |
$θ=nп, n∈Z$ $θ=(2n+1)\frac{п}{2},n∈Z$ $θ=2nп+\frac{п}{3},n∈Z$ none of these |
$θ=2nп+\frac{п}{3},n∈Z$ |
We have, $A^T+A=I_2$ $⇒\begin{bmatrix}\cos θ&\sin θ\\-\sin θ&\cos θ\end{bmatrix}+\begin{bmatrix}\cos θ&-\sin θ\\\sin θ&\cos θ\end{bmatrix}=\begin{bmatrix}1&0\\0&1\end{bmatrix}$ $⇒\cos θ=\frac{1}{2}=\cos \frac{п}{3}⇒θ=2nп±\frac{п}{3},n∈Z$ |
