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Miscellaneous Exercise 4(B) · Q194

Q.If Aα=[cos⁡αsin⁡α−sin⁡αcos⁡α]A_\alpha=\begin{bmatrix}\cos\alpha & \sin\alpha\\-\sin\alpha & \cos\alpha\end{bmatrix}, show that Aα⋅Aβ=Aα+βA_\alpha\cdot A_\beta=A_{\alpha+\beta}.

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Aα=[cos⁡αsin⁡α−sin⁡αcos⁡α]A_\alpha=\begin{bmatrix}\cos\alpha & \sin\alpha\\-\sin\alpha & \cos\alpha\end{bmatrix}, Aβ=[cos⁡βsin⁡β−sin⁡βcos⁡β]A_\beta=\begin{bmatrix}\cos\beta & \sin\beta\\-\sin\beta & \cos\beta\end{bmatrix}.

AαAβA_\alpha A_\beta entry(1,1) =cos⁡αcos⁡β−sin⁡αsin⁡β=cos⁡(α+β)=\cos\alpha\cos\beta-\sin\alpha\sin\beta=\cos(\alpha+\beta).

entry(1,2) =cos⁡αsin⁡β+sin⁡αcos⁡β=sin⁡(α+β)=\cos\alpha\sin\beta+\sin\alpha\cos\beta=\sin(\alpha+\beta).

entry(2,1) =−sin⁡αcos⁡β−cos⁡αsin⁡β=−sin⁡(α+β)=-\sin\alpha\cos\beta-\cos\alpha\sin\beta=-\sin(\alpha+\beta).

entry(2,2) =−sin⁡αsin⁡β+cos⁡αcos⁡β=cos⁡(α+β)=-\sin\alpha\sin\beta+\cos\alpha\cos\beta=\cos(\alpha+\beta). …

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