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  4. If f(α ) [cos α&sin α -sin α &cos α ] and if α, β, γ, are angle of a triangle, then f (α).f(β).f(γ) equals

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Q. If $f(\alpha ) \begin{bmatrix}cos\,\alpha&sin\,\alpha \\ -sin\,\alpha &cos\,\alpha \end{bmatrix}$ and if $\alpha, \beta, \gamma,$ are angle of a triangle, then $f \left(\alpha\right).f\left(\beta\right).f\left(\gamma\right)$ equals

Matrices

Solution:

Hence $f\left(\alpha\right)f\left(\beta\right)=\begin{bmatrix}cos\,\alpha&sin\,\alpha \\ -sin\,\alpha &cos\,\alpha \end{bmatrix}\begin{bmatrix}cos\,\beta&sin\,\beta\\ -sin\,\alpha \beta&cos\,\beta\end{bmatrix}$
$\begin{bmatrix}cos\,\alpha \,cos\,\beta-sin\,\alpha \,sin\,\beta&cos\,\alpha \,sin\,\beta+sin\,\alpha \,cos\,\beta\\ -sin\,\alpha\,cos\,\beta-cos\,\alpha\,sin\,\beta&-sin\,\alpha\,sin\,\beta+cos\,\alpha\,cos\,\beta\end{bmatrix}$
$=\begin{bmatrix}cos\left(\alpha+\beta\right)&sin\left(\alpha+\beta\right)\\ -sin\left(\alpha+\beta\right)&cos\left(\alpha+\beta\right)\end{bmatrix}$
similarly $f\left(\alpha\right)f\left(\beta\right)f\left(\gamma\right)=\begin{bmatrix}cos\left(\alpha+\beta+\gamma\right)&sin\left(\alpha+\beta+\gamma\right)\\ -sin\left(\alpha+\beta+\gamma \right)&cos\left(\alpha+\beta+\gamma \right)\end{bmatrix}$
$=\begin{bmatrix}cos\,\pi &sin\,\pi \\ -sin\,\pi &cos\,\pi \end{bmatrix}$ as $\alpha+\beta +\gamma =\pi$
$=\begin{bmatrix}-1&0\\ 0&-1\end{bmatrix}=-\begin{bmatrix}1&0\\ 0&1\end{bmatrix}=-I_{2}$