LA2.1.17

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Show that the determinant of the matrix {\begin{bmatrix}\cos(\theta +\alpha )&\sin(\theta +\alpha )&1\\\cos(\theta +\beta )&\sin(\theta +\beta )&1\\\cos(\theta +\gamma )&\sin(\theta +\gamma )&1\end{bmatrix}}\, is independent of theta.

Applying R_{2}-R_{1},R_{3}-R_{2}\,

Determinant is {\begin{vmatrix}\cos(\theta +\alpha )&\sin(\theta +\alpha )&1\\\cos(\theta +\beta )-\cos(\theta +\alpha )&\sin(\theta +\beta )-\sin(\theta +\alpha )&0\\\cos(\theta +\gamma )-\cos(\theta +\beta )&\sin(\theta +\gamma )-\sin(\theta +\beta )&0\end{vmatrix}}\,

Expanding along C3,we get -4\sin({\frac  {\beta -\alpha }{2}})\cdot \sin({\frac  {\gamma -\beta }{2}})[\sin({\frac  {2\theta +\alpha +\beta }{2}})\cdot \cos({\frac  {2\theta +\beta +\gamma }{2}})-\cos({\frac  {2\theta +\alpha +\beta }{2}})\cdot \sin({\frac  {2\theta +\beta +\gamma }{2}})]\,

-4\sin({\frac  {\beta -\alpha }{2}})\cdot \sin({\frac  {\gamma -\beta }{2}})\cdot \sin({\frac  {2\theta +\alpha +\beta -2\theta -\beta -\gamma }{2}})\, which is equal to

-4\sin({\frac  {\beta -\alpha }{2}})\cdot \sin({\frac  {\gamma -\beta }{2}})\sin({\frac  {\alpha -\gamma }{2}})\, which doesnot contain the theta.


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