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.

Main Page:Linear Algebra

LA2.1.17

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