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  • Prove that the determinant x sin theta cos theta -sin theta -x 1 cos theta 1 x is independent of theta.

Q : 1         Prove that the determinant  \begin{vmatrix} x & \sin \theta &\cos \theta \\ -\sin \theta &-x & 1\\ \cos \theta &1 &x \end{vmatrix}  is independent of \theta.

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Calculating the determinant value of \begin{vmatrix} x & \sin \theta &\cos \theta \\ -\sin \theta &-x & 1\\ \cos \theta &1 &x \end{vmatrix};

= x\begin{bmatrix} -x &1 \\ 1& x \end{bmatrix}-\sin \Theta\begin{bmatrix} -\sin \Theta &1 \\ \cos \Theta& x \end{bmatrix} + \cos \Theta \begin{bmatrix} -\sin \Theta &-x \\ \cos \Theta& 1 \end{bmatrix}

= x(-x^2-1)-\sin \Theta (-x\sin \Theta-\cos \Theta)+\cos\Theta(-\sin \Theta+x\cos\Theta)

= -x^3-x+x\sin^2 \Theta+ \sin \Theta\cos \Theta-\cos\Theta\sin \Theta+x\cos^2\Theta

= -x^3-x+x(\sin^2 \Theta+\cos^2\Theta)

= -x^3-x+x = -x^3

Clearly, the determinant is independent of \Theta.

Posted by

Divya Prakash Singh

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