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Question

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


Solution

$$\begin{vmatrix} x & \sin { \theta  }  & \cos { \theta  }  \\ -\sin { \theta  }  & -x & 1 \\ \cos { \theta  }  & 1 & x \end{vmatrix}$$
$$=x\left( -{ x }^{ 2 }-1 \right) \sin { \theta  } \left( -x\sin { \theta  } -\cos { \theta  }  \right) +\cos { \theta  } \left( -\sin { \theta  } +x\cos { \theta  }  \right) \\=-{ x }^{ 3 }-x+x\sin ^{ 2 }{ \theta  } +\sin { \theta  } \cos { \theta  } -\sin { \theta  } \cos { \theta  } +x\cos ^{ 2 }{ \theta  } $$
$$=-{ x }^{ 3 }-x+x\times 1=-{ x }^{ 3 }$$ which is independent of $$\theta$$

Mathematics

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