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Prove that the determinant is independent of $\theta$:

$$ \begin{vmatrix} x & \sin \theta & \cos \theta \\ - \sin \theta & -x &1 \\ \cos \theta& 1 & x \end{vmatrix} $$

I expanded it along row 1 and I got $-x^3 + 2x$. My question is, since we say that we can apply as many operation on a determinant like row-row , row-column , row 1 - row 2 etc. value of determinant remains the same.

Similarly for this question, will the value of determinant every time I solve using any operation yield the same answer?

Ankit Saha
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Rider
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  • Does this answer your question: https://math.stackexchange.com/questions/1904416/how-do-row-operations-affect-detu – Joe Apr 18 '21 at 15:01
  • @Joe Thanks a lot. I did check it . Still , have one Q that if I don’t do multiplication operation on the determinant . Would the answer as I got under addition , subtraction operation of row , column yield value independent of theta always ? – Rider Apr 18 '21 at 15:07
  • @Rider the determinant is uniquely defined by the matrix you're given. If you change the matrix you can change the determinant. – CyclotomicField Apr 18 '21 at 15:28
  • To second what @CyclotomicField has already said, this problem is independent of theta only because of this matrix. For a different matrix with theta terms, it will most likely not be invariant to changes in theta, but you would have to check. But if you get a matrix from row addition/subtraction some number of times from this matrix, then it will have the same determinant, which is invariant under changes to theta. – Joe Apr 18 '21 at 17:03

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