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Find the determinant of the following matrix: $$\begin{pmatrix}\cos\left(a_{1}-b_{1}\right) & \cos\left(a_{1}-b_{2}\right) & \cos\left(a_{1}-b_{3}\right)\\ \cos\left(a_{2}-b_{1}\right) & \cos\left(a_{2}-b_{2}\right) & \cos\left(a_{2}-b_{3}\right)\\ \cos\left(a_{3}-b_{1}\right) & \cos\left(a_{3}-b_{2}\right) & \cos\left(a_{3}-b_{3}\right) \end{pmatrix}$$For $a_1,\dots ,a_3,b_1,\dots, b_3\in\mathbb{R}$

I'm completely stumped honestly. I tried using the cosine addition identity to open the cosine, but I wasn't able to find how it helps me, and even for a $2\times2$ version of the matrix I wasn't really sure what to do. Any help?

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    $\begingroup$ I think you should try to express this determinant as a product of two matrices determinant. $\endgroup$ – Mann Jun 5 '15 at 8:02
  • $\begingroup$ Without solving!... The answer is $0$... (Just if you want the answer immediately without steps...) >>> {Wolfram Alpha} (goo.gl/jMXjod) Sorry $\endgroup$ – NeilRoy Jun 5 '15 at 8:12
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    $\begingroup$ @NeilRoy I did that myself as well :P Unfortunately it didn't help me too much... $\endgroup$ – Nescio Jun 5 '15 at 8:15
  • $\begingroup$ Wait, I will write an answer if you didn't get the hint yet. $\endgroup$ – Mann Jun 5 '15 at 8:16
  • $\begingroup$ @Mann I didn't get it honestly :/ Still thinking about it, I do see that there seems to be a pattern of matrix multiplication after expanding to $\cos(a_i)\cos(b_j)+\sin(a_i)\sin(b_j)$ but wasn't able to take it any further than that... $\endgroup$ – Nescio Jun 5 '15 at 8:18
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The matrix is the product \begin{eqnarray} \begin{pmatrix}\cos a_1& \sin a_1& 0\\ \cos a_2&\sin a_2& 0\\ \cos a_3&\sin a_3 &0\end{pmatrix}\begin{pmatrix}\cos b_1& \cos b_2&\cos b_3\\ \sin b_1&\sin b_2&\sin b_3\\ 0&0&0\end{pmatrix} \end{eqnarray} and thus the determinant is 0.

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  • $\begingroup$ Just simply augment the matrices with zero rows and columns. See my edited post. $\endgroup$ – Alex Fok Jun 5 '15 at 8:19
  • $\begingroup$ Well that makes sense, I failed to see it split like that myself ><". Thank you $\endgroup$ – Nescio Jun 5 '15 at 8:20
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    $\begingroup$ Nice proof!!!!!!!!!!! $\endgroup$ – Empty Jun 5 '15 at 8:20
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Notice that your determinant is same as , $$\begin{eqnarray} \begin{vmatrix}\cos a_1& \sin a_1& 0\\ \cos a_2&\sin a_2& 0\\ \cos a_3&\sin a_3 &0\end{vmatrix}\begin{vmatrix}\cos b_1& \sin b_1& 0\\ \cos b_2&\sin b_2& 0\\ \cos b_3& \sin b_3&0\end{vmatrix} \end{eqnarray}$$

In determinant multiplication , its is similar to matrix multiplication but the multiplication of $R_iC_j's$ is equivalent to $R_iR_j's$ since $det(A)=det(A^T)$.

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Find the determinant of the following trigonometric matrix(unit in radian):

sin1 sin2 sin3 sin4 sin5

sin2 sin3 sin4 sin5 sin6

sin3 sin4 sin5 sin6 sin7

sin4 sin5 sin6 sin7 sin8

sin5 sin6 sin7 sin8 sin9

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  • $\begingroup$ How does this answer the question? $\endgroup$ – theyaoster Dec 11 '17 at 21:45

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