I just faced this problem where i am asked to show this matrix has determinant = 0 and I got stuck and can't find a way out of this...would really appreciate if someone could help $$ \begin{pmatrix} \cos \alpha & \sin \alpha & \sin (\alpha + \theta) \\ \cos \beta & \sin \beta & \sin (\beta + \theta) \\ \cos \gamma & \sin \gamma & \sin (\gamma + \theta) \\ \end{pmatrix} $$

My attempt: my try

  • $\begingroup$ Is "$\operatorname{sen}$" what most of us would call "$\sin$"? $\endgroup$ – Arthur Apr 3 at 15:44
  • $\begingroup$ Yes, sorry, forgot to mention that $\endgroup$ – Tomás Lopes Apr 3 at 15:45

The third column is equal to $\sin\theta$ times the first column plus $\cos\theta$ times the second column, by the well-known formula $$ \sin(u)\cos(v) + \cos(u)\sin(v) = \sin(u + v) $$ This makes the columns linearly dependent and therefore the matrix is singular.

  • $\begingroup$ Really appreciate your fast reply....i was so focused in calculating the determinant I absolutely forgot about that... $\endgroup$ – Tomás Lopes Apr 3 at 15:50
  • $\begingroup$ If they had asked for any value other than $0$, then calculating the determinant and hoping to spot a whole lot of trigonometric identities to simplify it might have been the only way. As it happens, $0$ is a very special value when it comes to determinants, and therefore we have a lot of different ways to determine it. $\endgroup$ – Arthur Apr 3 at 16:01

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