# Show the following matrix has determinant = 0

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:

• Is "$\operatorname{sen}$" what most of us would call "$\sin$"? – Arthur Apr 3 at 15:44
• Yes, sorry, forgot to mention that – 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.
• 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. – Arthur Apr 3 at 16:01