# Given scalars $\alpha, \beta, \gamma \in \mathbb{F},$ prove that the following matrix is not invertible

Question:

Given scalars $\alpha, \beta, \gamma \in \mathbb{F},$ prove that the following matrix is not invertible: $$\begin{bmatrix} sin^2(\alpha) & sin^2(\beta) & sin^2(\gamma) \\ cos^2(\alpha) & cos^2(\beta) & cos^2(\gamma) \\ 1 & 1 & 1 \\ \end{bmatrix}$$

My Steps: I calculated the determinant which resulted in: $$sin^2(\alpha) \cdot (cos^2(\beta)-cos^2(\gamma))-sin^2(\beta) \cdot (cos^2(\alpha)-cos^2(\gamma))+sin^2(\gamma) \cdot (cos^2(\alpha) - cos^2(\gamma)$$ Since each term when distributed will contain a $sin^2 \cdot cos^2$ pair, it follows that values for $\alpha = \beta =\gamma = 0 + \frac{\pi}{2} \cdot k$ will result in a determinant that is equal to zero meaning that the matrix is not invertible.

However, if $\alpha, \beta, \gamma$ do not equal some variation of $0 + \frac{\pi}{2} \cdot k$, it seems that the determinant would not be equal to zero making the matrix not invertible for only a handful of special cases.

• How general is $\mathbb{F}$ when you can take sine and cosine of its numbers? – Jeppe Stig Nielsen Nov 7 '17 at 22:21

Notice that $$\sin^2(\alpha)+\cos^2(\alpha)=\sin^2(\beta)+\cos^2(\beta)=\sin^2(\gamma)+\cos^2(\gamma)=1.$$ Therefore the third row is the sum of the other two (is a linear combination) Hence the rows of the matrix are linearly dependent. That implies that the matrix is not invertible.