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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.

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

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add the first two rows, you get the third. The rows are dependent

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HINT

Apply simple linear transformations.

Add the second row to the first one.

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