Prove that the matrix is invertible for any value of $\beta$.
I've done several exercises of this type. But I'm not sure with this one:
$$\begin{bmatrix}\cos \beta & \sin \beta & 0\\ -\sin\beta & \cos \beta & 0\\0&0&1\end{bmatrix}$$
My knowledge of trigonometric functions is pretty poor, but anyway, let's see:
The matrix would be invertible if its determinant is $\not = 0$. The determinant of this matrix is equivalent to
$$(\cos\beta \cdot \cos \beta) – (\sin \beta \cdot -\sin\beta)$$
Alright, I'm not sure what am I doing. Surely there is a basic law I'm missing, but I can't quite grasp it.
Best Answer
The basic law that you're missing is $\cos^2 \beta + \sin^2 \beta = 1$ for any $\beta$.