Determinant of a matrix with trigonometry functions.

Question

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.

Answer 1

It is an orthogonal matrix, since the dot product of any two columns is zero and the euclidean norm of any column is one. Hence such a matrix represents an isometry of $\mathbb{R}^3$, hence it is invertible.

---

Someone thinks this is nuking mosquitoes. Ok, let us make things easier.

The determinant of such a matrix is $\cos^2\beta+\sin^2\beta$. Hence, by the Pythagorean theorem, the determinant of the matrix is always one, hence the matrix is invertible.

Answer 2

The basic law that you're missing is $\cos^2 \beta + \sin^2 \beta = 1$ for any $\beta$.