What's the best method to find the determinant of a trigonometry matrix?

Question

I'm facing this trigonometry matrix, it's painful to find it with elimination or co-factors. I wonder if there's any good method or trick to tackle this.

$$A=\begin{pmatrix}\sin\alpha \cos\theta &\beta \cos \alpha \cos \theta &-\beta \sin \alpha \sin \theta\\ \sin\alpha\sin\theta &\beta\cos\alpha\sin\theta & \beta\sin\alpha\cos\theta \\ \cos\alpha &-\beta\sin\alpha&0\\\end{pmatrix}$$

I'm not asking for the answer, I just want tips and hints.

Answer 1

Hints:

• The determinant of the transposed matrix is the same. That is, $|A| = |A^T|$ for any matrix $A$.
• The determinant of a $3 \times 3$ matrix with row vectors $\vec{a}, \vec{b}, \vec{c}$ is equal to the signed volume of the parallelepiped defined by these vectors.
• Can we not calculate the volume in a different, less painful manner? Recall the basic formula for calculating the volume of a parallelepiped: area of the base times the height with respect to the base. Can we express this using scalar and vector products?