Question

The value of the determinant $\Delta =\left|\begin{array}{ccc}\cos \alpha \cos \beta & \cos \alpha \sin \beta & -\sin \alpha \\ -\sin \beta & \cos \beta & 0\\ \sin \alpha \cos \beta & \sin \alpha \sin \beta & \cos \alpha \end{array}\right|$ is

Answer 1

$\Delta =\left|\begin{array}{ccc}\cos \alpha \cos \beta & \cos \alpha \sin \beta & -\sin \alpha \\ -\sin \beta & \cos \beta & 0\\ \sin \alpha \cos \beta & \sin \alpha \sin \beta & \cos \alpha \end{array}\right|$
Expanding along the second law
$|A|=\sin \beta \left|\begin{array}{cc}\cos \alpha \sin \beta & -\sin \alpha \\ \sin \alpha \sin \beta & \cos \alpha \end{array}\right|+\cos \beta \left|\begin{array}{cc}\cos \alpha \cos \beta & -\sin \alpha \\ \sin \alpha \cos \beta & \cos \alpha \end{array}\right|$
$-0\left|\begin{array}{cc}\cos \alpha \cos \beta & \cos \alpha \sin \beta \\ \sin \alpha \cos \beta & \sin \alpha \sin \beta \end{array}\right|$
$\Rightarrow |A|=\sin \beta (\cos \alpha \times \cos \alpha \sin \beta +\sin \alpha \times \sin \alpha \sin \beta )$
$+\cos \beta (\cos \alpha \cos \beta \times \cos \alpha +\sin \alpha \times \sin \alpha \cos \beta )-0$
$|A|={\sin }^2\beta ({\cos }^2\alpha +{\sin }^2\alpha )+{\cos }^2\beta ({\cos }^2\alpha +{\sin }^2\alpha )$
$|A|={\sin }^2\beta \left(1\right)+{\cos }^2\beta \left(1\right)$
$|A|={\sin }^2\beta +{\cos }^2\beta$
$|A|=1$