How to find the determinant of this matrix? (A spherical-Cartesian transformation Jacobian matrix)

If you were really clever (e.g., if you already knew the answer, or thought hard about what a Jacobian in a different coordinate system represents), you could compute $\det(A)$ by computing $\det(B)\det(A) = \det (BA)$, where $\det B$ was particularly easy.

Picking $$ B = \pmatrix{\cos \phi & \sin \phi & 0 \\ -\sin \phi & \cos \phi, & 0 \\ 0 & 0 & 1} $$ generates a matrix $BA$ whose form is rather simpler than that of $A$ (there's no $\phi$, for instance!), while $\det B$ is evidently $1$.

But I thing the question-asker's intent here is that you're just supposed to do the algebra and practice trig simplification.

By using the Rule of Sarrus, $$\begin{align} \det{(A)}&=(\sin{\theta}\cos{\phi})(r\cos{\theta}\sin{\phi})(0)\\ &\,\,\,+(\sin{\theta}\sin{\phi})(-r\sin{\theta})(-r\sin{\theta\sin{\phi}})\\ &\,\,\,+(\cos{\theta})(r\cos{\theta}\cos{\phi})(r\sin{\theta}\cos{\phi})\\ &\,\,\,-(-r\sin{\theta}\sin{\phi})(r\cos{\theta}\sin{\phi})(\cos{\theta})\\ &\,\,\,-(r\sin{\theta}\cos{\phi})(-r\sin{\theta})(\sin{\theta}\cos{\phi})\\ &\,\,\,-(0)(r\cos{\theta}\cos{\phi})(\sin{\theta}\sin{\phi})\\ &=0+r^2\sin^3{\theta}\sin^2{\phi}+r^2\sin{\theta}\cos^2{\theta}\cos^2{\phi}+r^2\sin{\theta}\sin^2{\phi}\cos^2{\theta}+r^2\sin^3{\theta}\cos^2{\phi}-0\\ &=r^2\sin^3{\theta}(\sin^2{\phi}+\cos^2{\phi})+r^2\sin{\theta}\cos^2{\theta}(\sin^2{\phi}+\cos^2{\phi})\\ &=r^2\sin^3{\theta}+r^2\sin{\theta}\cos^2{\theta}\\ &=r^2\sin{\theta}(\sin^2{\theta}+\cos^2{\theta})\\ &\boxed{=r^2\sin{\theta}}\\ \end{align}$$ Now in order to find $\det{(A^{-1})}$ and $\det{(A^2)}$ we can use the fact that $\det{(AB)}=\det{(A)}\cdot\det{(B)}$ to get $$\det{(I)}=\det{(AA^{-1})}=\det{(A)}\det{(A^{-1})}=r^2\sin{\theta}\det{(A^{-1})}=1$$ $$\therefore \det{(A^{-1})}=\frac{1}{r^2\sin{\theta}}$$ $$\det{(A^2)}=(\det{(A)})^2=(r^2\sin{\theta})^2=r^4\sin^2{\theta}$$