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

Question

I meet a difficult determinant question as the followings:
$$ \text{Matrix A is given as:} $$ $$ A=\begin{bmatrix}\frac{\partial x}{\partial r}&\frac{\partial x}{\partial\theta}&\frac{\partial x}{\partial\phi}\\\frac{\partial y}{\partial r}&\frac{\partial y}{\partial\theta}&\frac{\partial y}{\partial\phi}\\\frac{\partial z}{\partial r}&\frac{\partial z}{\partial\theta}&\frac{\partial z}{\partial\phi}\end{bmatrix} $$ $$ \text{where }x=r\sin\theta\cos\phi\text{, }y=r\sin\theta\sin\phi\text{, and }z=r\cos\theta.\text{ Find determinants }\det{(A)}\text{, }\det{(A^{-1})}\text{, and }\det{(A^2)}. $$ I tried to simplify it, but just got: $$ A=\begin{bmatrix}\sin\theta\cos\phi&r\cos\theta\cos\phi&-r\sin\theta\sin\phi\\\sin\theta\sin\phi&r\cos\theta\sin\phi&r\sin\theta\cos\phi\\\cos\theta&-r\sin\theta&0\end{bmatrix} $$ Because it is wired, I have also searched the Internet. But till now all I know is that this is just a spherical-Cartesian transformation formula using Jacobian matrix. (Maybe we can make a breakthough here?)
I can only solve $\det{(A)}$ by directly calculating it, $\det{(A)}=r^2\sin\theta$ .
However, I think it is still hard to find the inverse matrix, needless to say the huge calculation to get $A^2$. As I think, there must be some ways to simplify it.
Could anyone kindly teach me that whether there is any way to simplify $A$, so as to calculate the determinant?Thank you!

Answer 1

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.

Answer 2

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 $$\det{(I)}=\det{(AA^{-1})}=\det{(A)}\det{(A^{-1})}=r^2\sin{\theta}\det{(A^{-1})}=1$$ $$\therefore \det{(A^{-1})}=\frac{1}{r^2\sin{\theta}}$$ $$\det{(A^2)}=(\det{(A)})^2=(r^2\sin{\theta})^2=r^4\sin^2{\theta}$$