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

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!

• Why can't you "solve" (=calculate?) $\det A$? It's bare for calculation. – Saucy O'Path Apr 8 at 11:03
• Thank you @Saucy O'Path, it is true that we just need to calculate, then we could get det(A)=r^2sinθ. I initially thought that maybe we could simplify A by add/sub the row/cow of the matrix to make things easier...... – Peter Nova Apr 8 at 11:26
• Nah, it's largely a matter of grouping all the $\sin^2+\cos^2$ that appear when you make the products. – Saucy O'Path Apr 8 at 11:34
• Of course you can calculate the determinant by reducing it using properties of determinant (like this for example). This is a standard result. – StubbornAtom Apr 8 at 12:20

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