Kdtyjb

I meet a difficult determinant question as the followings:
$$
text{Matrix A is given as:}
$$
$$
A=begin{bmatrix}frac{partial x}{partial r}&frac{partial x}{partialtheta}&frac{partial x}{partialphi}\frac{partial y}{partial r}&frac{partial y}{partialtheta}&frac{partial y}{partialphi}\frac{partial z}{partial r}&frac{partial z}{partialtheta}&frac{partial z}{partialphi}end{bmatrix}
$$
$$
text{where }x=rsinthetacosphitext{, }y=rsinthetasinphitext{, and }z=rcostheta.text{ Find determinants }det{(A)}text{, }det{(A^{-1})}text{, and }det{(A^2)}.
$$
I tried to simplify it, but just got:
$$
A=begin{bmatrix}sinthetacosphi&rcosthetacosphi&-rsinthetasinphi\sinthetasinphi&rcosthetasinphi&rsinthetacosphi\costheta&-rsintheta&0end{bmatrix}
$$
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, $det{(A)}=r^2sintheta$ .

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!

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

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 $1$.

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