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I meet a difficult determinant question as the followings:
$$
textMatrix A is given as:
$$
$$
A=beginbmatrixfracpartial xpartial r&fracpartial xpartialtheta&fracpartial xpartialphi\fracpartial ypartial r&fracpartial ypartialtheta&fracpartial ypartialphi\fracpartial zpartial r&fracpartial zpartialtheta&fracpartial zpartialphiendbmatrix
$$
$$
textwhere x=rsinthetacosphitext, y=rsinthetasinphitext, and z=rcostheta.text Find determinants det(A)text, det(A^-1)text, and det(A^2).
$$
I tried to simplify it, but just got:
$$
A=beginbmatrixsinthetacosphi&rcosthetacosphi&-rsinthetasinphi\sinthetasinphi&rcosthetasinphi&rsinthetacosphi\costheta&-rsintheta&0endbmatrix
$$
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,
$$beginalign
det(A)&=(sinthetacosphi)(rcosthetasinphi)(0)\
&,,,+(sinthetasinphi)(-rsintheta)(-rsinthetasinphi)\
&,,,+(costheta)(rcosthetacosphi)(rsinthetacosphi)\
&,,,-(-rsinthetasinphi)(rcosthetasinphi)(costheta)\
&,,,-(rsinthetacosphi)(-rsintheta)(sinthetacosphi)\
&,,,-(0)(rcosthetacosphi)(sinthetasinphi)\
&=0+r^2sin^3thetasin^2phi+r^2sinthetacos^2thetacos^2phi+r^2sinthetasin^2phicos^2theta+r^2sin^3thetacos^2phi-0\
&=r^2sin^3theta(sin^2phi+cos^2phi)+r^2sinthetacos^2theta(sin^2phi+cos^2phi)\
&=r^2sin^3theta+r^2sinthetacos^2theta\
&=r^2sintheta(sin^2theta+cos^2theta)\
&boxed=r^2sintheta\
endalign$$
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
$$det(I)=det(AA^-1)=det(A)det(A^-1)=r^2sinthetadet(A^-1)=1$$
$$therefore det(A^-1)=frac1r^2sintheta$$
$$det(A^2)=(det(A))^2=(r^2sintheta)^2=r^4sin^2theta$$

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
$$
B = pmatrixcos phi & sin phi & 0 \
-sin phi & cos phi, & 0 \
0 & 0 & 1
$$
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.