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












5












$begingroup$


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!






linear-algebra matrices determinant







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  • $begingroup$
    Why can't you "solve" (=calculate?) $det A$? It's bare for calculation.
    $endgroup$
    – Saucy O'Path
    yesterday










  • $begingroup$
    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......
    $endgroup$
    – Peter Nova
    yesterday












  • $begingroup$
    Nah, it's largely a matter of grouping all the $sin^2+cos^2$ that appear when you make the products.
    $endgroup$
    – Saucy O'Path
    yesterday










  • $begingroup$
    Of course you can calculate the determinant by reducing it using properties of determinant (like this for example). This is a standard result.
    $endgroup$
    – StubbornAtom
    yesterday


















5












$begingroup$


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!






linear-algebra matrices determinant







share|cite|improve this question









New contributor




Peter Nova is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$












  • $begingroup$
    Why can't you "solve" (=calculate?) $det A$? It's bare for calculation.
    $endgroup$
    – Saucy O'Path
    yesterday










  • $begingroup$
    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......
    $endgroup$
    – Peter Nova
    yesterday












  • $begingroup$
    Nah, it's largely a matter of grouping all the $sin^2+cos^2$ that appear when you make the products.
    $endgroup$
    – Saucy O'Path
    yesterday










  • $begingroup$
    Of course you can calculate the determinant by reducing it using properties of determinant (like this for example). This is a standard result.
    $endgroup$
    – StubbornAtom
    yesterday
















5












5








5


1



$begingroup$


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!






linear-algebra matrices determinant







share|cite|improve this question









New contributor




Peter Nova is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$




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!






linear-algebra matrices determinant




linear-algebra matrices determinant






share|cite|improve this question









New contributor




Peter Nova is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|cite|improve this question









New contributor




Peter Nova is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|cite|improve this question




share|cite|improve this question








edited yesterday







Peter Nova













New contributor




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Check out our Code of Conduct.









asked yesterday









Peter NovaPeter Nova

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284




New contributor




Peter Nova is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.





New contributor





Peter Nova is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






Peter Nova is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.












  • $begingroup$
    Why can't you "solve" (=calculate?) $det A$? It's bare for calculation.
    $endgroup$
    – Saucy O'Path
    yesterday










  • $begingroup$
    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......
    $endgroup$
    – Peter Nova
    yesterday












  • $begingroup$
    Nah, it's largely a matter of grouping all the $sin^2+cos^2$ that appear when you make the products.
    $endgroup$
    – Saucy O'Path
    yesterday










  • $begingroup$
    Of course you can calculate the determinant by reducing it using properties of determinant (like this for example). This is a standard result.
    $endgroup$
    – StubbornAtom
    yesterday




















  • $begingroup$
    Why can't you "solve" (=calculate?) $det A$? It's bare for calculation.
    $endgroup$
    – Saucy O'Path
    yesterday










  • $begingroup$
    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......
    $endgroup$
    – Peter Nova
    yesterday












  • $begingroup$
    Nah, it's largely a matter of grouping all the $sin^2+cos^2$ that appear when you make the products.
    $endgroup$
    – Saucy O'Path
    yesterday










  • $begingroup$
    Of course you can calculate the determinant by reducing it using properties of determinant (like this for example). This is a standard result.
    $endgroup$
    – StubbornAtom
    yesterday


















$begingroup$
Why can't you "solve" (=calculate?) $det A$? It's bare for calculation.
$endgroup$
– Saucy O'Path
yesterday




$begingroup$
Why can't you "solve" (=calculate?) $det A$? It's bare for calculation.
$endgroup$
– Saucy O'Path
yesterday












$begingroup$
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......
$endgroup$
– Peter Nova
yesterday






$begingroup$
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......
$endgroup$
– Peter Nova
yesterday














$begingroup$
Nah, it's largely a matter of grouping all the $sin^2+cos^2$ that appear when you make the products.
$endgroup$
– Saucy O'Path
yesterday




$begingroup$
Nah, it's largely a matter of grouping all the $sin^2+cos^2$ that appear when you make the products.
$endgroup$
– Saucy O'Path
yesterday












$begingroup$
Of course you can calculate the determinant by reducing it using properties of determinant (like this for example). This is a standard result.
$endgroup$
– StubbornAtom
yesterday






$begingroup$
Of course you can calculate the determinant by reducing it using properties of determinant (like this for example). This is a standard result.
$endgroup$
– StubbornAtom
yesterday












2 Answers
2






active

oldest

votes


















2












$begingroup$

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






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you so much! I thought too much on determinant calculation and simplification, but ignored the most important thing -- A is also a matrix. Thank you! I will remember this lesson.
    $endgroup$
    – Peter Nova
    yesterday



















4












$begingroup$

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.






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$endgroup$













  • $begingroup$
    Yeah... I just only thought about the calculation method from a determinant prospect, just want to somehow simplify it. But I ignored A is also a matrix... Thank you for your answer too! (But I think maybe I cannot think up this B immediately =-O , I will fight in the future study)
    $endgroup$
    – Peter Nova
    yesterday












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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

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active

oldest

votes






active

oldest

votes









2












$begingroup$

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






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you so much! I thought too much on determinant calculation and simplification, but ignored the most important thing -- A is also a matrix. Thank you! I will remember this lesson.
    $endgroup$
    – Peter Nova
    yesterday
















2












$begingroup$

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






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you so much! I thought too much on determinant calculation and simplification, but ignored the most important thing -- A is also a matrix. Thank you! I will remember this lesson.
    $endgroup$
    – Peter Nova
    yesterday














2












2








2





$begingroup$

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






share|cite|improve this answer









$endgroup$



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







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answered yesterday









Peter ForemanPeter Foreman

6,6341318




6,6341318












  • $begingroup$
    Thank you so much! I thought too much on determinant calculation and simplification, but ignored the most important thing -- A is also a matrix. Thank you! I will remember this lesson.
    $endgroup$
    – Peter Nova
    yesterday


















  • $begingroup$
    Thank you so much! I thought too much on determinant calculation and simplification, but ignored the most important thing -- A is also a matrix. Thank you! I will remember this lesson.
    $endgroup$
    – Peter Nova
    yesterday
















$begingroup$
Thank you so much! I thought too much on determinant calculation and simplification, but ignored the most important thing -- A is also a matrix. Thank you! I will remember this lesson.
$endgroup$
– Peter Nova
yesterday




$begingroup$
Thank you so much! I thought too much on determinant calculation and simplification, but ignored the most important thing -- A is also a matrix. Thank you! I will remember this lesson.
$endgroup$
– Peter Nova
yesterday











4












$begingroup$

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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Yeah... I just only thought about the calculation method from a determinant prospect, just want to somehow simplify it. But I ignored A is also a matrix... Thank you for your answer too! (But I think maybe I cannot think up this B immediately =-O , I will fight in the future study)
    $endgroup$
    – Peter Nova
    yesterday
















4












$begingroup$

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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Yeah... I just only thought about the calculation method from a determinant prospect, just want to somehow simplify it. But I ignored A is also a matrix... Thank you for your answer too! (But I think maybe I cannot think up this B immediately =-O , I will fight in the future study)
    $endgroup$
    – Peter Nova
    yesterday














4












4








4





$begingroup$

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.






share|cite|improve this answer









$endgroup$



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.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered yesterday









John HughesJohn Hughes

65.4k24293




65.4k24293












  • $begingroup$
    Yeah... I just only thought about the calculation method from a determinant prospect, just want to somehow simplify it. But I ignored A is also a matrix... Thank you for your answer too! (But I think maybe I cannot think up this B immediately =-O , I will fight in the future study)
    $endgroup$
    – Peter Nova
    yesterday


















  • $begingroup$
    Yeah... I just only thought about the calculation method from a determinant prospect, just want to somehow simplify it. But I ignored A is also a matrix... Thank you for your answer too! (But I think maybe I cannot think up this B immediately =-O , I will fight in the future study)
    $endgroup$
    – Peter Nova
    yesterday
















$begingroup$
Yeah... I just only thought about the calculation method from a determinant prospect, just want to somehow simplify it. But I ignored A is also a matrix... Thank you for your answer too! (But I think maybe I cannot think up this B immediately =-O , I will fight in the future study)
$endgroup$
– Peter Nova
yesterday




$begingroup$
Yeah... I just only thought about the calculation method from a determinant prospect, just want to somehow simplify it. But I ignored A is also a matrix... Thank you for your answer too! (But I think maybe I cannot think up this B immediately =-O , I will fight in the future study)
$endgroup$
– Peter Nova
yesterday










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