How to find the determinant of this matrix? (A spherical-Cartesian transformation Jacobian matrix)
$begingroup$
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
asked yesterday
Peter NovaPeter Nova
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.
$endgroup$
add a comment |
$begingroup$
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
asked yesterday
Peter NovaPeter Nova
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.
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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
asked yesterday
Peter NovaPeter Nova
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.
$endgroup$
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
asked yesterday
Peter NovaPeter Nova
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.
asked yesterday
Peter NovaPeter Nova
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.
edited yesterday
Peter Nova
asked yesterday
Peter NovaPeter Nova
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.
asked yesterday
Peter NovaPeter Nova
284
asked yesterday
Peter NovaPeter Nova
284
284
New contributor
Peter Nova is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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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
add a comment |
$begingroup$
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
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
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
$$det{(I)}=det{(AA^{-1})}=det{(A)}det{(A^{-1})}=r^2sin{theta}det{(A^{-1})}=1$$
$$therefore det{(A^{-1})}=frac{1}{r^2sin{theta}}$$
$$det{(A^2)}=(det{(A)})^2=(r^2sin{theta})^2=r^4sin^2{theta}$$
answered yesterday
Peter ForemanPeter Foreman
6,6341318
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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.
answered yesterday
John HughesJohn Hughes
65.4k24293
$endgroup$
$begingroup$
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
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
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votes
active
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votes
$begingroup$
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
$$det{(I)}=det{(AA^{-1})}=det{(A)}det{(A^{-1})}=r^2sin{theta}det{(A^{-1})}=1$$
$$therefore det{(A^{-1})}=frac{1}{r^2sin{theta}}$$
$$det{(A^2)}=(det{(A)})^2=(r^2sin{theta})^2=r^4sin^2{theta}$$
answered yesterday
Peter ForemanPeter Foreman
6,6341318
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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
$$det{(I)}=det{(AA^{-1})}=det{(A)}det{(A^{-1})}=r^2sin{theta}det{(A^{-1})}=1$$
$$therefore det{(A^{-1})}=frac{1}{r^2sin{theta}}$$
$$det{(A^2)}=(det{(A)})^2=(r^2sin{theta})^2=r^4sin^2{theta}$$
answered yesterday
Peter ForemanPeter Foreman
6,6341318
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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
$$det{(I)}=det{(AA^{-1})}=det{(A)}det{(A^{-1})}=r^2sin{theta}det{(A^{-1})}=1$$
$$therefore det{(A^{-1})}=frac{1}{r^2sin{theta}}$$
$$det{(A^2)}=(det{(A)})^2=(r^2sin{theta})^2=r^4sin^2{theta}$$
answered yesterday
Peter ForemanPeter Foreman
6,6341318
$endgroup$
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
$$det{(I)}=det{(AA^{-1})}=det{(A)}det{(A^{-1})}=r^2sin{theta}det{(A^{-1})}=1$$
$$therefore det{(A^{-1})}=frac{1}{r^2sin{theta}}$$
$$det{(A^2)}=(det{(A)})^2=(r^2sin{theta})^2=r^4sin^2{theta}$$
answered yesterday
Peter ForemanPeter Foreman
6,6341318
answered yesterday
Peter ForemanPeter Foreman
6,6341318
answered yesterday
Peter ForemanPeter Foreman
6,6341318
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
add a comment |
$begingroup$
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
add a comment |
$begingroup$
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.
answered yesterday
John HughesJohn Hughes
65.4k24293
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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.
answered yesterday
John HughesJohn Hughes
65.4k24293
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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.
answered yesterday
John HughesJohn Hughes
65.4k24293
$endgroup$
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.
answered yesterday
John HughesJohn Hughes
65.4k24293
answered yesterday
John HughesJohn Hughes
65.4k24293
answered yesterday
John HughesJohn Hughes
65.4k24293
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
add a comment |
$begingroup$
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
add a comment |
Peter Nova is a new contributor. Be nice, and check out our Code of Conduct.
Peter Nova is a new contributor. Be nice, and check out our Code of Conduct.
Peter Nova is a new contributor. Be nice, and check out our Code of Conduct.
Peter Nova is a new contributor. Be nice, and check out our Code of Conduct.
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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