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How to find the determinant of this matrix? (A spherical-Cartesian transformation Jacobian matrix)
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)How to compute the following JacobianCalculating the Jacobian of inverse functionsMatrix calculation with sinusoidsUsing the Jacobian matrix to find surface area without a change of basis.Transforming vector field into spherical coordinates. Why and how does this method work?Compute the determinant of circulant matrix with entries $cos jtheta$Find the rotation/reflection angle for orthogonal matrix AJacobian matrix vs. Transformation matrixFurthest point in direction ellipsoid with Newton's methodCalculate the determinant of this $5 times 5$ matrixWhat is the correct matrix form for transforming spherical coordinate to Cartesian?
$begingroup$
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!
linear-algebra matrices determinant
asked Apr 8 at 10:58
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:
$$
textMatrix A is given as:
$$
$$
A=beginbmatrixfracpartial xpartial r&fracpartial xpartialtheta&fracpartial xpartialphi\fracpartial ypartial r&fracpartial ypartialtheta&fracpartial ypartialphi\fracpartial zpartial r&fracpartial zpartialtheta&fracpartial zpartialphiendbmatrix
$$
$$
textwhere x=rsinthetacosphitext, y=rsinthetasinphitext, and z=rcostheta.text Find determinants det(A)text, det(A^-1)text, and det(A^2).
$$
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!
linear-algebra matrices determinant
asked Apr 8 at 10:58
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
Apr 8 at 11:03
$begingroup$
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
Apr 8 at 11:26
$begingroup$
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
Apr 8 at 11:34
$begingroup$
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
Apr 8 at 12:20
add a comment |
$begingroup$
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!
linear-algebra matrices determinant
asked Apr 8 at 10:58
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:
$$
textMatrix A is given as:
$$
$$
A=beginbmatrixfracpartial xpartial r&fracpartial xpartialtheta&fracpartial xpartialphi\fracpartial ypartial r&fracpartial ypartialtheta&fracpartial ypartialphi\fracpartial zpartial r&fracpartial zpartialtheta&fracpartial zpartialphiendbmatrix
$$
$$
textwhere x=rsinthetacosphitext, y=rsinthetasinphitext, and z=rcostheta.text Find determinants det(A)text, det(A^-1)text, and det(A^2).
$$
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!
linear-algebra matrices determinant
linear-algebra matrices determinant
asked Apr 8 at 10:58
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 Apr 8 at 10:58
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 Apr 8 at 11:35
Peter Nova
asked Apr 8 at 10:58
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 Apr 8 at 10:58
Peter NovaPeter Nova
284
asked Apr 8 at 10:58
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.
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
Apr 8 at 11:03
$begingroup$
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
Apr 8 at 11:26
$begingroup$
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
Apr 8 at 11:34
$begingroup$
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
Apr 8 at 12:20
add a comment |
$begingroup$
Why can't you "solve" (=calculate?) $det A$? It's bare for calculation.
$endgroup$
– Saucy O'Path
Apr 8 at 11:03
$begingroup$
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
Apr 8 at 11:26
$begingroup$
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
Apr 8 at 11:34
$begingroup$
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
Apr 8 at 12:20
$begingroup$
Why can't you "solve" (=calculate?) $det A$? It's bare for calculation.
$endgroup$
– Saucy O'Path
Apr 8 at 11:03
$begingroup$
Why can't you "solve" (=calculate?) $det A$? It's bare for calculation.
$endgroup$
– Saucy O'Path
Apr 8 at 11:03
$begingroup$
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
Apr 8 at 11:26
$begingroup$
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
Apr 8 at 11:26
$begingroup$
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
Apr 8 at 11:34
$begingroup$
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
Apr 8 at 11:34
$begingroup$
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
Apr 8 at 12:20
$begingroup$
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
Apr 8 at 12:20
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
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
$$det(I)=det(AA^-1)=det(A)det(A^-1)=r^2sinthetadet(A^-1)=1$$
$$therefore det(A^-1)=frac1r^2sintheta$$
$$det(A^2)=(det(A))^2=(r^2sintheta)^2=r^4sin^2theta$$
answered Apr 8 at 11:40
Peter ForemanPeter Foreman
7,3181319
$endgroup$
$begingroup$
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
Apr 8 at 12:03
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
$$
B = pmatrixcos phi & sin phi & 0 \
-sin phi & cos phi, & 0 \
0 & 0 & 1
$$
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.
answered Apr 8 at 11:45
John HughesJohn Hughes
65.5k24293
$endgroup$
$begingroup$
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
Apr 8 at 12:08
add a comment |
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2 Answers
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2 Answers
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$begingroup$
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
$$det(I)=det(AA^-1)=det(A)det(A^-1)=r^2sinthetadet(A^-1)=1$$
$$therefore det(A^-1)=frac1r^2sintheta$$
$$det(A^2)=(det(A))^2=(r^2sintheta)^2=r^4sin^2theta$$
answered Apr 8 at 11:40
Peter ForemanPeter Foreman
7,3181319
$endgroup$
$begingroup$
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
Apr 8 at 12:03
add a comment |
$begingroup$
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
$$det(I)=det(AA^-1)=det(A)det(A^-1)=r^2sinthetadet(A^-1)=1$$
$$therefore det(A^-1)=frac1r^2sintheta$$
$$det(A^2)=(det(A))^2=(r^2sintheta)^2=r^4sin^2theta$$
answered Apr 8 at 11:40
Peter ForemanPeter Foreman
7,3181319
$endgroup$
$begingroup$
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
Apr 8 at 12:03
add a comment |
$begingroup$
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
$$det(I)=det(AA^-1)=det(A)det(A^-1)=r^2sinthetadet(A^-1)=1$$
$$therefore det(A^-1)=frac1r^2sintheta$$
$$det(A^2)=(det(A))^2=(r^2sintheta)^2=r^4sin^2theta$$
answered Apr 8 at 11:40
Peter ForemanPeter Foreman
7,3181319
$endgroup$
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
$$det(I)=det(AA^-1)=det(A)det(A^-1)=r^2sinthetadet(A^-1)=1$$
$$therefore det(A^-1)=frac1r^2sintheta$$
$$det(A^2)=(det(A))^2=(r^2sintheta)^2=r^4sin^2theta$$
answered Apr 8 at 11:40
Peter ForemanPeter Foreman
7,3181319
answered Apr 8 at 11:40
Peter ForemanPeter Foreman
7,3181319
answered Apr 8 at 11:40
Peter ForemanPeter Foreman
7,3181319
answered Apr 8 at 11:40
Peter ForemanPeter Foreman
7,3181319
7,3181319
$begingroup$
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
Apr 8 at 12:03
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
Apr 8 at 12:03
$begingroup$
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
Apr 8 at 12:03
$begingroup$
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.
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– Peter Nova
Apr 8 at 12:03
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
$$
B = pmatrixcos phi & sin phi & 0 \
-sin phi & cos phi, & 0 \
0 & 0 & 1
$$
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.
answered Apr 8 at 11:45
John HughesJohn Hughes
65.5k24293
$endgroup$
$begingroup$
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
Apr 8 at 12:08
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
$$
B = pmatrixcos phi & sin phi & 0 \
-sin phi & cos phi, & 0 \
0 & 0 & 1
$$
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.
answered Apr 8 at 11:45
John HughesJohn Hughes
65.5k24293
$endgroup$
$begingroup$
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
Apr 8 at 12:08
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
$$
B = pmatrixcos phi & sin phi & 0 \
-sin phi & cos phi, & 0 \
0 & 0 & 1
$$
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.
answered Apr 8 at 11:45
John HughesJohn Hughes
65.5k24293
$endgroup$
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
$$
B = pmatrixcos phi & sin phi & 0 \
-sin phi & cos phi, & 0 \
0 & 0 & 1
$$
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.
answered Apr 8 at 11:45
John HughesJohn Hughes
65.5k24293
answered Apr 8 at 11:45
John HughesJohn Hughes
65.5k24293
answered Apr 8 at 11:45
John HughesJohn Hughes
65.5k24293
answered Apr 8 at 11:45
John HughesJohn Hughes
65.5k24293
65.5k24293
$begingroup$
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
Apr 8 at 12:08
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
Apr 8 at 12:08
$begingroup$
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
Apr 8 at 12:08
$begingroup$
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
Apr 8 at 12:08
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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Why can't you "solve" (=calculate?) $det A$? It's bare for calculation.
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– Saucy O'Path
Apr 8 at 11:03
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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......
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– Peter Nova
Apr 8 at 11:26
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Nah, it's largely a matter of grouping all the $sin^2+cos^2$ that appear when you make the products.
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– Saucy O'Path
Apr 8 at 11:34
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Of course you can calculate the determinant by reducing it using properties of determinant (like this for example). This is a standard result.
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– StubbornAtom
Apr 8 at 12:20