Proving the that inverse of the rotation matrix is equal to the transformation.If $AB = I$ then $BA =...
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$begingroup$
I am trying to prove that:
If I have the matrix Mrot = [[cosx,-sinx],[sinx,cosx]] , the inverse and the transpose are the same.
I understand the inverse acts as a sort of "inverse button" but,I cannot see how the inverse and transpose matrices are the same. I am currently in freshman linear algebra.
Thanks
linear-algebra proof-writing
asked Mar 12 at 4:34
MCCMCC
215
$endgroup$
add a comment |
$begingroup$
I am trying to prove that:
If I have the matrix Mrot = [[cosx,-sinx],[sinx,cosx]] , the inverse and the transpose are the same.
I understand the inverse acts as a sort of "inverse button" but,I cannot see how the inverse and transpose matrices are the same. I am currently in freshman linear algebra.
Thanks
linear-algebra proof-writing
asked Mar 12 at 4:34
MCCMCC
215
$endgroup$
$begingroup$
What methods do you know to find the inverse of a matrix?
$endgroup$
– Aniruddha Deshmukh
Mar 12 at 4:37
$begingroup$
I can do guassian elimination or use the methods of cofactors and minors. I am also permitted to use the inversion formula for 2 by 2.
$endgroup$
– MCC
Mar 12 at 4:38
$begingroup$
It's much easier to take the transpose than to compute the inverse. Take the transpose and prove it is the inverse by matrix multiplication (see my answer below).
$endgroup$
– parsiad
Mar 12 at 4:39
add a comment |
$begingroup$
I am trying to prove that:
If I have the matrix Mrot = [[cosx,-sinx],[sinx,cosx]] , the inverse and the transpose are the same.
I understand the inverse acts as a sort of "inverse button" but,I cannot see how the inverse and transpose matrices are the same. I am currently in freshman linear algebra.
Thanks
linear-algebra proof-writing
asked Mar 12 at 4:34
MCCMCC
215
$endgroup$
I am trying to prove that:
If I have the matrix Mrot = [[cosx,-sinx],[sinx,cosx]] , the inverse and the transpose are the same.
I understand the inverse acts as a sort of "inverse button" but,I cannot see how the inverse and transpose matrices are the same. I am currently in freshman linear algebra.
Thanks
linear-algebra proof-writing
linear-algebra proof-writing
asked Mar 12 at 4:34
MCCMCC
215
asked Mar 12 at 4:34
MCCMCC
215
asked Mar 12 at 4:34
MCCMCC
215
asked Mar 12 at 4:34
MCCMCC
215
asked Mar 12 at 4:34
MCCMCC
215
215
$begingroup$
What methods do you know to find the inverse of a matrix?
$endgroup$
– Aniruddha Deshmukh
Mar 12 at 4:37
$begingroup$
I can do guassian elimination or use the methods of cofactors and minors. I am also permitted to use the inversion formula for 2 by 2.
$endgroup$
– MCC
Mar 12 at 4:38
$begingroup$
It's much easier to take the transpose than to compute the inverse. Take the transpose and prove it is the inverse by matrix multiplication (see my answer below).
$endgroup$
– parsiad
Mar 12 at 4:39
add a comment |
$begingroup$
What methods do you know to find the inverse of a matrix?
$endgroup$
– Aniruddha Deshmukh
Mar 12 at 4:37
$begingroup$
I can do guassian elimination or use the methods of cofactors and minors. I am also permitted to use the inversion formula for 2 by 2.
$endgroup$
– MCC
Mar 12 at 4:38
$begingroup$
It's much easier to take the transpose than to compute the inverse. Take the transpose and prove it is the inverse by matrix multiplication (see my answer below).
$endgroup$
– parsiad
Mar 12 at 4:39
$begingroup$
What methods do you know to find the inverse of a matrix?
$endgroup$
– Aniruddha Deshmukh
Mar 12 at 4:37
$begingroup$
What methods do you know to find the inverse of a matrix?
$endgroup$
– Aniruddha Deshmukh
Mar 12 at 4:37
$begingroup$
I can do guassian elimination or use the methods of cofactors and minors. I am also permitted to use the inversion formula for 2 by 2.
$endgroup$
– MCC
Mar 12 at 4:38
$begingroup$
I can do guassian elimination or use the methods of cofactors and minors. I am also permitted to use the inversion formula for 2 by 2.
$endgroup$
– MCC
Mar 12 at 4:38
$begingroup$
It's much easier to take the transpose than to compute the inverse. Take the transpose and prove it is the inverse by matrix multiplication (see my answer below).
$endgroup$
– parsiad
Mar 12 at 4:39
$begingroup$
It's much easier to take the transpose than to compute the inverse. Take the transpose and prove it is the inverse by matrix multiplication (see my answer below).
$endgroup$
– parsiad
Mar 12 at 4:39
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
You can prove this by direct computation.
Take the transpose of $M$ to get
$$
M^{intercal}=begin{pmatrix}cos x & sin x\
-sin x & cos x
end{pmatrix}.
$$
Multiply this by $M$ to get
$$
MM^{intercal}=begin{pmatrix}cos(x)^{2}+sin(x)^{2}\
& cos(x)^{2}+sin(x)^{2}
end{pmatrix}
$$
and apply a certain famous identity.
answered Mar 12 at 4:38
parsiadparsiad
18.5k32453
$endgroup$
$begingroup$
We will also need to prove that $M^TM = I$ so that $M^T = M^{-1}$.
$endgroup$
– Aniruddha Deshmukh
Mar 12 at 4:42
$begingroup$
Indeed. See here.
$endgroup$
– parsiad
Mar 12 at 4:45
add a comment |
Your Answer
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
You can prove this by direct computation.
Take the transpose of $M$ to get
$$
M^{intercal}=begin{pmatrix}cos x & sin x\
-sin x & cos x
end{pmatrix}.
$$
Multiply this by $M$ to get
$$
MM^{intercal}=begin{pmatrix}cos(x)^{2}+sin(x)^{2}\
& cos(x)^{2}+sin(x)^{2}
end{pmatrix}
$$
and apply a certain famous identity.
answered Mar 12 at 4:38
parsiadparsiad
18.5k32453
$endgroup$
$begingroup$
We will also need to prove that $M^TM = I$ so that $M^T = M^{-1}$.
$endgroup$
– Aniruddha Deshmukh
Mar 12 at 4:42
$begingroup$
Indeed. See here.
$endgroup$
– parsiad
Mar 12 at 4:45
add a comment |
$begingroup$
You can prove this by direct computation.
Take the transpose of $M$ to get
$$
M^{intercal}=begin{pmatrix}cos x & sin x\
-sin x & cos x
end{pmatrix}.
$$
Multiply this by $M$ to get
$$
MM^{intercal}=begin{pmatrix}cos(x)^{2}+sin(x)^{2}\
& cos(x)^{2}+sin(x)^{2}
end{pmatrix}
$$
and apply a certain famous identity.
answered Mar 12 at 4:38
parsiadparsiad
18.5k32453
$endgroup$
$begingroup$
We will also need to prove that $M^TM = I$ so that $M^T = M^{-1}$.
$endgroup$
– Aniruddha Deshmukh
Mar 12 at 4:42
$begingroup$
Indeed. See here.
$endgroup$
– parsiad
Mar 12 at 4:45
add a comment |
$begingroup$
You can prove this by direct computation.
Take the transpose of $M$ to get
$$
M^{intercal}=begin{pmatrix}cos x & sin x\
-sin x & cos x
end{pmatrix}.
$$
Multiply this by $M$ to get
$$
MM^{intercal}=begin{pmatrix}cos(x)^{2}+sin(x)^{2}\
& cos(x)^{2}+sin(x)^{2}
end{pmatrix}
$$
and apply a certain famous identity.
answered Mar 12 at 4:38
parsiadparsiad
18.5k32453
$endgroup$
You can prove this by direct computation.
Take the transpose of $M$ to get
$$
M^{intercal}=begin{pmatrix}cos x & sin x\
-sin x & cos x
end{pmatrix}.
$$
Multiply this by $M$ to get
$$
MM^{intercal}=begin{pmatrix}cos(x)^{2}+sin(x)^{2}\
& cos(x)^{2}+sin(x)^{2}
end{pmatrix}
$$
and apply a certain famous identity.
answered Mar 12 at 4:38
parsiadparsiad
18.5k32453
answered Mar 12 at 4:38
parsiadparsiad
18.5k32453
answered Mar 12 at 4:38
parsiadparsiad
18.5k32453
answered Mar 12 at 4:38
parsiadparsiad
18.5k32453
18.5k32453
$begingroup$
We will also need to prove that $M^TM = I$ so that $M^T = M^{-1}$.
$endgroup$
– Aniruddha Deshmukh
Mar 12 at 4:42
$begingroup$
Indeed. See here.
$endgroup$
– parsiad
Mar 12 at 4:45
add a comment |
$begingroup$
We will also need to prove that $M^TM = I$ so that $M^T = M^{-1}$.
$endgroup$
– Aniruddha Deshmukh
Mar 12 at 4:42
$begingroup$
Indeed. See here.
$endgroup$
– parsiad
Mar 12 at 4:45
$begingroup$
We will also need to prove that $M^TM = I$ so that $M^T = M^{-1}$.
$endgroup$
– Aniruddha Deshmukh
Mar 12 at 4:42
$begingroup$
We will also need to prove that $M^TM = I$ so that $M^T = M^{-1}$.
$endgroup$
– Aniruddha Deshmukh
Mar 12 at 4:42
$begingroup$
Indeed. See here.
$endgroup$
– parsiad
Mar 12 at 4:45
$begingroup$
Indeed. See here.
$endgroup$
– parsiad
Mar 12 at 4:45
add a comment |
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$begingroup$
What methods do you know to find the inverse of a matrix?
$endgroup$
– Aniruddha Deshmukh
Mar 12 at 4:37
$begingroup$
I can do guassian elimination or use the methods of cofactors and minors. I am also permitted to use the inversion formula for 2 by 2.
$endgroup$
– MCC
Mar 12 at 4:38
$begingroup$
It's much easier to take the transpose than to compute the inverse. Take the transpose and prove it is the inverse by matrix multiplication (see my answer below).
$endgroup$
– parsiad
Mar 12 at 4:39