Mixing
How is this matrix expression equivalent?
$begingroup$
I have two matrix expression that are supposed to be the same which I don't get.
First one
$begin{pmatrix}N_sE & \N_pE &end{pmatrix} begin{pmatrix} wend{pmatrix} = begin{pmatrix} -v_s\-v_pend{pmatrix} $
Second one
$begin{pmatrix}N_sE & -v_s\N_pE &-v_pend{pmatrix} begin{pmatrix} w \ 1end{pmatrix} = begin{pmatrix} 0\ 0end{pmatrix} $
How is this the same?
Thanks!
matrices
asked Jan 18 at 11:04
Noob ProgrammerNoob Programmer
103
$endgroup$
add a comment |
$begingroup$
I have two matrix expression that are supposed to be the same which I don't get.
First one
$begin{pmatrix}N_sE & \N_pE &end{pmatrix} begin{pmatrix} wend{pmatrix} = begin{pmatrix} -v_s\-v_pend{pmatrix} $
Second one
$begin{pmatrix}N_sE & -v_s\N_pE &-v_pend{pmatrix} begin{pmatrix} w \ 1end{pmatrix} = begin{pmatrix} 0\ 0end{pmatrix} $
How is this the same?
Thanks!
matrices
asked Jan 18 at 11:04
Noob ProgrammerNoob Programmer
103
$endgroup$
2
$begingroup$
As currently written, these equations aren't equivalent - you need to drop the minus sign from one copy of the "v"s. Macrophage's answer drops it from the second time they come up, but either works.
$endgroup$
– jmerry
Jan 18 at 11:21
add a comment |
$begingroup$
I have two matrix expression that are supposed to be the same which I don't get.
First one
$begin{pmatrix}N_sE & \N_pE &end{pmatrix} begin{pmatrix} wend{pmatrix} = begin{pmatrix} -v_s\-v_pend{pmatrix} $
Second one
$begin{pmatrix}N_sE & -v_s\N_pE &-v_pend{pmatrix} begin{pmatrix} w \ 1end{pmatrix} = begin{pmatrix} 0\ 0end{pmatrix} $
How is this the same?
Thanks!
matrices
asked Jan 18 at 11:04
Noob ProgrammerNoob Programmer
103
$endgroup$
I have two matrix expression that are supposed to be the same which I don't get.
First one
$begin{pmatrix}N_sE & \N_pE &end{pmatrix} begin{pmatrix} wend{pmatrix} = begin{pmatrix} -v_s\-v_pend{pmatrix} $
Second one
$begin{pmatrix}N_sE & -v_s\N_pE &-v_pend{pmatrix} begin{pmatrix} w \ 1end{pmatrix} = begin{pmatrix} 0\ 0end{pmatrix} $
How is this the same?
Thanks!
matrices
matrices
asked Jan 18 at 11:04
Noob ProgrammerNoob Programmer
103
asked Jan 18 at 11:04
Noob ProgrammerNoob Programmer
103
asked Jan 18 at 11:04
Noob ProgrammerNoob Programmer
103
asked Jan 18 at 11:04
Noob ProgrammerNoob Programmer
103
asked Jan 18 at 11:04
Noob ProgrammerNoob Programmer
103
103
2
$begingroup$
As currently written, these equations aren't equivalent - you need to drop the minus sign from one copy of the "v"s. Macrophage's answer drops it from the second time they come up, but either works.
$endgroup$
– jmerry
Jan 18 at 11:21
add a comment |
2
$begingroup$
As currently written, these equations aren't equivalent - you need to drop the minus sign from one copy of the "v"s. Macrophage's answer drops it from the second time they come up, but either works.
$endgroup$
– jmerry
Jan 18 at 11:21
2
2
$begingroup$
As currently written, these equations aren't equivalent - you need to drop the minus sign from one copy of the "v"s. Macrophage's answer drops it from the second time they come up, but either works.
$endgroup$
– jmerry
Jan 18 at 11:21
$begingroup$
As currently written, these equations aren't equivalent - you need to drop the minus sign from one copy of the "v"s. Macrophage's answer drops it from the second time they come up, but either works.
$endgroup$
– jmerry
Jan 18 at 11:21
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The left hand side of the first equation is a matrix multiplied by a number so you multiply each entry of the matrix. Hence, the systems of equations represented is
$$begin{cases}
&wcdot N_sE=-v_s\
&wcdot N_pE=-v_p\
end{cases}
$$
The left hand side of the second equation is multiplication of two matrices. So you multiply row by column and get the following system of equations
$$begin{cases}
&wcdot N_sE+1cdot v_s=0\
&wcdot N_pE+1cdot v_p=0\
end{cases}
$$
Simple rearrangment shows that the two systems of equations above are equivalent.
answered Jan 18 at 11:17
MacrophageMacrophage
1,181115
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add a comment |
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1 Answer
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1 Answer
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active
oldest
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votes
$begingroup$
The left hand side of the first equation is a matrix multiplied by a number so you multiply each entry of the matrix. Hence, the systems of equations represented is
$$begin{cases}
&wcdot N_sE=-v_s\
&wcdot N_pE=-v_p\
end{cases}
$$
The left hand side of the second equation is multiplication of two matrices. So you multiply row by column and get the following system of equations
$$begin{cases}
&wcdot N_sE+1cdot v_s=0\
&wcdot N_pE+1cdot v_p=0\
end{cases}
$$
Simple rearrangment shows that the two systems of equations above are equivalent.
answered Jan 18 at 11:17
MacrophageMacrophage
1,181115
$endgroup$
add a comment |
$begingroup$
The left hand side of the first equation is a matrix multiplied by a number so you multiply each entry of the matrix. Hence, the systems of equations represented is
$$begin{cases}
&wcdot N_sE=-v_s\
&wcdot N_pE=-v_p\
end{cases}
$$
The left hand side of the second equation is multiplication of two matrices. So you multiply row by column and get the following system of equations
$$begin{cases}
&wcdot N_sE+1cdot v_s=0\
&wcdot N_pE+1cdot v_p=0\
end{cases}
$$
Simple rearrangment shows that the two systems of equations above are equivalent.
answered Jan 18 at 11:17
MacrophageMacrophage
1,181115
$endgroup$
add a comment |
$begingroup$
The left hand side of the first equation is a matrix multiplied by a number so you multiply each entry of the matrix. Hence, the systems of equations represented is
$$begin{cases}
&wcdot N_sE=-v_s\
&wcdot N_pE=-v_p\
end{cases}
$$
The left hand side of the second equation is multiplication of two matrices. So you multiply row by column and get the following system of equations
$$begin{cases}
&wcdot N_sE+1cdot v_s=0\
&wcdot N_pE+1cdot v_p=0\
end{cases}
$$
Simple rearrangment shows that the two systems of equations above are equivalent.
answered Jan 18 at 11:17
MacrophageMacrophage
1,181115
$endgroup$
The left hand side of the first equation is a matrix multiplied by a number so you multiply each entry of the matrix. Hence, the systems of equations represented is
$$begin{cases}
&wcdot N_sE=-v_s\
&wcdot N_pE=-v_p\
end{cases}
$$
The left hand side of the second equation is multiplication of two matrices. So you multiply row by column and get the following system of equations
$$begin{cases}
&wcdot N_sE+1cdot v_s=0\
&wcdot N_pE+1cdot v_p=0\
end{cases}
$$
Simple rearrangment shows that the two systems of equations above are equivalent.
answered Jan 18 at 11:17
MacrophageMacrophage
1,181115
answered Jan 18 at 11:17
MacrophageMacrophage
1,181115
answered Jan 18 at 11:17
MacrophageMacrophage
1,181115
answered Jan 18 at 11:17
MacrophageMacrophage
1,181115
1,181115
add a comment |
add a comment |
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2
$begingroup$
As currently written, these equations aren't equivalent - you need to drop the minus sign from one copy of the "v"s. Macrophage's answer drops it from the second time they come up, but either works.
$endgroup$
– jmerry
Jan 18 at 11:21