Mixing
find the basis of the kernel of matrix A
$begingroup$
$$
A= begin{pmatrix} 1 & -2 & 1 & 3 & 0\ 2 & -4 & 4 & 6 & 4\ -2 & 4 & -1 & -6 & 2 \ 1 &-2 & -3 & 3 & -8
end{pmatrix}
$$
(I was given this matrix and i am assuming this is an augmented matrix, the variables we are using are a,b,c,d and e that correspond to the columns)
So obviously the first step that i took was to find the kernel of A
I did row reduction and i got the matrix below
$$
A= begin{pmatrix} 1 & -2 & 0 & 3 & 0\ 0 & 0 & 0 & 0 & 0\ 0 & 0 & 1 & 0 & 2 \ 0 &0 & 0 & 0 & 0
end{pmatrix}
$$
if i wrote down a system of linear equations that correspond to the matrix above, i would get the equations below
a - 2b + 3d = 0
c + 2e = 0
if we let b = t, d = m and e = n(t, m and n are any real numbers) then we get...
a = 2t - 3m
b = t
c = -2n
d = m
e = n
so if we collect the arbitrary terms in the equations above, we get the kernel, which is written below
kernel of A = { (2t, t, 0, 0, 0) , (-3m, 0, 0, m, 0) , (0, 0, -2n, 0 ,n): t, m and n are real numbers }
is this the kernel of A? did i make a mistake somewhere. I would appreciate it if someone checked my work.
linear-algebra matrices vectors
asked Nov 4 '17 at 22:41
Soon_to_be_code_masterSoon_to_be_code_master
387310
$endgroup$
add a comment |
$begingroup$
$$
A= begin{pmatrix} 1 & -2 & 1 & 3 & 0\ 2 & -4 & 4 & 6 & 4\ -2 & 4 & -1 & -6 & 2 \ 1 &-2 & -3 & 3 & -8
end{pmatrix}
$$
(I was given this matrix and i am assuming this is an augmented matrix, the variables we are using are a,b,c,d and e that correspond to the columns)
So obviously the first step that i took was to find the kernel of A
I did row reduction and i got the matrix below
$$
A= begin{pmatrix} 1 & -2 & 0 & 3 & 0\ 0 & 0 & 0 & 0 & 0\ 0 & 0 & 1 & 0 & 2 \ 0 &0 & 0 & 0 & 0
end{pmatrix}
$$
if i wrote down a system of linear equations that correspond to the matrix above, i would get the equations below
a - 2b + 3d = 0
c + 2e = 0
if we let b = t, d = m and e = n(t, m and n are any real numbers) then we get...
a = 2t - 3m
b = t
c = -2n
d = m
e = n
so if we collect the arbitrary terms in the equations above, we get the kernel, which is written below
kernel of A = { (2t, t, 0, 0, 0) , (-3m, 0, 0, m, 0) , (0, 0, -2n, 0 ,n): t, m and n are real numbers }
is this the kernel of A? did i make a mistake somewhere. I would appreciate it if someone checked my work.
linear-algebra matrices vectors
asked Nov 4 '17 at 22:41
Soon_to_be_code_masterSoon_to_be_code_master
387310
$endgroup$
$begingroup$
The domain of $A$ is $mathbb R^5$, but you’ve got elements of $mathbb R^4$ in your kernel, so that’s clearly wrong. See this answer for a guide to reading a kernel basis from the RREF of a matrix.
$endgroup$
– amd
Nov 4 '17 at 22:58
$begingroup$
Finding the kernel of a $4 times 5$ matrix is a different problem than solving a $4 times 4$ system of equations with a $4 times 1$ target vector.
$endgroup$
– Hurkyl
Nov 4 '17 at 23:46
$begingroup$
Your formula for the kernel of the $4 times 4$ matrix is incorrect; you presumably meant to write $(2t, t, 0 ,0) + (-3m, 0, 0, m)$.
$endgroup$
– Hurkyl
Nov 4 '17 at 23:46
$begingroup$
Finally, you don't have to find the kernel before you find a basis for the kernel; it is quite possible (and common!) to do it the other way around. In fact, you've done something extremely similar in the process of obtaining your formula for the kernel of the $4 times 4$ matrix: you identified $(2,1,0,0)$ and $(-3,0,0,1)$ as a basis for its kernel, and used that fact to help you write down the kernel!
$endgroup$
– Hurkyl
Nov 4 '17 at 23:48
$begingroup$
I have made the appropriate corrections in my original post, please double check everything just in case i made an error........ so i am assuming that the basis of the kernel are the following vectors (2,1,0,0,0), (-3,0,0,1,0),(0,0,-2,0,1), is this correct?
$endgroup$
– Soon_to_be_code_master
Nov 5 '17 at 0:10
add a comment |
$begingroup$
$$
A= begin{pmatrix} 1 & -2 & 1 & 3 & 0\ 2 & -4 & 4 & 6 & 4\ -2 & 4 & -1 & -6 & 2 \ 1 &-2 & -3 & 3 & -8
end{pmatrix}
$$
(I was given this matrix and i am assuming this is an augmented matrix, the variables we are using are a,b,c,d and e that correspond to the columns)
So obviously the first step that i took was to find the kernel of A
I did row reduction and i got the matrix below
$$
A= begin{pmatrix} 1 & -2 & 0 & 3 & 0\ 0 & 0 & 0 & 0 & 0\ 0 & 0 & 1 & 0 & 2 \ 0 &0 & 0 & 0 & 0
end{pmatrix}
$$
if i wrote down a system of linear equations that correspond to the matrix above, i would get the equations below
a - 2b + 3d = 0
c + 2e = 0
if we let b = t, d = m and e = n(t, m and n are any real numbers) then we get...
a = 2t - 3m
b = t
c = -2n
d = m
e = n
so if we collect the arbitrary terms in the equations above, we get the kernel, which is written below
kernel of A = { (2t, t, 0, 0, 0) , (-3m, 0, 0, m, 0) , (0, 0, -2n, 0 ,n): t, m and n are real numbers }
is this the kernel of A? did i make a mistake somewhere. I would appreciate it if someone checked my work.
linear-algebra matrices vectors
asked Nov 4 '17 at 22:41
Soon_to_be_code_masterSoon_to_be_code_master
387310
$endgroup$
$$
A= begin{pmatrix} 1 & -2 & 1 & 3 & 0\ 2 & -4 & 4 & 6 & 4\ -2 & 4 & -1 & -6 & 2 \ 1 &-2 & -3 & 3 & -8
end{pmatrix}
$$
(I was given this matrix and i am assuming this is an augmented matrix, the variables we are using are a,b,c,d and e that correspond to the columns)
So obviously the first step that i took was to find the kernel of A
I did row reduction and i got the matrix below
$$
A= begin{pmatrix} 1 & -2 & 0 & 3 & 0\ 0 & 0 & 0 & 0 & 0\ 0 & 0 & 1 & 0 & 2 \ 0 &0 & 0 & 0 & 0
end{pmatrix}
$$
if i wrote down a system of linear equations that correspond to the matrix above, i would get the equations below
a - 2b + 3d = 0
c + 2e = 0
if we let b = t, d = m and e = n(t, m and n are any real numbers) then we get...
a = 2t - 3m
b = t
c = -2n
d = m
e = n
so if we collect the arbitrary terms in the equations above, we get the kernel, which is written below
kernel of A = { (2t, t, 0, 0, 0) , (-3m, 0, 0, m, 0) , (0, 0, -2n, 0 ,n): t, m and n are real numbers }
is this the kernel of A? did i make a mistake somewhere. I would appreciate it if someone checked my work.
linear-algebra matrices vectors
linear-algebra matrices vectors
asked Nov 4 '17 at 22:41
Soon_to_be_code_masterSoon_to_be_code_master
387310
asked Nov 4 '17 at 22:41
Soon_to_be_code_masterSoon_to_be_code_master
387310
edited Nov 5 '17 at 0:19
Soon_to_be_code_master
asked Nov 4 '17 at 22:41
Soon_to_be_code_masterSoon_to_be_code_master
387310
asked Nov 4 '17 at 22:41
Soon_to_be_code_masterSoon_to_be_code_master
387310
asked Nov 4 '17 at 22:41
Soon_to_be_code_masterSoon_to_be_code_master
387310
387310
$begingroup$
The domain of $A$ is $mathbb R^5$, but you’ve got elements of $mathbb R^4$ in your kernel, so that’s clearly wrong. See this answer for a guide to reading a kernel basis from the RREF of a matrix.
$endgroup$
– amd
Nov 4 '17 at 22:58
$begingroup$
Finding the kernel of a $4 times 5$ matrix is a different problem than solving a $4 times 4$ system of equations with a $4 times 1$ target vector.
$endgroup$
– Hurkyl
Nov 4 '17 at 23:46
$begingroup$
Your formula for the kernel of the $4 times 4$ matrix is incorrect; you presumably meant to write $(2t, t, 0 ,0) + (-3m, 0, 0, m)$.
$endgroup$
– Hurkyl
Nov 4 '17 at 23:46
$begingroup$
Finally, you don't have to find the kernel before you find a basis for the kernel; it is quite possible (and common!) to do it the other way around. In fact, you've done something extremely similar in the process of obtaining your formula for the kernel of the $4 times 4$ matrix: you identified $(2,1,0,0)$ and $(-3,0,0,1)$ as a basis for its kernel, and used that fact to help you write down the kernel!
$endgroup$
– Hurkyl
Nov 4 '17 at 23:48
$begingroup$
I have made the appropriate corrections in my original post, please double check everything just in case i made an error........ so i am assuming that the basis of the kernel are the following vectors (2,1,0,0,0), (-3,0,0,1,0),(0,0,-2,0,1), is this correct?
$endgroup$
– Soon_to_be_code_master
Nov 5 '17 at 0:10
add a comment |
$begingroup$
The domain of $A$ is $mathbb R^5$, but you’ve got elements of $mathbb R^4$ in your kernel, so that’s clearly wrong. See this answer for a guide to reading a kernel basis from the RREF of a matrix.
$endgroup$
– amd
Nov 4 '17 at 22:58
$begingroup$
Finding the kernel of a $4 times 5$ matrix is a different problem than solving a $4 times 4$ system of equations with a $4 times 1$ target vector.
$endgroup$
– Hurkyl
Nov 4 '17 at 23:46
$begingroup$
Your formula for the kernel of the $4 times 4$ matrix is incorrect; you presumably meant to write $(2t, t, 0 ,0) + (-3m, 0, 0, m)$.
$endgroup$
– Hurkyl
Nov 4 '17 at 23:46
$begingroup$
Finally, you don't have to find the kernel before you find a basis for the kernel; it is quite possible (and common!) to do it the other way around. In fact, you've done something extremely similar in the process of obtaining your formula for the kernel of the $4 times 4$ matrix: you identified $(2,1,0,0)$ and $(-3,0,0,1)$ as a basis for its kernel, and used that fact to help you write down the kernel!
$endgroup$
– Hurkyl
Nov 4 '17 at 23:48
$begingroup$
I have made the appropriate corrections in my original post, please double check everything just in case i made an error........ so i am assuming that the basis of the kernel are the following vectors (2,1,0,0,0), (-3,0,0,1,0),(0,0,-2,0,1), is this correct?
$endgroup$
– Soon_to_be_code_master
Nov 5 '17 at 0:10
$begingroup$
The domain of $A$ is $mathbb R^5$, but you’ve got elements of $mathbb R^4$ in your kernel, so that’s clearly wrong. See this answer for a guide to reading a kernel basis from the RREF of a matrix.
$endgroup$
– amd
Nov 4 '17 at 22:58
$begingroup$
The domain of $A$ is $mathbb R^5$, but you’ve got elements of $mathbb R^4$ in your kernel, so that’s clearly wrong. See this answer for a guide to reading a kernel basis from the RREF of a matrix.
$endgroup$
– amd
Nov 4 '17 at 22:58
$begingroup$
Finding the kernel of a $4 times 5$ matrix is a different problem than solving a $4 times 4$ system of equations with a $4 times 1$ target vector.
$endgroup$
– Hurkyl
Nov 4 '17 at 23:46
$begingroup$
Finding the kernel of a $4 times 5$ matrix is a different problem than solving a $4 times 4$ system of equations with a $4 times 1$ target vector.
$endgroup$
– Hurkyl
Nov 4 '17 at 23:46
$begingroup$
Your formula for the kernel of the $4 times 4$ matrix is incorrect; you presumably meant to write $(2t, t, 0 ,0) + (-3m, 0, 0, m)$.
$endgroup$
– Hurkyl
Nov 4 '17 at 23:46
$begingroup$
Your formula for the kernel of the $4 times 4$ matrix is incorrect; you presumably meant to write $(2t, t, 0 ,0) + (-3m, 0, 0, m)$.
$endgroup$
– Hurkyl
Nov 4 '17 at 23:46
$begingroup$
Finally, you don't have to find the kernel before you find a basis for the kernel; it is quite possible (and common!) to do it the other way around. In fact, you've done something extremely similar in the process of obtaining your formula for the kernel of the $4 times 4$ matrix: you identified $(2,1,0,0)$ and $(-3,0,0,1)$ as a basis for its kernel, and used that fact to help you write down the kernel!
$endgroup$
– Hurkyl
Nov 4 '17 at 23:48
$begingroup$
Finally, you don't have to find the kernel before you find a basis for the kernel; it is quite possible (and common!) to do it the other way around. In fact, you've done something extremely similar in the process of obtaining your formula for the kernel of the $4 times 4$ matrix: you identified $(2,1,0,0)$ and $(-3,0,0,1)$ as a basis for its kernel, and used that fact to help you write down the kernel!
$endgroup$
– Hurkyl
Nov 4 '17 at 23:48
$begingroup$
I have made the appropriate corrections in my original post, please double check everything just in case i made an error........ so i am assuming that the basis of the kernel are the following vectors (2,1,0,0,0), (-3,0,0,1,0),(0,0,-2,0,1), is this correct?
$endgroup$
– Soon_to_be_code_master
Nov 5 '17 at 0:10
$begingroup$
I have made the appropriate corrections in my original post, please double check everything just in case i made an error........ so i am assuming that the basis of the kernel are the following vectors (2,1,0,0,0), (-3,0,0,1,0),(0,0,-2,0,1), is this correct?
$endgroup$
– Soon_to_be_code_master
Nov 5 '17 at 0:10
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
You have the solution
$$begin{pmatrix}a\ b\ c\ d\ eend{pmatrix}=begin{pmatrix} 2b-3d\ b\ c
-2e\ d\ eend{pmatrix}=bbegin{pmatrix}2\ 1\ 0\ 0\ 0end{pmatrix}+dbegin{pmatrix}-3\ 0\ 0\ 1\ 0end{pmatrix}+ebegin{pmatrix}0\ 0\ -2\ 0\ 1end{pmatrix}$$
So the vectors
$$begin{pmatrix}2\ 1\ 0\ 0\ 0end{pmatrix},begin{pmatrix}-3\ 0\ 0\ 1\ 0end{pmatrix},begin{pmatrix}0\ 0\ -2\ 0\ 1end{pmatrix}$$ are a basis for the kernel.
edited Jan 12 at 18:39
amWhy
1
answered Nov 5 '17 at 0:28
Rene SchipperusRene Schipperus
32.3k11960
$endgroup$
$begingroup$
@Marine Galantin: If suspect an error, particularly in an Accepted Answer, the better approach is to broach a correction in a Comment to the author, even if the suspicion is more of a certainty (as here, where $c$ has disappeared from the right hand side). Fortunately you have plenty enough reputation to leave such a Comment, and if the author does not respond after a few days, then it would be more justified to make a substantive edit.
$endgroup$
– hardmath
Jan 12 at 18:29
add a comment |
Your Answer
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1 Answer
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1 Answer
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$begingroup$
You have the solution
$$begin{pmatrix}a\ b\ c\ d\ eend{pmatrix}=begin{pmatrix} 2b-3d\ b\ c
-2e\ d\ eend{pmatrix}=bbegin{pmatrix}2\ 1\ 0\ 0\ 0end{pmatrix}+dbegin{pmatrix}-3\ 0\ 0\ 1\ 0end{pmatrix}+ebegin{pmatrix}0\ 0\ -2\ 0\ 1end{pmatrix}$$
So the vectors
$$begin{pmatrix}2\ 1\ 0\ 0\ 0end{pmatrix},begin{pmatrix}-3\ 0\ 0\ 1\ 0end{pmatrix},begin{pmatrix}0\ 0\ -2\ 0\ 1end{pmatrix}$$ are a basis for the kernel.
edited Jan 12 at 18:39
amWhy
1
answered Nov 5 '17 at 0:28
Rene SchipperusRene Schipperus
32.3k11960
$endgroup$
$begingroup$
@Marine Galantin: If suspect an error, particularly in an Accepted Answer, the better approach is to broach a correction in a Comment to the author, even if the suspicion is more of a certainty (as here, where $c$ has disappeared from the right hand side). Fortunately you have plenty enough reputation to leave such a Comment, and if the author does not respond after a few days, then it would be more justified to make a substantive edit.
$endgroup$
– hardmath
Jan 12 at 18:29
add a comment |
$begingroup$
You have the solution
$$begin{pmatrix}a\ b\ c\ d\ eend{pmatrix}=begin{pmatrix} 2b-3d\ b\ c
-2e\ d\ eend{pmatrix}=bbegin{pmatrix}2\ 1\ 0\ 0\ 0end{pmatrix}+dbegin{pmatrix}-3\ 0\ 0\ 1\ 0end{pmatrix}+ebegin{pmatrix}0\ 0\ -2\ 0\ 1end{pmatrix}$$
So the vectors
$$begin{pmatrix}2\ 1\ 0\ 0\ 0end{pmatrix},begin{pmatrix}-3\ 0\ 0\ 1\ 0end{pmatrix},begin{pmatrix}0\ 0\ -2\ 0\ 1end{pmatrix}$$ are a basis for the kernel.
edited Jan 12 at 18:39
amWhy
1
answered Nov 5 '17 at 0:28
Rene SchipperusRene Schipperus
32.3k11960
$endgroup$
$begingroup$
@Marine Galantin: If suspect an error, particularly in an Accepted Answer, the better approach is to broach a correction in a Comment to the author, even if the suspicion is more of a certainty (as here, where $c$ has disappeared from the right hand side). Fortunately you have plenty enough reputation to leave such a Comment, and if the author does not respond after a few days, then it would be more justified to make a substantive edit.
$endgroup$
– hardmath
Jan 12 at 18:29
add a comment |
$begingroup$
You have the solution
$$begin{pmatrix}a\ b\ c\ d\ eend{pmatrix}=begin{pmatrix} 2b-3d\ b\ c
-2e\ d\ eend{pmatrix}=bbegin{pmatrix}2\ 1\ 0\ 0\ 0end{pmatrix}+dbegin{pmatrix}-3\ 0\ 0\ 1\ 0end{pmatrix}+ebegin{pmatrix}0\ 0\ -2\ 0\ 1end{pmatrix}$$
So the vectors
$$begin{pmatrix}2\ 1\ 0\ 0\ 0end{pmatrix},begin{pmatrix}-3\ 0\ 0\ 1\ 0end{pmatrix},begin{pmatrix}0\ 0\ -2\ 0\ 1end{pmatrix}$$ are a basis for the kernel.
edited Jan 12 at 18:39
amWhy
1
answered Nov 5 '17 at 0:28
Rene SchipperusRene Schipperus
32.3k11960
$endgroup$
You have the solution
$$begin{pmatrix}a\ b\ c\ d\ eend{pmatrix}=begin{pmatrix} 2b-3d\ b\ c
-2e\ d\ eend{pmatrix}=bbegin{pmatrix}2\ 1\ 0\ 0\ 0end{pmatrix}+dbegin{pmatrix}-3\ 0\ 0\ 1\ 0end{pmatrix}+ebegin{pmatrix}0\ 0\ -2\ 0\ 1end{pmatrix}$$
So the vectors
$$begin{pmatrix}2\ 1\ 0\ 0\ 0end{pmatrix},begin{pmatrix}-3\ 0\ 0\ 1\ 0end{pmatrix},begin{pmatrix}0\ 0\ -2\ 0\ 1end{pmatrix}$$ are a basis for the kernel.
edited Jan 12 at 18:39
amWhy
1
answered Nov 5 '17 at 0:28
Rene SchipperusRene Schipperus
32.3k11960
edited Jan 12 at 18:39
amWhy
1
edited Jan 12 at 18:39
amWhy
1
edited Jan 12 at 18:39
amWhy
1
1
answered Nov 5 '17 at 0:28
Rene SchipperusRene Schipperus
32.3k11960
answered Nov 5 '17 at 0:28
Rene SchipperusRene Schipperus
32.3k11960
answered Nov 5 '17 at 0:28
Rene SchipperusRene Schipperus
32.3k11960
32.3k11960
$begingroup$
@Marine Galantin: If suspect an error, particularly in an Accepted Answer, the better approach is to broach a correction in a Comment to the author, even if the suspicion is more of a certainty (as here, where $c$ has disappeared from the right hand side). Fortunately you have plenty enough reputation to leave such a Comment, and if the author does not respond after a few days, then it would be more justified to make a substantive edit.
$endgroup$
– hardmath
Jan 12 at 18:29
add a comment |
$begingroup$
@Marine Galantin: If suspect an error, particularly in an Accepted Answer, the better approach is to broach a correction in a Comment to the author, even if the suspicion is more of a certainty (as here, where $c$ has disappeared from the right hand side). Fortunately you have plenty enough reputation to leave such a Comment, and if the author does not respond after a few days, then it would be more justified to make a substantive edit.
$endgroup$
– hardmath
Jan 12 at 18:29
$begingroup$
@Marine Galantin: If suspect an error, particularly in an Accepted Answer, the better approach is to broach a correction in a Comment to the author, even if the suspicion is more of a certainty (as here, where $c$ has disappeared from the right hand side). Fortunately you have plenty enough reputation to leave such a Comment, and if the author does not respond after a few days, then it would be more justified to make a substantive edit.
$endgroup$
– hardmath
Jan 12 at 18:29
$begingroup$
@Marine Galantin: If suspect an error, particularly in an Accepted Answer, the better approach is to broach a correction in a Comment to the author, even if the suspicion is more of a certainty (as here, where $c$ has disappeared from the right hand side). Fortunately you have plenty enough reputation to leave such a Comment, and if the author does not respond after a few days, then it would be more justified to make a substantive edit.
$endgroup$
– hardmath
Jan 12 at 18:29
add a comment |
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$begingroup$
The domain of $A$ is $mathbb R^5$, but you’ve got elements of $mathbb R^4$ in your kernel, so that’s clearly wrong. See this answer for a guide to reading a kernel basis from the RREF of a matrix.
$endgroup$
– amd
Nov 4 '17 at 22:58
$begingroup$
Finding the kernel of a $4 times 5$ matrix is a different problem than solving a $4 times 4$ system of equations with a $4 times 1$ target vector.
$endgroup$
– Hurkyl
Nov 4 '17 at 23:46
$begingroup$
Your formula for the kernel of the $4 times 4$ matrix is incorrect; you presumably meant to write $(2t, t, 0 ,0) + (-3m, 0, 0, m)$.
$endgroup$
– Hurkyl
Nov 4 '17 at 23:46
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Finally, you don't have to find the kernel before you find a basis for the kernel; it is quite possible (and common!) to do it the other way around. In fact, you've done something extremely similar in the process of obtaining your formula for the kernel of the $4 times 4$ matrix: you identified $(2,1,0,0)$ and $(-3,0,0,1)$ as a basis for its kernel, and used that fact to help you write down the kernel!
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– Hurkyl
Nov 4 '17 at 23:48
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I have made the appropriate corrections in my original post, please double check everything just in case i made an error........ so i am assuming that the basis of the kernel are the following vectors (2,1,0,0,0), (-3,0,0,1,0),(0,0,-2,0,1), is this correct?
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– Soon_to_be_code_master
Nov 5 '17 at 0:10