Is echelon form a requirement to show a matrix has no solutions?
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Do I need to get a matrix into echelon form to prove that it has no solutions, or do I only need a pivot in the last column. For example, would the following row operation show that there are no solutions to the linear system represented that by the augmented matrix below?
$begin{bmatrix}1&-3&0&5\-1&1&5&2\-1&1&5&3end{bmatrix}$ $Rightarrow$ $begin{bmatrix}1&-3&0&5\-1&1&5&2\0&0&0&1end{bmatrix}$
linear-algebra matrices
asked Dec 25 '18 at 22:17
Gaussian EliminationGaussian Elimination
204
$endgroup$
add a comment |
$begingroup$
Do I need to get a matrix into echelon form to prove that it has no solutions, or do I only need a pivot in the last column. For example, would the following row operation show that there are no solutions to the linear system represented that by the augmented matrix below?
$begin{bmatrix}1&-3&0&5\-1&1&5&2\-1&1&5&3end{bmatrix}$ $Rightarrow$ $begin{bmatrix}1&-3&0&5\-1&1&5&2\0&0&0&1end{bmatrix}$
linear-algebra matrices
asked Dec 25 '18 at 22:17
Gaussian EliminationGaussian Elimination
204
$endgroup$
$begingroup$
What entry did you pivot on?
$endgroup$
– coffeemath
Dec 25 '18 at 22:21
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I'm not sure what definition of pivot you are using. By pivot, I mean the "leading entry in a row".
$endgroup$
– Gaussian Elimination
Dec 25 '18 at 22:24
add a comment |
$begingroup$
Do I need to get a matrix into echelon form to prove that it has no solutions, or do I only need a pivot in the last column. For example, would the following row operation show that there are no solutions to the linear system represented that by the augmented matrix below?
$begin{bmatrix}1&-3&0&5\-1&1&5&2\-1&1&5&3end{bmatrix}$ $Rightarrow$ $begin{bmatrix}1&-3&0&5\-1&1&5&2\0&0&0&1end{bmatrix}$
linear-algebra matrices
asked Dec 25 '18 at 22:17
Gaussian EliminationGaussian Elimination
204
$endgroup$
Do I need to get a matrix into echelon form to prove that it has no solutions, or do I only need a pivot in the last column. For example, would the following row operation show that there are no solutions to the linear system represented that by the augmented matrix below?
$begin{bmatrix}1&-3&0&5\-1&1&5&2\-1&1&5&3end{bmatrix}$ $Rightarrow$ $begin{bmatrix}1&-3&0&5\-1&1&5&2\0&0&0&1end{bmatrix}$
linear-algebra matrices
linear-algebra matrices
asked Dec 25 '18 at 22:17
Gaussian EliminationGaussian Elimination
204
asked Dec 25 '18 at 22:17
Gaussian EliminationGaussian Elimination
204
asked Dec 25 '18 at 22:17
Gaussian EliminationGaussian Elimination
204
asked Dec 25 '18 at 22:17
Gaussian EliminationGaussian Elimination
204
asked Dec 25 '18 at 22:17
Gaussian EliminationGaussian Elimination
204
204
$begingroup$
What entry did you pivot on?
$endgroup$
– coffeemath
Dec 25 '18 at 22:21
$begingroup$
I'm not sure what definition of pivot you are using. By pivot, I mean the "leading entry in a row".
$endgroup$
– Gaussian Elimination
Dec 25 '18 at 22:24
add a comment |
$begingroup$
What entry did you pivot on?
$endgroup$
– coffeemath
Dec 25 '18 at 22:21
$begingroup$
I'm not sure what definition of pivot you are using. By pivot, I mean the "leading entry in a row".
$endgroup$
– Gaussian Elimination
Dec 25 '18 at 22:24
$begingroup$
What entry did you pivot on?
$endgroup$
– coffeemath
Dec 25 '18 at 22:21
$begingroup$
What entry did you pivot on?
$endgroup$
– coffeemath
Dec 25 '18 at 22:21
$begingroup$
I'm not sure what definition of pivot you are using. By pivot, I mean the "leading entry in a row".
$endgroup$
– Gaussian Elimination
Dec 25 '18 at 22:24
$begingroup$
I'm not sure what definition of pivot you are using. By pivot, I mean the "leading entry in a row".
$endgroup$
– Gaussian Elimination
Dec 25 '18 at 22:24
add a comment |
3 Answers
3
active
oldest
votes
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Your transformation suffices; if you translate your matrix back into a system of linear equations, the last row gives the equation $0 cdot x + 0 cdot y + 0 cdot z = 1$.
This is indeed the beauty of the representation of systems of linear equations as matrices: you perform simple operations on the rows of the matrix but preserve the solution set of the original system. At the end, you just read off the solutions.
answered Dec 25 '18 at 22:24
MacRanceMacRance
1826
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add a comment |
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No, it's not required. Once you've performed row operations (as you have) to a point where you reach an inconsistency, you can conclude the system has no solutions. Going all the way to echelon form, or rref, won't change that.
answered Dec 25 '18 at 22:25
pwerthpwerth
3,340417
$endgroup$
add a comment |
$begingroup$
You only have to prove the matrix of the homogeneous part and the augmented matrix do not have the same rank. In the present case, what you've done is enough: the matrix of the homogeneous part (first $3$ columns) has rank $2$ and the augmented matrix has rank $3$.
answered Dec 25 '18 at 22:26
BernardBernard
124k741117
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add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Your transformation suffices; if you translate your matrix back into a system of linear equations, the last row gives the equation $0 cdot x + 0 cdot y + 0 cdot z = 1$.
This is indeed the beauty of the representation of systems of linear equations as matrices: you perform simple operations on the rows of the matrix but preserve the solution set of the original system. At the end, you just read off the solutions.
answered Dec 25 '18 at 22:24
MacRanceMacRance
1826
$endgroup$
add a comment |
$begingroup$
Your transformation suffices; if you translate your matrix back into a system of linear equations, the last row gives the equation $0 cdot x + 0 cdot y + 0 cdot z = 1$.
This is indeed the beauty of the representation of systems of linear equations as matrices: you perform simple operations on the rows of the matrix but preserve the solution set of the original system. At the end, you just read off the solutions.
answered Dec 25 '18 at 22:24
MacRanceMacRance
1826
$endgroup$
add a comment |
$begingroup$
Your transformation suffices; if you translate your matrix back into a system of linear equations, the last row gives the equation $0 cdot x + 0 cdot y + 0 cdot z = 1$.
This is indeed the beauty of the representation of systems of linear equations as matrices: you perform simple operations on the rows of the matrix but preserve the solution set of the original system. At the end, you just read off the solutions.
answered Dec 25 '18 at 22:24
MacRanceMacRance
1826
$endgroup$
Your transformation suffices; if you translate your matrix back into a system of linear equations, the last row gives the equation $0 cdot x + 0 cdot y + 0 cdot z = 1$.
This is indeed the beauty of the representation of systems of linear equations as matrices: you perform simple operations on the rows of the matrix but preserve the solution set of the original system. At the end, you just read off the solutions.
answered Dec 25 '18 at 22:24
MacRanceMacRance
1826
answered Dec 25 '18 at 22:24
MacRanceMacRance
1826
answered Dec 25 '18 at 22:24
MacRanceMacRance
1826
answered Dec 25 '18 at 22:24
MacRanceMacRance
1826
1826
add a comment |
add a comment |
$begingroup$
No, it's not required. Once you've performed row operations (as you have) to a point where you reach an inconsistency, you can conclude the system has no solutions. Going all the way to echelon form, or rref, won't change that.
answered Dec 25 '18 at 22:25
pwerthpwerth
3,340417
$endgroup$
add a comment |
$begingroup$
No, it's not required. Once you've performed row operations (as you have) to a point where you reach an inconsistency, you can conclude the system has no solutions. Going all the way to echelon form, or rref, won't change that.
answered Dec 25 '18 at 22:25
pwerthpwerth
3,340417
$endgroup$
add a comment |
$begingroup$
No, it's not required. Once you've performed row operations (as you have) to a point where you reach an inconsistency, you can conclude the system has no solutions. Going all the way to echelon form, or rref, won't change that.
answered Dec 25 '18 at 22:25
pwerthpwerth
3,340417
$endgroup$
No, it's not required. Once you've performed row operations (as you have) to a point where you reach an inconsistency, you can conclude the system has no solutions. Going all the way to echelon form, or rref, won't change that.
answered Dec 25 '18 at 22:25
pwerthpwerth
3,340417
answered Dec 25 '18 at 22:25
pwerthpwerth
3,340417
answered Dec 25 '18 at 22:25
pwerthpwerth
3,340417
answered Dec 25 '18 at 22:25
pwerthpwerth
3,340417
3,340417
add a comment |
add a comment |
$begingroup$
You only have to prove the matrix of the homogeneous part and the augmented matrix do not have the same rank. In the present case, what you've done is enough: the matrix of the homogeneous part (first $3$ columns) has rank $2$ and the augmented matrix has rank $3$.
answered Dec 25 '18 at 22:26
BernardBernard
124k741117
$endgroup$
add a comment |
$begingroup$
You only have to prove the matrix of the homogeneous part and the augmented matrix do not have the same rank. In the present case, what you've done is enough: the matrix of the homogeneous part (first $3$ columns) has rank $2$ and the augmented matrix has rank $3$.
answered Dec 25 '18 at 22:26
BernardBernard
124k741117
$endgroup$
add a comment |
$begingroup$
You only have to prove the matrix of the homogeneous part and the augmented matrix do not have the same rank. In the present case, what you've done is enough: the matrix of the homogeneous part (first $3$ columns) has rank $2$ and the augmented matrix has rank $3$.
answered Dec 25 '18 at 22:26
BernardBernard
124k741117
$endgroup$
You only have to prove the matrix of the homogeneous part and the augmented matrix do not have the same rank. In the present case, what you've done is enough: the matrix of the homogeneous part (first $3$ columns) has rank $2$ and the augmented matrix has rank $3$.
answered Dec 25 '18 at 22:26
BernardBernard
124k741117
edited Dec 26 '18 at 9:27
answered Dec 25 '18 at 22:26
BernardBernard
124k741117
answered Dec 25 '18 at 22:26
BernardBernard
124k741117
answered Dec 25 '18 at 22:26
BernardBernard
124k741117
124k741117
add a comment |
add a comment |
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$begingroup$
What entry did you pivot on?
$endgroup$
– coffeemath
Dec 25 '18 at 22:21
$begingroup$
I'm not sure what definition of pivot you are using. By pivot, I mean the "leading entry in a row".
$endgroup$
– Gaussian Elimination
Dec 25 '18 at 22:24