Is echelon form a requirement to show a matrix has no solutions?
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 at 22:17
Gaussian Elimination
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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 at 22:17
Gaussian Elimination
204
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Gaussian Elimination is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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What entry did you pivot on?
– coffeemath
Dec 25 at 22:21
I'm not sure what definition of pivot you are using. By pivot, I mean the "leading entry in a row".
– Gaussian Elimination
Dec 25 at 22:24
add a comment |
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 at 22:17
Gaussian Elimination
204
New contributor
Gaussian Elimination is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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 at 22:17
Gaussian Elimination
204
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Gaussian Elimination is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Dec 25 at 22:17
Gaussian Elimination
204
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Gaussian Elimination is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Dec 25 at 22:17
Gaussian Elimination
204
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asked Dec 25 at 22:17
Gaussian Elimination
204
asked Dec 25 at 22:17
Gaussian Elimination
204
204
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Gaussian Elimination is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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What entry did you pivot on?
– coffeemath
Dec 25 at 22:21
I'm not sure what definition of pivot you are using. By pivot, I mean the "leading entry in a row".
– Gaussian Elimination
Dec 25 at 22:24
add a comment |
What entry did you pivot on?
– coffeemath
Dec 25 at 22:21
I'm not sure what definition of pivot you are using. By pivot, I mean the "leading entry in a row".
– Gaussian Elimination
Dec 25 at 22:24
What entry did you pivot on?
– coffeemath
Dec 25 at 22:21
What entry did you pivot on?
– coffeemath
Dec 25 at 22:21
I'm not sure what definition of pivot you are using. By pivot, I mean the "leading entry in a row".
– Gaussian Elimination
Dec 25 at 22:24
I'm not sure what definition of pivot you are using. By pivot, I mean the "leading entry in a row".
– Gaussian Elimination
Dec 25 at 22:24
add a comment |
3 Answers
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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 at 22:24
MacRance
1206
add a comment |
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 at 22:25
pwerth
1,618411
add a comment |
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 at 22:26
Bernard
118k639112
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
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votes
active
oldest
votes
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 at 22:24
MacRance
1206
add a comment |
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 at 22:24
MacRance
1206
add a comment |
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 at 22:24
MacRance
1206
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 at 22:24
MacRance
1206
answered Dec 25 at 22:24
MacRance
1206
answered Dec 25 at 22:24
MacRance
1206
answered Dec 25 at 22:24
MacRance
1206
1206
add a comment |
add a comment |
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 at 22:25
pwerth
1,618411
add a comment |
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 at 22:25
pwerth
1,618411
add a comment |
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 at 22:25
pwerth
1,618411
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 at 22:25
pwerth
1,618411
answered Dec 25 at 22:25
pwerth
1,618411
answered Dec 25 at 22:25
pwerth
1,618411
answered Dec 25 at 22:25
pwerth
1,618411
1,618411
add a comment |
add a comment |
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 at 22:26
Bernard
118k639112
add a comment |
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 at 22:26
Bernard
118k639112
add a comment |
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 at 22:26
Bernard
118k639112
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 at 22:26
Bernard
118k639112
edited Dec 26 at 9:27
answered Dec 25 at 22:26
Bernard
118k639112
answered Dec 25 at 22:26
Bernard
118k639112
answered Dec 25 at 22:26
Bernard
118k639112
118k639112
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add a comment |
Gaussian Elimination is a new contributor. Be nice, and check out our Code of Conduct.
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What entry did you pivot on?
– coffeemath
Dec 25 at 22:21
I'm not sure what definition of pivot you are using. By pivot, I mean the "leading entry in a row".
– Gaussian Elimination
Dec 25 at 22:24