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 7 hours ago
Gaussian Elimination
112
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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 7 hours ago
Gaussian Elimination
112
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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
7 hours ago
I'm not sure what definition of pivot you are using. By pivot, I mean the "leading entry in a row".
– Gaussian Elimination
7 hours ago
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 7 hours ago
Gaussian Elimination
112
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 7 hours ago
Gaussian Elimination
112
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.
asked 7 hours ago
Gaussian Elimination
112
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.
asked 7 hours ago
Gaussian Elimination
112
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.
asked 7 hours ago
Gaussian Elimination
112
asked 7 hours ago
Gaussian Elimination
112
112
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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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New contributor
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
7 hours ago
I'm not sure what definition of pivot you are using. By pivot, I mean the "leading entry in a row".
– Gaussian Elimination
7 hours ago
add a comment |
What entry did you pivot on?
– coffeemath
7 hours ago
I'm not sure what definition of pivot you are using. By pivot, I mean the "leading entry in a row".
– Gaussian Elimination
7 hours ago
What entry did you pivot on?
– coffeemath
7 hours ago
What entry did you pivot on?
– coffeemath
7 hours ago
I'm not sure what definition of pivot you are using. By pivot, I mean the "leading entry in a row".
– Gaussian Elimination
7 hours ago
I'm not sure what definition of pivot you are using. By pivot, I mean the "leading entry in a row".
– Gaussian Elimination
7 hours ago
add a comment |
3 Answers
3
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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 7 hours ago
MacRance
956
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 7 hours ago
pwerth
1,609411
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, it is enough: the matrix of the homogeneous part has rank $2$ and the augmented matrix has rank $3$.
answered 7 hours ago
Bernard
118k638112
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
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 7 hours ago
MacRance
956
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 7 hours ago
MacRance
956
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 7 hours ago
MacRance
956
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 7 hours ago
MacRance
956
answered 7 hours ago
MacRance
956
answered 7 hours ago
MacRance
956
answered 7 hours ago
MacRance
956
956
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 7 hours ago
pwerth
1,609411
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 7 hours ago
pwerth
1,609411
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 7 hours ago
pwerth
1,609411
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 7 hours ago
pwerth
1,609411
answered 7 hours ago
pwerth
1,609411
answered 7 hours ago
pwerth
1,609411
answered 7 hours ago
pwerth
1,609411
1,609411
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, it is enough: the matrix of the homogeneous part has rank $2$ and the augmented matrix has rank $3$.
answered 7 hours ago
Bernard
118k638112
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, it is enough: the matrix of the homogeneous part has rank $2$ and the augmented matrix has rank $3$.
answered 7 hours ago
Bernard
118k638112
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, it is enough: the matrix of the homogeneous part has rank $2$ and the augmented matrix has rank $3$.
answered 7 hours ago
Bernard
118k638112
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, it is enough: the matrix of the homogeneous part has rank $2$ and the augmented matrix has rank $3$.
answered 7 hours ago
Bernard
118k638112
answered 7 hours ago
Bernard
118k638112
answered 7 hours ago
Bernard
118k638112
answered 7 hours ago
Bernard
118k638112
118k638112
add a comment |
add a comment |
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What entry did you pivot on?
– coffeemath
7 hours ago
I'm not sure what definition of pivot you are using. By pivot, I mean the "leading entry in a row".
– Gaussian Elimination
7 hours ago