Is echelon form a requirement to show a matrix has no solutions?












3














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







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  • What entry did you pivot on?
    – coffeemath
    14 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
    14 hours ago
















3














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







share|cite|improve this question







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.




















  • What entry did you pivot on?
    – coffeemath
    14 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
    14 hours ago














3












3








3


1





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







share|cite|improve this question







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






share|cite|improve this question







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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.











share|cite|improve this question







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share|cite|improve this question




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asked 14 hours ago









Gaussian Elimination

162




162




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  • What entry did you pivot on?
    – coffeemath
    14 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
    14 hours ago


















  • What entry did you pivot on?
    – coffeemath
    14 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
    14 hours ago
















What entry did you pivot on?
– coffeemath
14 hours ago




What entry did you pivot on?
– coffeemath
14 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
14 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
14 hours ago










3 Answers
3






active

oldest

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4














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.






share|cite|improve this answer





























    2














    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.






    share|cite|improve this answer





























      2














      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 $3$.






      share|cite|improve this answer























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        3 Answers
        3






        active

        oldest

        votes








        3 Answers
        3






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        4














        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.






        share|cite|improve this answer


























          4














          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.






          share|cite|improve this answer
























            4












            4








            4






            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.






            share|cite|improve this answer












            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.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered 14 hours ago









            MacRance

            1056




            1056























                2














                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.






                share|cite|improve this answer


























                  2














                  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.






                  share|cite|improve this answer
























                    2












                    2








                    2






                    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.






                    share|cite|improve this answer












                    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.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered 14 hours ago









                    pwerth

                    1,609411




                    1,609411























                        2














                        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 $3$.






                        share|cite|improve this answer




























                          2














                          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 $3$.






                          share|cite|improve this answer


























                            2












                            2








                            2






                            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 $3$.






                            share|cite|improve this answer














                            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 $3$.







                            share|cite|improve this answer














                            share|cite|improve this answer



                            share|cite|improve this answer








                            edited 3 hours ago

























                            answered 14 hours ago









                            Bernard

                            118k638112




                            118k638112






















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