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












3












$begingroup$


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







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$endgroup$












  • $begingroup$
    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
















3












$begingroup$


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







share|cite|improve this question









$endgroup$












  • $begingroup$
    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














3












3








3


0



$begingroup$


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







share|cite|improve this question









$endgroup$




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






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




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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


















  • $begingroup$
    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










3 Answers
3






active

oldest

votes


















4












$begingroup$

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.






share|cite|improve this answer









$endgroup$





















    2












    $begingroup$

    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.






    share|cite|improve this answer









    $endgroup$





















      2












      $begingroup$

      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 $3$.






      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












        $begingroup$

        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.






        share|cite|improve this answer









        $endgroup$


















          4












          $begingroup$

          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.






          share|cite|improve this answer









          $endgroup$
















            4












            4








            4





            $begingroup$

            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.






            share|cite|improve this answer









            $endgroup$



            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.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Dec 25 '18 at 22:24









            MacRanceMacRance

            1826




            1826























                2












                $begingroup$

                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.






                share|cite|improve this answer









                $endgroup$


















                  2












                  $begingroup$

                  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.






                  share|cite|improve this answer









                  $endgroup$
















                    2












                    2








                    2





                    $begingroup$

                    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.






                    share|cite|improve this answer









                    $endgroup$



                    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.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Dec 25 '18 at 22:25









                    pwerthpwerth

                    3,340417




                    3,340417























                        2












                        $begingroup$

                        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 $3$.






                        share|cite|improve this answer











                        $endgroup$


















                          2












                          $begingroup$

                          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 $3$.






                          share|cite|improve this answer











                          $endgroup$
















                            2












                            2








                            2





                            $begingroup$

                            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 $3$.






                            share|cite|improve this answer











                            $endgroup$



                            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 $3$.







                            share|cite|improve this answer














                            share|cite|improve this answer



                            share|cite|improve this answer








                            edited Dec 26 '18 at 9:27

























                            answered Dec 25 '18 at 22:26









                            BernardBernard

                            124k741117




                            124k741117






























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