Gauss elimination. Where did I go wrong?

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enter image description here



Gaussian elimination with back sub:



So my starting matrix:



beginbmatrix
1 & -1 & 1 & -1
\2 & 1 & -3 & 4
\2 & 0 & 2 & 2
endbmatrix



multiply the 2nd and 3rd row by -1 * (first row):



beginbmatrix
1 & -1 & 1 & -1
\0 & 3 & -5 & 6
\0 & 2 & 0 & 4
endbmatrix



then add -1(third row) to the 2nd row->



beginbmatrix
1 & -1 & 1 & -1
\0 & 1 & -5 & 2
\0 & 2 & 0 & 4
endbmatrix



add -2(2nd row) to the third row ->
beginbmatrix
1 & -1 & 1 & -1
\0 & 1 & -5 & 2
\0 & 0 & 10 & 0
endbmatrix



But then this seems to have no solution because $10z = 0$.... ugh



EDIT



As I was writing this, it occurred to me that $z = 0$, $y = 2$, $x = 1$. Is that right?




linear-algebra




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  • 3




    Your edit is correct
    – Rushabh Mehta
    Aug 1 at 15:30






  • 1




    In general, if you get an incorrect answer, find your error by "bisection" ... take one of your steps from the middle, plug in your incorrect answer, and see if it works. This will tell you whether your mistake is before or after that point.
    – GEdgar
    Aug 1 at 15:40














up vote
3
down vote

favorite












enter image description here



Gaussian elimination with back sub:



So my starting matrix:



beginbmatrix
1 & -1 & 1 & -1
\2 & 1 & -3 & 4
\2 & 0 & 2 & 2
endbmatrix



multiply the 2nd and 3rd row by -1 * (first row):



beginbmatrix
1 & -1 & 1 & -1
\0 & 3 & -5 & 6
\0 & 2 & 0 & 4
endbmatrix



then add -1(third row) to the 2nd row->



beginbmatrix
1 & -1 & 1 & -1
\0 & 1 & -5 & 2
\0 & 2 & 0 & 4
endbmatrix



add -2(2nd row) to the third row ->
beginbmatrix
1 & -1 & 1 & -1
\0 & 1 & -5 & 2
\0 & 0 & 10 & 0
endbmatrix



But then this seems to have no solution because $10z = 0$.... ugh



EDIT



As I was writing this, it occurred to me that $z = 0$, $y = 2$, $x = 1$. Is that right?




linear-algebra




share|cite|improve this question

















  • 3




    Your edit is correct
    – Rushabh Mehta
    Aug 1 at 15:30






  • 1




    In general, if you get an incorrect answer, find your error by "bisection" ... take one of your steps from the middle, plug in your incorrect answer, and see if it works. This will tell you whether your mistake is before or after that point.
    – GEdgar
    Aug 1 at 15:40












up vote
3
down vote

favorite









up vote
3
down vote

favorite











enter image description here



Gaussian elimination with back sub:



So my starting matrix:



beginbmatrix
1 & -1 & 1 & -1
\2 & 1 & -3 & 4
\2 & 0 & 2 & 2
endbmatrix



multiply the 2nd and 3rd row by -1 * (first row):



beginbmatrix
1 & -1 & 1 & -1
\0 & 3 & -5 & 6
\0 & 2 & 0 & 4
endbmatrix



then add -1(third row) to the 2nd row->



beginbmatrix
1 & -1 & 1 & -1
\0 & 1 & -5 & 2
\0 & 2 & 0 & 4
endbmatrix



add -2(2nd row) to the third row ->
beginbmatrix
1 & -1 & 1 & -1
\0 & 1 & -5 & 2
\0 & 0 & 10 & 0
endbmatrix



But then this seems to have no solution because $10z = 0$.... ugh



EDIT



As I was writing this, it occurred to me that $z = 0$, $y = 2$, $x = 1$. Is that right?




linear-algebra




share|cite|improve this question













enter image description here



Gaussian elimination with back sub:



So my starting matrix:



beginbmatrix
1 & -1 & 1 & -1
\2 & 1 & -3 & 4
\2 & 0 & 2 & 2
endbmatrix



multiply the 2nd and 3rd row by -1 * (first row):



beginbmatrix
1 & -1 & 1 & -1
\0 & 3 & -5 & 6
\0 & 2 & 0 & 4
endbmatrix



then add -1(third row) to the 2nd row->



beginbmatrix
1 & -1 & 1 & -1
\0 & 1 & -5 & 2
\0 & 2 & 0 & 4
endbmatrix



add -2(2nd row) to the third row ->
beginbmatrix
1 & -1 & 1 & -1
\0 & 1 & -5 & 2
\0 & 0 & 10 & 0
endbmatrix



But then this seems to have no solution because $10z = 0$.... ugh



EDIT



As I was writing this, it occurred to me that $z = 0$, $y = 2$, $x = 1$. Is that right?





linear-algebra





share|cite|improve this question












share|cite|improve this question




share|cite|improve this question








edited Aug 1 at 15:47









gimusi

63.9k73480




63.9k73480









asked Aug 1 at 15:29









Jwan622

1,60111224




1,60111224







  • 3




    Your edit is correct
    – Rushabh Mehta
    Aug 1 at 15:30






  • 1




    In general, if you get an incorrect answer, find your error by "bisection" ... take one of your steps from the middle, plug in your incorrect answer, and see if it works. This will tell you whether your mistake is before or after that point.
    – GEdgar
    Aug 1 at 15:40












  • 3




    Your edit is correct
    – Rushabh Mehta
    Aug 1 at 15:30






  • 1




    In general, if you get an incorrect answer, find your error by "bisection" ... take one of your steps from the middle, plug in your incorrect answer, and see if it works. This will tell you whether your mistake is before or after that point.
    – GEdgar
    Aug 1 at 15:40







3




3




Your edit is correct
– Rushabh Mehta
Aug 1 at 15:30




Your edit is correct
– Rushabh Mehta
Aug 1 at 15:30




1




1




In general, if you get an incorrect answer, find your error by "bisection" ... take one of your steps from the middle, plug in your incorrect answer, and see if it works. This will tell you whether your mistake is before or after that point.
– GEdgar
Aug 1 at 15:40




In general, if you get an incorrect answer, find your error by "bisection" ... take one of your steps from the middle, plug in your incorrect answer, and see if it works. This will tell you whether your mistake is before or after that point.
– GEdgar
Aug 1 at 15:40










3 Answers
3






active

oldest

votes

















up vote
2
down vote



accepted










Yes, your answer is correct.



Check that $(1,2,0)$ is a solution and also since the rank is $3$, there is a unique solution.



$10z=0 implies z=0$, substitute that to other equations, we easily get $y=2$ and then $x-y+0 = -1 implies x-2=-1 implies x=1$.






share|cite|improve this answer





















  • What do you mean the rank is 3 there is a unique solution? can you clarify?
    – Jwan622
    Aug 1 at 15:36










  • you do not have any non-pivot columns, every column ($3$ of them) are pivot columns hence there is a unique solution.
    – Siong Thye Goh
    Aug 1 at 15:38










  • what's a pivot column
    – Jwan622
    Aug 1 at 15:39










  • In the case of Gaussian elimination, those leading non-zero entries are the pivot elements, those columns corresponding to the pivot elements are pivot columns. might be of interest
    – Siong Thye Goh
    Aug 1 at 15:41


















up vote
2
down vote













I don't understand your way to obtain the RREF, we can proceed as follow



$$beginbmatrix
1 & -1 & 1 & -1
\2 & 1 & -3 & 4
\2 & 0 & 2 & 2
endbmatrixstackrelR3-R2to beginbmatrix
1 & -1 & 1 & -1
\2 & 1 & -3 & 4
\0 & -1 & 5 & -2
endbmatrixstackrelR2-2cdot R1to beginbmatrix
1 & -1 & 1 & -1
\0 & 3 & -5 & 6
\0 & -1 & 5 & -2
endbmatrixstackrel3cdot R3+R2to beginbmatrix
1 & -1 & 1 & -1
\0 & 3 & -5 & 6
\0 & 0 & 10 & 0
endbmatrix$$



and since the matrix is full rank (we have three pivots) we have an unique solution that is



  • from the third row: $z=0$

  • from the second row: $y=2$

  • from the first row: $x=1$





share|cite|improve this answer























  • its wrong, you made a mistake in the first step when you tried to add -1(2nd row) to the third row
    – Jwan622
    Aug 1 at 15:37










  • @Jwan622 Yes just noticed and fixed meanwhile! Thanks
    – gimusi
    Aug 1 at 15:38






  • 1




    @Isham Yes, now it should be ok. Thanks
    – gimusi
    Aug 1 at 15:40










  • @Isham Ok that's was really a typo! Thanks again
    – gimusi
    Aug 1 at 15:45

















up vote
0
down vote













I'd use a more systematic method:
beginalign
beginbmatrix
1 & -1 & 1 & -1\
2 & 1 & -3 & 4\
2 & 0 & 2 & 2
endbmatrix
&to
beginbmatrix
1 & -1 & 1 & -1\
0 & 3 & -5 & 6\
0 & 2 & 0 & 4
endbmatrix
&&beginaligned R_2&gets R_2-2R_1 \ R_3&gets R_3-2R_1 endaligned
\ &to
beginbmatrix
1 & -1 & 1 & -1\
0 & 1 & -5/3 & 2\
0 & 2 & 0 & 4
endbmatrix
&& R_2getstfrac13R_2
\ &to
beginbmatrix
1 & -1 & 1 & -1\
0 & 1 & -5/3 & 2\
0 & 0 & 10/3 & 0
endbmatrix
&& R_3gets R_3-2R_2
\ &to
beginbmatrix
1 & -1 & 1 & -1\
0 & 1 & -5/3 & 2\
0 & 0 & 1 & 0
endbmatrix
&& R_3getstfrac310R_3
\ &to
beginbmatrix
1 & -1 & 0 & -1\
0 & 1 & 0 & 2\
0 & 0 & 1 & 0
endbmatrix
&& beginaligned R_2 &gets R_2+tfrac53R_3 \ R_1&gets R_1-R_3endaligned
\ &to
beginbmatrix
1 & 0 & 0 & 1\
0 & 1 & 0 & 2\
0 & 0 & 1 & 0
endbmatrix
&& R_1gets R_1+R_2
endalign
The solution, which is explicit when the RREF is reached, is
beginbmatrix 1 \ 2 \ 0 endbmatrix






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    3 Answers
    3






    active

    oldest

    votes








    3 Answers
    3






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes








    up vote
    2
    down vote



    accepted










    Yes, your answer is correct.



    Check that $(1,2,0)$ is a solution and also since the rank is $3$, there is a unique solution.



    $10z=0 implies z=0$, substitute that to other equations, we easily get $y=2$ and then $x-y+0 = -1 implies x-2=-1 implies x=1$.






    share|cite|improve this answer





















    • What do you mean the rank is 3 there is a unique solution? can you clarify?
      – Jwan622
      Aug 1 at 15:36










    • you do not have any non-pivot columns, every column ($3$ of them) are pivot columns hence there is a unique solution.
      – Siong Thye Goh
      Aug 1 at 15:38










    • what's a pivot column
      – Jwan622
      Aug 1 at 15:39










    • In the case of Gaussian elimination, those leading non-zero entries are the pivot elements, those columns corresponding to the pivot elements are pivot columns. might be of interest
      – Siong Thye Goh
      Aug 1 at 15:41















    up vote
    2
    down vote



    accepted










    Yes, your answer is correct.



    Check that $(1,2,0)$ is a solution and also since the rank is $3$, there is a unique solution.



    $10z=0 implies z=0$, substitute that to other equations, we easily get $y=2$ and then $x-y+0 = -1 implies x-2=-1 implies x=1$.






    share|cite|improve this answer





















    • What do you mean the rank is 3 there is a unique solution? can you clarify?
      – Jwan622
      Aug 1 at 15:36










    • you do not have any non-pivot columns, every column ($3$ of them) are pivot columns hence there is a unique solution.
      – Siong Thye Goh
      Aug 1 at 15:38










    • what's a pivot column
      – Jwan622
      Aug 1 at 15:39










    • In the case of Gaussian elimination, those leading non-zero entries are the pivot elements, those columns corresponding to the pivot elements are pivot columns. might be of interest
      – Siong Thye Goh
      Aug 1 at 15:41













    up vote
    2
    down vote



    accepted







    up vote
    2
    down vote



    accepted






    Yes, your answer is correct.



    Check that $(1,2,0)$ is a solution and also since the rank is $3$, there is a unique solution.



    $10z=0 implies z=0$, substitute that to other equations, we easily get $y=2$ and then $x-y+0 = -1 implies x-2=-1 implies x=1$.






    share|cite|improve this answer













    Yes, your answer is correct.



    Check that $(1,2,0)$ is a solution and also since the rank is $3$, there is a unique solution.



    $10z=0 implies z=0$, substitute that to other equations, we easily get $y=2$ and then $x-y+0 = -1 implies x-2=-1 implies x=1$.







    share|cite|improve this answer













    share|cite|improve this answer



    share|cite|improve this answer











    answered Aug 1 at 15:33









    Siong Thye Goh

    76.7k134794




    76.7k134794











    • What do you mean the rank is 3 there is a unique solution? can you clarify?
      – Jwan622
      Aug 1 at 15:36










    • you do not have any non-pivot columns, every column ($3$ of them) are pivot columns hence there is a unique solution.
      – Siong Thye Goh
      Aug 1 at 15:38










    • what's a pivot column
      – Jwan622
      Aug 1 at 15:39










    • In the case of Gaussian elimination, those leading non-zero entries are the pivot elements, those columns corresponding to the pivot elements are pivot columns. might be of interest
      – Siong Thye Goh
      Aug 1 at 15:41

















    • What do you mean the rank is 3 there is a unique solution? can you clarify?
      – Jwan622
      Aug 1 at 15:36










    • you do not have any non-pivot columns, every column ($3$ of them) are pivot columns hence there is a unique solution.
      – Siong Thye Goh
      Aug 1 at 15:38










    • what's a pivot column
      – Jwan622
      Aug 1 at 15:39










    • In the case of Gaussian elimination, those leading non-zero entries are the pivot elements, those columns corresponding to the pivot elements are pivot columns. might be of interest
      – Siong Thye Goh
      Aug 1 at 15:41
















    What do you mean the rank is 3 there is a unique solution? can you clarify?
    – Jwan622
    Aug 1 at 15:36




    What do you mean the rank is 3 there is a unique solution? can you clarify?
    – Jwan622
    Aug 1 at 15:36












    you do not have any non-pivot columns, every column ($3$ of them) are pivot columns hence there is a unique solution.
    – Siong Thye Goh
    Aug 1 at 15:38




    you do not have any non-pivot columns, every column ($3$ of them) are pivot columns hence there is a unique solution.
    – Siong Thye Goh
    Aug 1 at 15:38












    what's a pivot column
    – Jwan622
    Aug 1 at 15:39




    what's a pivot column
    – Jwan622
    Aug 1 at 15:39












    In the case of Gaussian elimination, those leading non-zero entries are the pivot elements, those columns corresponding to the pivot elements are pivot columns. might be of interest
    – Siong Thye Goh
    Aug 1 at 15:41





    In the case of Gaussian elimination, those leading non-zero entries are the pivot elements, those columns corresponding to the pivot elements are pivot columns. might be of interest
    – Siong Thye Goh
    Aug 1 at 15:41











    up vote
    2
    down vote













    I don't understand your way to obtain the RREF, we can proceed as follow



    $$beginbmatrix
    1 & -1 & 1 & -1
    \2 & 1 & -3 & 4
    \2 & 0 & 2 & 2
    endbmatrixstackrelR3-R2to beginbmatrix
    1 & -1 & 1 & -1
    \2 & 1 & -3 & 4
    \0 & -1 & 5 & -2
    endbmatrixstackrelR2-2cdot R1to beginbmatrix
    1 & -1 & 1 & -1
    \0 & 3 & -5 & 6
    \0 & -1 & 5 & -2
    endbmatrixstackrel3cdot R3+R2to beginbmatrix
    1 & -1 & 1 & -1
    \0 & 3 & -5 & 6
    \0 & 0 & 10 & 0
    endbmatrix$$



    and since the matrix is full rank (we have three pivots) we have an unique solution that is



    • from the third row: $z=0$

    • from the second row: $y=2$

    • from the first row: $x=1$





    share|cite|improve this answer























    • its wrong, you made a mistake in the first step when you tried to add -1(2nd row) to the third row
      – Jwan622
      Aug 1 at 15:37










    • @Jwan622 Yes just noticed and fixed meanwhile! Thanks
      – gimusi
      Aug 1 at 15:38






    • 1




      @Isham Yes, now it should be ok. Thanks
      – gimusi
      Aug 1 at 15:40










    • @Isham Ok that's was really a typo! Thanks again
      – gimusi
      Aug 1 at 15:45














    up vote
    2
    down vote













    I don't understand your way to obtain the RREF, we can proceed as follow



    $$beginbmatrix
    1 & -1 & 1 & -1
    \2 & 1 & -3 & 4
    \2 & 0 & 2 & 2
    endbmatrixstackrelR3-R2to beginbmatrix
    1 & -1 & 1 & -1
    \2 & 1 & -3 & 4
    \0 & -1 & 5 & -2
    endbmatrixstackrelR2-2cdot R1to beginbmatrix
    1 & -1 & 1 & -1
    \0 & 3 & -5 & 6
    \0 & -1 & 5 & -2
    endbmatrixstackrel3cdot R3+R2to beginbmatrix
    1 & -1 & 1 & -1
    \0 & 3 & -5 & 6
    \0 & 0 & 10 & 0
    endbmatrix$$



    and since the matrix is full rank (we have three pivots) we have an unique solution that is



    • from the third row: $z=0$

    • from the second row: $y=2$

    • from the first row: $x=1$





    share|cite|improve this answer























    • its wrong, you made a mistake in the first step when you tried to add -1(2nd row) to the third row
      – Jwan622
      Aug 1 at 15:37










    • @Jwan622 Yes just noticed and fixed meanwhile! Thanks
      – gimusi
      Aug 1 at 15:38






    • 1




      @Isham Yes, now it should be ok. Thanks
      – gimusi
      Aug 1 at 15:40










    • @Isham Ok that's was really a typo! Thanks again
      – gimusi
      Aug 1 at 15:45












    up vote
    2
    down vote










    up vote
    2
    down vote









    I don't understand your way to obtain the RREF, we can proceed as follow



    $$beginbmatrix
    1 & -1 & 1 & -1
    \2 & 1 & -3 & 4
    \2 & 0 & 2 & 2
    endbmatrixstackrelR3-R2to beginbmatrix
    1 & -1 & 1 & -1
    \2 & 1 & -3 & 4
    \0 & -1 & 5 & -2
    endbmatrixstackrelR2-2cdot R1to beginbmatrix
    1 & -1 & 1 & -1
    \0 & 3 & -5 & 6
    \0 & -1 & 5 & -2
    endbmatrixstackrel3cdot R3+R2to beginbmatrix
    1 & -1 & 1 & -1
    \0 & 3 & -5 & 6
    \0 & 0 & 10 & 0
    endbmatrix$$



    and since the matrix is full rank (we have three pivots) we have an unique solution that is



    • from the third row: $z=0$

    • from the second row: $y=2$

    • from the first row: $x=1$





    share|cite|improve this answer















    I don't understand your way to obtain the RREF, we can proceed as follow



    $$beginbmatrix
    1 & -1 & 1 & -1
    \2 & 1 & -3 & 4
    \2 & 0 & 2 & 2
    endbmatrixstackrelR3-R2to beginbmatrix
    1 & -1 & 1 & -1
    \2 & 1 & -3 & 4
    \0 & -1 & 5 & -2
    endbmatrixstackrelR2-2cdot R1to beginbmatrix
    1 & -1 & 1 & -1
    \0 & 3 & -5 & 6
    \0 & -1 & 5 & -2
    endbmatrixstackrel3cdot R3+R2to beginbmatrix
    1 & -1 & 1 & -1
    \0 & 3 & -5 & 6
    \0 & 0 & 10 & 0
    endbmatrix$$



    and since the matrix is full rank (we have three pivots) we have an unique solution that is



    • from the third row: $z=0$

    • from the second row: $y=2$

    • from the first row: $x=1$






    share|cite|improve this answer















    share|cite|improve this answer



    share|cite|improve this answer








    edited Aug 1 at 15:45


























    answered Aug 1 at 15:35









    gimusi

    63.9k73480




    63.9k73480











    • its wrong, you made a mistake in the first step when you tried to add -1(2nd row) to the third row
      – Jwan622
      Aug 1 at 15:37










    • @Jwan622 Yes just noticed and fixed meanwhile! Thanks
      – gimusi
      Aug 1 at 15:38






    • 1




      @Isham Yes, now it should be ok. Thanks
      – gimusi
      Aug 1 at 15:40










    • @Isham Ok that's was really a typo! Thanks again
      – gimusi
      Aug 1 at 15:45
















    • its wrong, you made a mistake in the first step when you tried to add -1(2nd row) to the third row
      – Jwan622
      Aug 1 at 15:37










    • @Jwan622 Yes just noticed and fixed meanwhile! Thanks
      – gimusi
      Aug 1 at 15:38






    • 1




      @Isham Yes, now it should be ok. Thanks
      – gimusi
      Aug 1 at 15:40










    • @Isham Ok that's was really a typo! Thanks again
      – gimusi
      Aug 1 at 15:45















    its wrong, you made a mistake in the first step when you tried to add -1(2nd row) to the third row
    – Jwan622
    Aug 1 at 15:37




    its wrong, you made a mistake in the first step when you tried to add -1(2nd row) to the third row
    – Jwan622
    Aug 1 at 15:37












    @Jwan622 Yes just noticed and fixed meanwhile! Thanks
    – gimusi
    Aug 1 at 15:38




    @Jwan622 Yes just noticed and fixed meanwhile! Thanks
    – gimusi
    Aug 1 at 15:38




    1




    1




    @Isham Yes, now it should be ok. Thanks
    – gimusi
    Aug 1 at 15:40




    @Isham Yes, now it should be ok. Thanks
    – gimusi
    Aug 1 at 15:40












    @Isham Ok that's was really a typo! Thanks again
    – gimusi
    Aug 1 at 15:45




    @Isham Ok that's was really a typo! Thanks again
    – gimusi
    Aug 1 at 15:45










    up vote
    0
    down vote













    I'd use a more systematic method:
    beginalign
    beginbmatrix
    1 & -1 & 1 & -1\
    2 & 1 & -3 & 4\
    2 & 0 & 2 & 2
    endbmatrix
    &to
    beginbmatrix
    1 & -1 & 1 & -1\
    0 & 3 & -5 & 6\
    0 & 2 & 0 & 4
    endbmatrix
    &&beginaligned R_2&gets R_2-2R_1 \ R_3&gets R_3-2R_1 endaligned
    \ &to
    beginbmatrix
    1 & -1 & 1 & -1\
    0 & 1 & -5/3 & 2\
    0 & 2 & 0 & 4
    endbmatrix
    && R_2getstfrac13R_2
    \ &to
    beginbmatrix
    1 & -1 & 1 & -1\
    0 & 1 & -5/3 & 2\
    0 & 0 & 10/3 & 0
    endbmatrix
    && R_3gets R_3-2R_2
    \ &to
    beginbmatrix
    1 & -1 & 1 & -1\
    0 & 1 & -5/3 & 2\
    0 & 0 & 1 & 0
    endbmatrix
    && R_3getstfrac310R_3
    \ &to
    beginbmatrix
    1 & -1 & 0 & -1\
    0 & 1 & 0 & 2\
    0 & 0 & 1 & 0
    endbmatrix
    && beginaligned R_2 &gets R_2+tfrac53R_3 \ R_1&gets R_1-R_3endaligned
    \ &to
    beginbmatrix
    1 & 0 & 0 & 1\
    0 & 1 & 0 & 2\
    0 & 0 & 1 & 0
    endbmatrix
    && R_1gets R_1+R_2
    endalign
    The solution, which is explicit when the RREF is reached, is
    beginbmatrix 1 \ 2 \ 0 endbmatrix






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      up vote
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      down vote













      I'd use a more systematic method:
      beginalign
      beginbmatrix
      1 & -1 & 1 & -1\
      2 & 1 & -3 & 4\
      2 & 0 & 2 & 2
      endbmatrix
      &to
      beginbmatrix
      1 & -1 & 1 & -1\
      0 & 3 & -5 & 6\
      0 & 2 & 0 & 4
      endbmatrix
      &&beginaligned R_2&gets R_2-2R_1 \ R_3&gets R_3-2R_1 endaligned
      \ &to
      beginbmatrix
      1 & -1 & 1 & -1\
      0 & 1 & -5/3 & 2\
      0 & 2 & 0 & 4
      endbmatrix
      && R_2getstfrac13R_2
      \ &to
      beginbmatrix
      1 & -1 & 1 & -1\
      0 & 1 & -5/3 & 2\
      0 & 0 & 10/3 & 0
      endbmatrix
      && R_3gets R_3-2R_2
      \ &to
      beginbmatrix
      1 & -1 & 1 & -1\
      0 & 1 & -5/3 & 2\
      0 & 0 & 1 & 0
      endbmatrix
      && R_3getstfrac310R_3
      \ &to
      beginbmatrix
      1 & -1 & 0 & -1\
      0 & 1 & 0 & 2\
      0 & 0 & 1 & 0
      endbmatrix
      && beginaligned R_2 &gets R_2+tfrac53R_3 \ R_1&gets R_1-R_3endaligned
      \ &to
      beginbmatrix
      1 & 0 & 0 & 1\
      0 & 1 & 0 & 2\
      0 & 0 & 1 & 0
      endbmatrix
      && R_1gets R_1+R_2
      endalign
      The solution, which is explicit when the RREF is reached, is
      beginbmatrix 1 \ 2 \ 0 endbmatrix






      share|cite|improve this answer























        up vote
        0
        down vote










        up vote
        0
        down vote









        I'd use a more systematic method:
        beginalign
        beginbmatrix
        1 & -1 & 1 & -1\
        2 & 1 & -3 & 4\
        2 & 0 & 2 & 2
        endbmatrix
        &to
        beginbmatrix
        1 & -1 & 1 & -1\
        0 & 3 & -5 & 6\
        0 & 2 & 0 & 4
        endbmatrix
        &&beginaligned R_2&gets R_2-2R_1 \ R_3&gets R_3-2R_1 endaligned
        \ &to
        beginbmatrix
        1 & -1 & 1 & -1\
        0 & 1 & -5/3 & 2\
        0 & 2 & 0 & 4
        endbmatrix
        && R_2getstfrac13R_2
        \ &to
        beginbmatrix
        1 & -1 & 1 & -1\
        0 & 1 & -5/3 & 2\
        0 & 0 & 10/3 & 0
        endbmatrix
        && R_3gets R_3-2R_2
        \ &to
        beginbmatrix
        1 & -1 & 1 & -1\
        0 & 1 & -5/3 & 2\
        0 & 0 & 1 & 0
        endbmatrix
        && R_3getstfrac310R_3
        \ &to
        beginbmatrix
        1 & -1 & 0 & -1\
        0 & 1 & 0 & 2\
        0 & 0 & 1 & 0
        endbmatrix
        && beginaligned R_2 &gets R_2+tfrac53R_3 \ R_1&gets R_1-R_3endaligned
        \ &to
        beginbmatrix
        1 & 0 & 0 & 1\
        0 & 1 & 0 & 2\
        0 & 0 & 1 & 0
        endbmatrix
        && R_1gets R_1+R_2
        endalign
        The solution, which is explicit when the RREF is reached, is
        beginbmatrix 1 \ 2 \ 0 endbmatrix






        share|cite|improve this answer













        I'd use a more systematic method:
        beginalign
        beginbmatrix
        1 & -1 & 1 & -1\
        2 & 1 & -3 & 4\
        2 & 0 & 2 & 2
        endbmatrix
        &to
        beginbmatrix
        1 & -1 & 1 & -1\
        0 & 3 & -5 & 6\
        0 & 2 & 0 & 4
        endbmatrix
        &&beginaligned R_2&gets R_2-2R_1 \ R_3&gets R_3-2R_1 endaligned
        \ &to
        beginbmatrix
        1 & -1 & 1 & -1\
        0 & 1 & -5/3 & 2\
        0 & 2 & 0 & 4
        endbmatrix
        && R_2getstfrac13R_2
        \ &to
        beginbmatrix
        1 & -1 & 1 & -1\
        0 & 1 & -5/3 & 2\
        0 & 0 & 10/3 & 0
        endbmatrix
        && R_3gets R_3-2R_2
        \ &to
        beginbmatrix
        1 & -1 & 1 & -1\
        0 & 1 & -5/3 & 2\
        0 & 0 & 1 & 0
        endbmatrix
        && R_3getstfrac310R_3
        \ &to
        beginbmatrix
        1 & -1 & 0 & -1\
        0 & 1 & 0 & 2\
        0 & 0 & 1 & 0
        endbmatrix
        && beginaligned R_2 &gets R_2+tfrac53R_3 \ R_1&gets R_1-R_3endaligned
        \ &to
        beginbmatrix
        1 & 0 & 0 & 1\
        0 & 1 & 0 & 2\
        0 & 0 & 1 & 0
        endbmatrix
        && R_1gets R_1+R_2
        endalign
        The solution, which is explicit when the RREF is reached, is
        beginbmatrix 1 \ 2 \ 0 endbmatrix







        share|cite|improve this answer













        share|cite|improve this answer



        share|cite|improve this answer











        answered Aug 1 at 16:52









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