What am I doing wrong solving this system of equations?











up vote
10
down vote

favorite
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$$begin{cases}
2x_1+5x_2-8x_3=8\
4x_1+3x_2-9x_3=9\
2x_1+3x_2-5x_3=7\
x_1+8x_2-7x_3=12
end{cases}$$



From my elementary row operations, I get that it has no solution. (Row operations are to be read from top to bottom.)



$$left[begin{array}{ccc|c}
2 & 5 & -8 & 8 \
4 & 3 & -9 & 9 \
2 & 3 & -5 & 7 \
1 & 8 & -7 & 12
end{array}right]
overset{overset{large{R_1to R_1-R_3}}{{R_2to R_2-2R_3}}}{overset{R_3to R_3-2R_4}{largelongrightarrow}}
left[begin{array}{ccc|c}
0 & 2 & -3 & 1 \
0 & -3 & 1 & -5 \
0 & -13 & 9 & -17 \
1 & 8 & -7 & 12
end{array}right]
overset{overset{large{R_3,leftrightarrow, R_4}}{R_2,leftrightarrow, R_3}}{overset{R_1,leftrightarrow,R_2}{largelongrightarrow}}
left[begin{array}{ccc|c}
1 & 8 & -7 & 12 \
0 & 2 & -3 & 1 \
0 & -3 & 1 & -5 \
0 & -13 & 9 & -17
end{array}right]$$



$$overset{R_4to R_4-R_3}{largelongrightarrow}
left[begin{array}{ccc|c}
1 & 8 & -7 & 12 \
0 & 2 & -3 & 1 \
0 & -3 & 1 & -5 \
0 & 10 & 8 & -12
end{array}right]
overset{overset{large{R_3to R_3+R_2}}{R_4to R_4-5R_2}}{largelongrightarrow}
left[begin{array}{ccc|c}
1 & 8 & -7 & 12 \
0 & 2 & -3 & 1 \
0 & -1 & -2 & -4 \
0 & 0 & 23 & -17
end{array}right]
overset{overset{large{R_2to R_2+2R_3}}{R_3to-R_3}}{largelongrightarrow}$$



$$left[begin{array}{ccc|c}
1 & 8 & -7 & 12 \
0 & 0 & -7 & -7 \
0 & 1 & 2 & 4 \
0 & 0 & 23 & -17 \
end{array}right]
overset{R_2,leftrightarrow,R_3}{largelongrightarrow}
left[begin{array}{ccc|c}
1 & 8 & -7 & 12 \
0 & 1 & 2 & 4 \
0 & 0 & -7 & -7 \
0 & 0 & 23 & -17 \
end{array}right]$$



However, the answer in the book $(3, 2, 1)$ fits the system.



Was there an arithmetical mistake, or do I misunderstand something fundamentally?










share|cite|improve this question




















  • 12




    It is in your best interest that you type your questions (using MathJax) instead of posting links to pictures.
    – José Carlos Santos
    Nov 7 at 9:59






  • 5




    I would also recommend simply checking answers via a computer, you can quickly see your arithmetic mistake matrix.reshish.com/gauss-jordanElimination.php . Even the most experienced mathematicians still make minus sign errors or simple multiplication errors. We're mathematicians after all, not primary school teachers :P
    – WesleyGroupshaveFeelingsToo
    Nov 7 at 10:23















up vote
10
down vote

favorite
1












$$begin{cases}
2x_1+5x_2-8x_3=8\
4x_1+3x_2-9x_3=9\
2x_1+3x_2-5x_3=7\
x_1+8x_2-7x_3=12
end{cases}$$



From my elementary row operations, I get that it has no solution. (Row operations are to be read from top to bottom.)



$$left[begin{array}{ccc|c}
2 & 5 & -8 & 8 \
4 & 3 & -9 & 9 \
2 & 3 & -5 & 7 \
1 & 8 & -7 & 12
end{array}right]
overset{overset{large{R_1to R_1-R_3}}{{R_2to R_2-2R_3}}}{overset{R_3to R_3-2R_4}{largelongrightarrow}}
left[begin{array}{ccc|c}
0 & 2 & -3 & 1 \
0 & -3 & 1 & -5 \
0 & -13 & 9 & -17 \
1 & 8 & -7 & 12
end{array}right]
overset{overset{large{R_3,leftrightarrow, R_4}}{R_2,leftrightarrow, R_3}}{overset{R_1,leftrightarrow,R_2}{largelongrightarrow}}
left[begin{array}{ccc|c}
1 & 8 & -7 & 12 \
0 & 2 & -3 & 1 \
0 & -3 & 1 & -5 \
0 & -13 & 9 & -17
end{array}right]$$



$$overset{R_4to R_4-R_3}{largelongrightarrow}
left[begin{array}{ccc|c}
1 & 8 & -7 & 12 \
0 & 2 & -3 & 1 \
0 & -3 & 1 & -5 \
0 & 10 & 8 & -12
end{array}right]
overset{overset{large{R_3to R_3+R_2}}{R_4to R_4-5R_2}}{largelongrightarrow}
left[begin{array}{ccc|c}
1 & 8 & -7 & 12 \
0 & 2 & -3 & 1 \
0 & -1 & -2 & -4 \
0 & 0 & 23 & -17
end{array}right]
overset{overset{large{R_2to R_2+2R_3}}{R_3to-R_3}}{largelongrightarrow}$$



$$left[begin{array}{ccc|c}
1 & 8 & -7 & 12 \
0 & 0 & -7 & -7 \
0 & 1 & 2 & 4 \
0 & 0 & 23 & -17 \
end{array}right]
overset{R_2,leftrightarrow,R_3}{largelongrightarrow}
left[begin{array}{ccc|c}
1 & 8 & -7 & 12 \
0 & 1 & 2 & 4 \
0 & 0 & -7 & -7 \
0 & 0 & 23 & -17 \
end{array}right]$$



However, the answer in the book $(3, 2, 1)$ fits the system.



Was there an arithmetical mistake, or do I misunderstand something fundamentally?










share|cite|improve this question




















  • 12




    It is in your best interest that you type your questions (using MathJax) instead of posting links to pictures.
    – José Carlos Santos
    Nov 7 at 9:59






  • 5




    I would also recommend simply checking answers via a computer, you can quickly see your arithmetic mistake matrix.reshish.com/gauss-jordanElimination.php . Even the most experienced mathematicians still make minus sign errors or simple multiplication errors. We're mathematicians after all, not primary school teachers :P
    – WesleyGroupshaveFeelingsToo
    Nov 7 at 10:23













up vote
10
down vote

favorite
1









up vote
10
down vote

favorite
1






1





$$begin{cases}
2x_1+5x_2-8x_3=8\
4x_1+3x_2-9x_3=9\
2x_1+3x_2-5x_3=7\
x_1+8x_2-7x_3=12
end{cases}$$



From my elementary row operations, I get that it has no solution. (Row operations are to be read from top to bottom.)



$$left[begin{array}{ccc|c}
2 & 5 & -8 & 8 \
4 & 3 & -9 & 9 \
2 & 3 & -5 & 7 \
1 & 8 & -7 & 12
end{array}right]
overset{overset{large{R_1to R_1-R_3}}{{R_2to R_2-2R_3}}}{overset{R_3to R_3-2R_4}{largelongrightarrow}}
left[begin{array}{ccc|c}
0 & 2 & -3 & 1 \
0 & -3 & 1 & -5 \
0 & -13 & 9 & -17 \
1 & 8 & -7 & 12
end{array}right]
overset{overset{large{R_3,leftrightarrow, R_4}}{R_2,leftrightarrow, R_3}}{overset{R_1,leftrightarrow,R_2}{largelongrightarrow}}
left[begin{array}{ccc|c}
1 & 8 & -7 & 12 \
0 & 2 & -3 & 1 \
0 & -3 & 1 & -5 \
0 & -13 & 9 & -17
end{array}right]$$



$$overset{R_4to R_4-R_3}{largelongrightarrow}
left[begin{array}{ccc|c}
1 & 8 & -7 & 12 \
0 & 2 & -3 & 1 \
0 & -3 & 1 & -5 \
0 & 10 & 8 & -12
end{array}right]
overset{overset{large{R_3to R_3+R_2}}{R_4to R_4-5R_2}}{largelongrightarrow}
left[begin{array}{ccc|c}
1 & 8 & -7 & 12 \
0 & 2 & -3 & 1 \
0 & -1 & -2 & -4 \
0 & 0 & 23 & -17
end{array}right]
overset{overset{large{R_2to R_2+2R_3}}{R_3to-R_3}}{largelongrightarrow}$$



$$left[begin{array}{ccc|c}
1 & 8 & -7 & 12 \
0 & 0 & -7 & -7 \
0 & 1 & 2 & 4 \
0 & 0 & 23 & -17 \
end{array}right]
overset{R_2,leftrightarrow,R_3}{largelongrightarrow}
left[begin{array}{ccc|c}
1 & 8 & -7 & 12 \
0 & 1 & 2 & 4 \
0 & 0 & -7 & -7 \
0 & 0 & 23 & -17 \
end{array}right]$$



However, the answer in the book $(3, 2, 1)$ fits the system.



Was there an arithmetical mistake, or do I misunderstand something fundamentally?










share|cite|improve this question















$$begin{cases}
2x_1+5x_2-8x_3=8\
4x_1+3x_2-9x_3=9\
2x_1+3x_2-5x_3=7\
x_1+8x_2-7x_3=12
end{cases}$$



From my elementary row operations, I get that it has no solution. (Row operations are to be read from top to bottom.)



$$left[begin{array}{ccc|c}
2 & 5 & -8 & 8 \
4 & 3 & -9 & 9 \
2 & 3 & -5 & 7 \
1 & 8 & -7 & 12
end{array}right]
overset{overset{large{R_1to R_1-R_3}}{{R_2to R_2-2R_3}}}{overset{R_3to R_3-2R_4}{largelongrightarrow}}
left[begin{array}{ccc|c}
0 & 2 & -3 & 1 \
0 & -3 & 1 & -5 \
0 & -13 & 9 & -17 \
1 & 8 & -7 & 12
end{array}right]
overset{overset{large{R_3,leftrightarrow, R_4}}{R_2,leftrightarrow, R_3}}{overset{R_1,leftrightarrow,R_2}{largelongrightarrow}}
left[begin{array}{ccc|c}
1 & 8 & -7 & 12 \
0 & 2 & -3 & 1 \
0 & -3 & 1 & -5 \
0 & -13 & 9 & -17
end{array}right]$$



$$overset{R_4to R_4-R_3}{largelongrightarrow}
left[begin{array}{ccc|c}
1 & 8 & -7 & 12 \
0 & 2 & -3 & 1 \
0 & -3 & 1 & -5 \
0 & 10 & 8 & -12
end{array}right]
overset{overset{large{R_3to R_3+R_2}}{R_4to R_4-5R_2}}{largelongrightarrow}
left[begin{array}{ccc|c}
1 & 8 & -7 & 12 \
0 & 2 & -3 & 1 \
0 & -1 & -2 & -4 \
0 & 0 & 23 & -17
end{array}right]
overset{overset{large{R_2to R_2+2R_3}}{R_3to-R_3}}{largelongrightarrow}$$



$$left[begin{array}{ccc|c}
1 & 8 & -7 & 12 \
0 & 0 & -7 & -7 \
0 & 1 & 2 & 4 \
0 & 0 & 23 & -17 \
end{array}right]
overset{R_2,leftrightarrow,R_3}{largelongrightarrow}
left[begin{array}{ccc|c}
1 & 8 & -7 & 12 \
0 & 1 & 2 & 4 \
0 & 0 & -7 & -7 \
0 & 0 & 23 & -17 \
end{array}right]$$



However, the answer in the book $(3, 2, 1)$ fits the system.



Was there an arithmetical mistake, or do I misunderstand something fundamentally?







linear-algebra systems-of-equations






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share|cite|improve this question













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edited 2 days ago









Abcd

2,82811130




2,82811130










asked Nov 7 at 9:58









fragileradius

196111




196111








  • 12




    It is in your best interest that you type your questions (using MathJax) instead of posting links to pictures.
    – José Carlos Santos
    Nov 7 at 9:59






  • 5




    I would also recommend simply checking answers via a computer, you can quickly see your arithmetic mistake matrix.reshish.com/gauss-jordanElimination.php . Even the most experienced mathematicians still make minus sign errors or simple multiplication errors. We're mathematicians after all, not primary school teachers :P
    – WesleyGroupshaveFeelingsToo
    Nov 7 at 10:23














  • 12




    It is in your best interest that you type your questions (using MathJax) instead of posting links to pictures.
    – José Carlos Santos
    Nov 7 at 9:59






  • 5




    I would also recommend simply checking answers via a computer, you can quickly see your arithmetic mistake matrix.reshish.com/gauss-jordanElimination.php . Even the most experienced mathematicians still make minus sign errors or simple multiplication errors. We're mathematicians after all, not primary school teachers :P
    – WesleyGroupshaveFeelingsToo
    Nov 7 at 10:23








12




12




It is in your best interest that you type your questions (using MathJax) instead of posting links to pictures.
– José Carlos Santos
Nov 7 at 9:59




It is in your best interest that you type your questions (using MathJax) instead of posting links to pictures.
– José Carlos Santos
Nov 7 at 9:59




5




5




I would also recommend simply checking answers via a computer, you can quickly see your arithmetic mistake matrix.reshish.com/gauss-jordanElimination.php . Even the most experienced mathematicians still make minus sign errors or simple multiplication errors. We're mathematicians after all, not primary school teachers :P
– WesleyGroupshaveFeelingsToo
Nov 7 at 10:23




I would also recommend simply checking answers via a computer, you can quickly see your arithmetic mistake matrix.reshish.com/gauss-jordanElimination.php . Even the most experienced mathematicians still make minus sign errors or simple multiplication errors. We're mathematicians after all, not primary school teachers :P
– WesleyGroupshaveFeelingsToo
Nov 7 at 10:23










4 Answers
4






active

oldest

votes

















up vote
73
down vote



accepted










Hint: Try inputing the solution $(3,2,1)$ into every step. That will allow you to identify the step where you went wrong.






share|cite|improve this answer

















  • 15




    Note: this is probably the simplest answer I ever gave that recieved a score of 10 :)
    – 5xum
    Nov 7 at 11:56






  • 6




    it's also a great and remarkably salient answer that fully solves the problem and many others like it. In other words: it's the best kind of answer. Well-done.
    – The Count
    Nov 7 at 22:45






  • 4




    In addition, it is a nice little case where it makes sense to do a binary search...
    – Sebastian Schoennenbeck
    2 days ago






  • 2




    Hint: There is only one equation in the last set that does not work with that solution, so it is not necessary to input the solution in EVERY equation, just working backwards from the wrong one reveals the problem relatively quickly.
    – kutschkem
    2 days ago


















up vote
19
down vote













You do (in the third matrix): $$L3-L4=(0, -3, 1 mid -5)-(0, -13, 9 mid -19)=(0, 10, -8 mid 12)$$ but you have $(0, 10, 8 mid -12)$ instead.






share|cite|improve this answer






























    up vote
    11
    down vote













    Note 1: Referring to 5xum's answer, in case you don't know the final answer, any solution will do, e.g.:
    $$(x_1,x_2,x_3)=(0,0,-1);(0,0,-1);(1,0,-1);(12,0,0) text{(respectively)}$$
    Note 2: In step $3$, you can reduce column $3$ instead of column $2$:
    $$left[
    begin{array}{ccc|c}
    1&8&-7&12\
    0&2&-3&1\
    0&-3&1&-5\
    0&-13&9&-17\
    end{array}
    right] Rightarrow left[
    begin{array}{ccc|c}
    1&8&-7&12\
    0&-7&0&-14\
    0&-3&1&-5\
    0&14&0&28\
    end{array}
    right] stackrel{frac{-R_2}{7};\ frac{-3R_2}{7}+R_3}{=}Rightarrow left[
    begin{array}{ccc|c}
    1&8&-7&12\
    0&1&0&2\
    0&0&1&1\
    0&14&0&28\
    end{array}
    right]$$

    The second and fourth equations are dependent, therefore they produce the same solution $x_2=2$. You can finish the problem now.



    Note 3. When you can not find your mistake (sometimes the brain gets blocked/accustommed and can not see obvious mistakes), leave it for some time (1 hour, 1 day) and return with fresh mind. You will be surprised to easily spot the error.






    share|cite|improve this answer

















    • 1




      +1 for the last paragraph: Keep calm & try again later is a good approach
      – Barranka
      Nov 7 at 18:20










    • can you explain Note 1? How can I use this four tuples you show us to find the error?
      – miracle173
      2 days ago










    • @miracle173, unfortunately all four cannot be used at once, but pairwise solutions (which is relatively simple to find by putting $0$ into one of three unknowns) must be checked. For example, equations $1$ and $2$ have common solution $(0,0,-1)$, so any row operation of the two equations must be satisfied by it. Or $(8/3,0,-1/3)$ is a common solution of the equations $1$ and $3$, so the OP's first result $[0 2 -3 | 1]$ must be satisfied by it.
      – farruhota
      2 days ago










    • I understand now, but this seems not to be useful to me.
      – miracle173
      2 days ago


















    up vote
    2
    down vote













    Usually you don't know the solution of your equation. So you have to use other methods for check your calculation. One way is to add checksums to your calculation.



    Don't start your calculations with the original matrix



    $$left[begin{array}{ccc|c}
    2 & 5 & -8 & 8 \
    4 & 3 & -9 & 9 \
    2 & 3 & -5 & 7 \
    1 & 8 & -7 & 12
    end{array}right]
    $$



    but start with the matrix augmented by a column at the right. An element of this column is equal to the sum of the other elements of this row. So the augmented matrix is



    $$left[begin{array}{ccc|c|c}
    2 & 5 & -8 & 8 & 7 \
    4 & 3 & -9 & 9 & 7 \
    2 & 3 & -5 & 7 & 7 \
    1 & 8 & -7 & 12 & 14
    end{array}right]
    $$



    You do the same row operations as with the original matrices.



    So instead of



    $$left[begin{array}{ccc|c}
    2 & 5 & -8 & 8 \
    4 & 3 & -9 & 9 \
    2 & 3 & -5 & 7 \
    1 & 8 & -7 & 12
    end{array}right]
    overset{overset{large{R_1to R_1-R_3}}{{R_2to R_2-2R_3}}}{overset{R_3to R_3-2R_4}{largelongrightarrow}}
    left[begin{array}{ccc|c}
    0 & 2 & -3 & 1 \
    0 & -3 & 1 & -5 \
    0 & -13 & 9 & -17 \
    1 & 8 & -7 & 12
    end{array}right]$$



    you do



    $$left[begin{array}{ccc|c|c}
    2 & 5 & -8 & 8 & 7 \
    4 & 3 & -9 & 9 & 7 \
    2 & 3 & -5 & 7 & 7 \
    1 & 8 & -7 & 12 & 14
    end{array}right]
    overset{overset{large{R_1to R_1-R_3}}{{R_2to R_2-2R_3}}}{overset{R_3to R_3-2R_4}{largelongrightarrow}}
    left[begin{array}{ccc|c}
    0 & 2 & -3 & 1 & 0 \
    0 & -3 & 1 & -5 & -7 \
    0 & -13 & 9 & -17 & -21 \
    1 & 8 & -7 & 12 & 14
    end{array}right]$$



    Then you check if for the matrix you calculated the checksum is still valid.
    Here the checksum is correct.



    Now let's investigate the step where the error occurs. We get



    $$left[begin{array}{ccc|c|c}
    1 & 8 & -7 & 12 & 14 \
    0 & 2 & -3 & 1 & 0 \
    0 & -3 & 1 & -5 & -7 \
    0 & -13 & 9 & -17 & -21
    end{array}right]
    overset{R_4to R_4-R_3}{largelongrightarrow}
    left[begin{array}{ccc|c | c}
    1 & 8 & -7 & 12 & 14\
    0 & 2 & -3 & 1 & 0\
    0 & -3 & 1 & -5 & -7\
    0 & 10 & 8 & -12 & -14
    end{array}right]
    $$



    For this result matrix the checksum of row 4 is not valid, so an error must have occurred.





    Actually the checksum property means $(1,1,1,1)^T$ is a solution of the following system of equations that corresponds to the augmented matrix:



    $$begin{eqnarray}
    2x_1&+&5x_2&-&8x_3&+&8x_4&=&7\
    4x_1&+&3x_2&-&9x_3&+&9x_4&=&7\
    2x_1&+&3x_2&-&5x_3&+&7x_4&=&7\
    x_1&+&8x_2&-&7x_3&+&12x_4&=&14
    end{eqnarray}$$



    So this method is similar to the method in the accepted answer, where the solution given by the textbook is used.



    What if this method does not uncover the errors. This can happen if you made more then one error in a step- Then you can use a weighted checksum to try to uncover the error.






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      4 Answers
      4






      active

      oldest

      votes








      4 Answers
      4






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes








      up vote
      73
      down vote



      accepted










      Hint: Try inputing the solution $(3,2,1)$ into every step. That will allow you to identify the step where you went wrong.






      share|cite|improve this answer

















      • 15




        Note: this is probably the simplest answer I ever gave that recieved a score of 10 :)
        – 5xum
        Nov 7 at 11:56






      • 6




        it's also a great and remarkably salient answer that fully solves the problem and many others like it. In other words: it's the best kind of answer. Well-done.
        – The Count
        Nov 7 at 22:45






      • 4




        In addition, it is a nice little case where it makes sense to do a binary search...
        – Sebastian Schoennenbeck
        2 days ago






      • 2




        Hint: There is only one equation in the last set that does not work with that solution, so it is not necessary to input the solution in EVERY equation, just working backwards from the wrong one reveals the problem relatively quickly.
        – kutschkem
        2 days ago















      up vote
      73
      down vote



      accepted










      Hint: Try inputing the solution $(3,2,1)$ into every step. That will allow you to identify the step where you went wrong.






      share|cite|improve this answer

















      • 15




        Note: this is probably the simplest answer I ever gave that recieved a score of 10 :)
        – 5xum
        Nov 7 at 11:56






      • 6




        it's also a great and remarkably salient answer that fully solves the problem and many others like it. In other words: it's the best kind of answer. Well-done.
        – The Count
        Nov 7 at 22:45






      • 4




        In addition, it is a nice little case where it makes sense to do a binary search...
        – Sebastian Schoennenbeck
        2 days ago






      • 2




        Hint: There is only one equation in the last set that does not work with that solution, so it is not necessary to input the solution in EVERY equation, just working backwards from the wrong one reveals the problem relatively quickly.
        – kutschkem
        2 days ago













      up vote
      73
      down vote



      accepted







      up vote
      73
      down vote



      accepted






      Hint: Try inputing the solution $(3,2,1)$ into every step. That will allow you to identify the step where you went wrong.






      share|cite|improve this answer












      Hint: Try inputing the solution $(3,2,1)$ into every step. That will allow you to identify the step where you went wrong.







      share|cite|improve this answer












      share|cite|improve this answer



      share|cite|improve this answer










      answered Nov 7 at 10:07









      5xum

      87.4k391159




      87.4k391159








      • 15




        Note: this is probably the simplest answer I ever gave that recieved a score of 10 :)
        – 5xum
        Nov 7 at 11:56






      • 6




        it's also a great and remarkably salient answer that fully solves the problem and many others like it. In other words: it's the best kind of answer. Well-done.
        – The Count
        Nov 7 at 22:45






      • 4




        In addition, it is a nice little case where it makes sense to do a binary search...
        – Sebastian Schoennenbeck
        2 days ago






      • 2




        Hint: There is only one equation in the last set that does not work with that solution, so it is not necessary to input the solution in EVERY equation, just working backwards from the wrong one reveals the problem relatively quickly.
        – kutschkem
        2 days ago














      • 15




        Note: this is probably the simplest answer I ever gave that recieved a score of 10 :)
        – 5xum
        Nov 7 at 11:56






      • 6




        it's also a great and remarkably salient answer that fully solves the problem and many others like it. In other words: it's the best kind of answer. Well-done.
        – The Count
        Nov 7 at 22:45






      • 4




        In addition, it is a nice little case where it makes sense to do a binary search...
        – Sebastian Schoennenbeck
        2 days ago






      • 2




        Hint: There is only one equation in the last set that does not work with that solution, so it is not necessary to input the solution in EVERY equation, just working backwards from the wrong one reveals the problem relatively quickly.
        – kutschkem
        2 days ago








      15




      15




      Note: this is probably the simplest answer I ever gave that recieved a score of 10 :)
      – 5xum
      Nov 7 at 11:56




      Note: this is probably the simplest answer I ever gave that recieved a score of 10 :)
      – 5xum
      Nov 7 at 11:56




      6




      6




      it's also a great and remarkably salient answer that fully solves the problem and many others like it. In other words: it's the best kind of answer. Well-done.
      – The Count
      Nov 7 at 22:45




      it's also a great and remarkably salient answer that fully solves the problem and many others like it. In other words: it's the best kind of answer. Well-done.
      – The Count
      Nov 7 at 22:45




      4




      4




      In addition, it is a nice little case where it makes sense to do a binary search...
      – Sebastian Schoennenbeck
      2 days ago




      In addition, it is a nice little case where it makes sense to do a binary search...
      – Sebastian Schoennenbeck
      2 days ago




      2




      2




      Hint: There is only one equation in the last set that does not work with that solution, so it is not necessary to input the solution in EVERY equation, just working backwards from the wrong one reveals the problem relatively quickly.
      – kutschkem
      2 days ago




      Hint: There is only one equation in the last set that does not work with that solution, so it is not necessary to input the solution in EVERY equation, just working backwards from the wrong one reveals the problem relatively quickly.
      – kutschkem
      2 days ago










      up vote
      19
      down vote













      You do (in the third matrix): $$L3-L4=(0, -3, 1 mid -5)-(0, -13, 9 mid -19)=(0, 10, -8 mid 12)$$ but you have $(0, 10, 8 mid -12)$ instead.






      share|cite|improve this answer



























        up vote
        19
        down vote













        You do (in the third matrix): $$L3-L4=(0, -3, 1 mid -5)-(0, -13, 9 mid -19)=(0, 10, -8 mid 12)$$ but you have $(0, 10, 8 mid -12)$ instead.






        share|cite|improve this answer

























          up vote
          19
          down vote










          up vote
          19
          down vote









          You do (in the third matrix): $$L3-L4=(0, -3, 1 mid -5)-(0, -13, 9 mid -19)=(0, 10, -8 mid 12)$$ but you have $(0, 10, 8 mid -12)$ instead.






          share|cite|improve this answer














          You do (in the third matrix): $$L3-L4=(0, -3, 1 mid -5)-(0, -13, 9 mid -19)=(0, 10, -8 mid 12)$$ but you have $(0, 10, 8 mid -12)$ instead.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Nov 7 at 10:23

























          answered Nov 7 at 10:16









          Jimmy R.

          32.5k42156




          32.5k42156






















              up vote
              11
              down vote













              Note 1: Referring to 5xum's answer, in case you don't know the final answer, any solution will do, e.g.:
              $$(x_1,x_2,x_3)=(0,0,-1);(0,0,-1);(1,0,-1);(12,0,0) text{(respectively)}$$
              Note 2: In step $3$, you can reduce column $3$ instead of column $2$:
              $$left[
              begin{array}{ccc|c}
              1&8&-7&12\
              0&2&-3&1\
              0&-3&1&-5\
              0&-13&9&-17\
              end{array}
              right] Rightarrow left[
              begin{array}{ccc|c}
              1&8&-7&12\
              0&-7&0&-14\
              0&-3&1&-5\
              0&14&0&28\
              end{array}
              right] stackrel{frac{-R_2}{7};\ frac{-3R_2}{7}+R_3}{=}Rightarrow left[
              begin{array}{ccc|c}
              1&8&-7&12\
              0&1&0&2\
              0&0&1&1\
              0&14&0&28\
              end{array}
              right]$$

              The second and fourth equations are dependent, therefore they produce the same solution $x_2=2$. You can finish the problem now.



              Note 3. When you can not find your mistake (sometimes the brain gets blocked/accustommed and can not see obvious mistakes), leave it for some time (1 hour, 1 day) and return with fresh mind. You will be surprised to easily spot the error.






              share|cite|improve this answer

















              • 1




                +1 for the last paragraph: Keep calm & try again later is a good approach
                – Barranka
                Nov 7 at 18:20










              • can you explain Note 1? How can I use this four tuples you show us to find the error?
                – miracle173
                2 days ago










              • @miracle173, unfortunately all four cannot be used at once, but pairwise solutions (which is relatively simple to find by putting $0$ into one of three unknowns) must be checked. For example, equations $1$ and $2$ have common solution $(0,0,-1)$, so any row operation of the two equations must be satisfied by it. Or $(8/3,0,-1/3)$ is a common solution of the equations $1$ and $3$, so the OP's first result $[0 2 -3 | 1]$ must be satisfied by it.
                – farruhota
                2 days ago










              • I understand now, but this seems not to be useful to me.
                – miracle173
                2 days ago















              up vote
              11
              down vote













              Note 1: Referring to 5xum's answer, in case you don't know the final answer, any solution will do, e.g.:
              $$(x_1,x_2,x_3)=(0,0,-1);(0,0,-1);(1,0,-1);(12,0,0) text{(respectively)}$$
              Note 2: In step $3$, you can reduce column $3$ instead of column $2$:
              $$left[
              begin{array}{ccc|c}
              1&8&-7&12\
              0&2&-3&1\
              0&-3&1&-5\
              0&-13&9&-17\
              end{array}
              right] Rightarrow left[
              begin{array}{ccc|c}
              1&8&-7&12\
              0&-7&0&-14\
              0&-3&1&-5\
              0&14&0&28\
              end{array}
              right] stackrel{frac{-R_2}{7};\ frac{-3R_2}{7}+R_3}{=}Rightarrow left[
              begin{array}{ccc|c}
              1&8&-7&12\
              0&1&0&2\
              0&0&1&1\
              0&14&0&28\
              end{array}
              right]$$

              The second and fourth equations are dependent, therefore they produce the same solution $x_2=2$. You can finish the problem now.



              Note 3. When you can not find your mistake (sometimes the brain gets blocked/accustommed and can not see obvious mistakes), leave it for some time (1 hour, 1 day) and return with fresh mind. You will be surprised to easily spot the error.






              share|cite|improve this answer

















              • 1




                +1 for the last paragraph: Keep calm & try again later is a good approach
                – Barranka
                Nov 7 at 18:20










              • can you explain Note 1? How can I use this four tuples you show us to find the error?
                – miracle173
                2 days ago










              • @miracle173, unfortunately all four cannot be used at once, but pairwise solutions (which is relatively simple to find by putting $0$ into one of three unknowns) must be checked. For example, equations $1$ and $2$ have common solution $(0,0,-1)$, so any row operation of the two equations must be satisfied by it. Or $(8/3,0,-1/3)$ is a common solution of the equations $1$ and $3$, so the OP's first result $[0 2 -3 | 1]$ must be satisfied by it.
                – farruhota
                2 days ago










              • I understand now, but this seems not to be useful to me.
                – miracle173
                2 days ago













              up vote
              11
              down vote










              up vote
              11
              down vote









              Note 1: Referring to 5xum's answer, in case you don't know the final answer, any solution will do, e.g.:
              $$(x_1,x_2,x_3)=(0,0,-1);(0,0,-1);(1,0,-1);(12,0,0) text{(respectively)}$$
              Note 2: In step $3$, you can reduce column $3$ instead of column $2$:
              $$left[
              begin{array}{ccc|c}
              1&8&-7&12\
              0&2&-3&1\
              0&-3&1&-5\
              0&-13&9&-17\
              end{array}
              right] Rightarrow left[
              begin{array}{ccc|c}
              1&8&-7&12\
              0&-7&0&-14\
              0&-3&1&-5\
              0&14&0&28\
              end{array}
              right] stackrel{frac{-R_2}{7};\ frac{-3R_2}{7}+R_3}{=}Rightarrow left[
              begin{array}{ccc|c}
              1&8&-7&12\
              0&1&0&2\
              0&0&1&1\
              0&14&0&28\
              end{array}
              right]$$

              The second and fourth equations are dependent, therefore they produce the same solution $x_2=2$. You can finish the problem now.



              Note 3. When you can not find your mistake (sometimes the brain gets blocked/accustommed and can not see obvious mistakes), leave it for some time (1 hour, 1 day) and return with fresh mind. You will be surprised to easily spot the error.






              share|cite|improve this answer












              Note 1: Referring to 5xum's answer, in case you don't know the final answer, any solution will do, e.g.:
              $$(x_1,x_2,x_3)=(0,0,-1);(0,0,-1);(1,0,-1);(12,0,0) text{(respectively)}$$
              Note 2: In step $3$, you can reduce column $3$ instead of column $2$:
              $$left[
              begin{array}{ccc|c}
              1&8&-7&12\
              0&2&-3&1\
              0&-3&1&-5\
              0&-13&9&-17\
              end{array}
              right] Rightarrow left[
              begin{array}{ccc|c}
              1&8&-7&12\
              0&-7&0&-14\
              0&-3&1&-5\
              0&14&0&28\
              end{array}
              right] stackrel{frac{-R_2}{7};\ frac{-3R_2}{7}+R_3}{=}Rightarrow left[
              begin{array}{ccc|c}
              1&8&-7&12\
              0&1&0&2\
              0&0&1&1\
              0&14&0&28\
              end{array}
              right]$$

              The second and fourth equations are dependent, therefore they produce the same solution $x_2=2$. You can finish the problem now.



              Note 3. When you can not find your mistake (sometimes the brain gets blocked/accustommed and can not see obvious mistakes), leave it for some time (1 hour, 1 day) and return with fresh mind. You will be surprised to easily spot the error.







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered Nov 7 at 15:47









              farruhota

              17.1k2736




              17.1k2736








              • 1




                +1 for the last paragraph: Keep calm & try again later is a good approach
                – Barranka
                Nov 7 at 18:20










              • can you explain Note 1? How can I use this four tuples you show us to find the error?
                – miracle173
                2 days ago










              • @miracle173, unfortunately all four cannot be used at once, but pairwise solutions (which is relatively simple to find by putting $0$ into one of three unknowns) must be checked. For example, equations $1$ and $2$ have common solution $(0,0,-1)$, so any row operation of the two equations must be satisfied by it. Or $(8/3,0,-1/3)$ is a common solution of the equations $1$ and $3$, so the OP's first result $[0 2 -3 | 1]$ must be satisfied by it.
                – farruhota
                2 days ago










              • I understand now, but this seems not to be useful to me.
                – miracle173
                2 days ago














              • 1




                +1 for the last paragraph: Keep calm & try again later is a good approach
                – Barranka
                Nov 7 at 18:20










              • can you explain Note 1? How can I use this four tuples you show us to find the error?
                – miracle173
                2 days ago










              • @miracle173, unfortunately all four cannot be used at once, but pairwise solutions (which is relatively simple to find by putting $0$ into one of three unknowns) must be checked. For example, equations $1$ and $2$ have common solution $(0,0,-1)$, so any row operation of the two equations must be satisfied by it. Or $(8/3,0,-1/3)$ is a common solution of the equations $1$ and $3$, so the OP's first result $[0 2 -3 | 1]$ must be satisfied by it.
                – farruhota
                2 days ago










              • I understand now, but this seems not to be useful to me.
                – miracle173
                2 days ago








              1




              1




              +1 for the last paragraph: Keep calm & try again later is a good approach
              – Barranka
              Nov 7 at 18:20




              +1 for the last paragraph: Keep calm & try again later is a good approach
              – Barranka
              Nov 7 at 18:20












              can you explain Note 1? How can I use this four tuples you show us to find the error?
              – miracle173
              2 days ago




              can you explain Note 1? How can I use this four tuples you show us to find the error?
              – miracle173
              2 days ago












              @miracle173, unfortunately all four cannot be used at once, but pairwise solutions (which is relatively simple to find by putting $0$ into one of three unknowns) must be checked. For example, equations $1$ and $2$ have common solution $(0,0,-1)$, so any row operation of the two equations must be satisfied by it. Or $(8/3,0,-1/3)$ is a common solution of the equations $1$ and $3$, so the OP's first result $[0 2 -3 | 1]$ must be satisfied by it.
              – farruhota
              2 days ago




              @miracle173, unfortunately all four cannot be used at once, but pairwise solutions (which is relatively simple to find by putting $0$ into one of three unknowns) must be checked. For example, equations $1$ and $2$ have common solution $(0,0,-1)$, so any row operation of the two equations must be satisfied by it. Or $(8/3,0,-1/3)$ is a common solution of the equations $1$ and $3$, so the OP's first result $[0 2 -3 | 1]$ must be satisfied by it.
              – farruhota
              2 days ago












              I understand now, but this seems not to be useful to me.
              – miracle173
              2 days ago




              I understand now, but this seems not to be useful to me.
              – miracle173
              2 days ago










              up vote
              2
              down vote













              Usually you don't know the solution of your equation. So you have to use other methods for check your calculation. One way is to add checksums to your calculation.



              Don't start your calculations with the original matrix



              $$left[begin{array}{ccc|c}
              2 & 5 & -8 & 8 \
              4 & 3 & -9 & 9 \
              2 & 3 & -5 & 7 \
              1 & 8 & -7 & 12
              end{array}right]
              $$



              but start with the matrix augmented by a column at the right. An element of this column is equal to the sum of the other elements of this row. So the augmented matrix is



              $$left[begin{array}{ccc|c|c}
              2 & 5 & -8 & 8 & 7 \
              4 & 3 & -9 & 9 & 7 \
              2 & 3 & -5 & 7 & 7 \
              1 & 8 & -7 & 12 & 14
              end{array}right]
              $$



              You do the same row operations as with the original matrices.



              So instead of



              $$left[begin{array}{ccc|c}
              2 & 5 & -8 & 8 \
              4 & 3 & -9 & 9 \
              2 & 3 & -5 & 7 \
              1 & 8 & -7 & 12
              end{array}right]
              overset{overset{large{R_1to R_1-R_3}}{{R_2to R_2-2R_3}}}{overset{R_3to R_3-2R_4}{largelongrightarrow}}
              left[begin{array}{ccc|c}
              0 & 2 & -3 & 1 \
              0 & -3 & 1 & -5 \
              0 & -13 & 9 & -17 \
              1 & 8 & -7 & 12
              end{array}right]$$



              you do



              $$left[begin{array}{ccc|c|c}
              2 & 5 & -8 & 8 & 7 \
              4 & 3 & -9 & 9 & 7 \
              2 & 3 & -5 & 7 & 7 \
              1 & 8 & -7 & 12 & 14
              end{array}right]
              overset{overset{large{R_1to R_1-R_3}}{{R_2to R_2-2R_3}}}{overset{R_3to R_3-2R_4}{largelongrightarrow}}
              left[begin{array}{ccc|c}
              0 & 2 & -3 & 1 & 0 \
              0 & -3 & 1 & -5 & -7 \
              0 & -13 & 9 & -17 & -21 \
              1 & 8 & -7 & 12 & 14
              end{array}right]$$



              Then you check if for the matrix you calculated the checksum is still valid.
              Here the checksum is correct.



              Now let's investigate the step where the error occurs. We get



              $$left[begin{array}{ccc|c|c}
              1 & 8 & -7 & 12 & 14 \
              0 & 2 & -3 & 1 & 0 \
              0 & -3 & 1 & -5 & -7 \
              0 & -13 & 9 & -17 & -21
              end{array}right]
              overset{R_4to R_4-R_3}{largelongrightarrow}
              left[begin{array}{ccc|c | c}
              1 & 8 & -7 & 12 & 14\
              0 & 2 & -3 & 1 & 0\
              0 & -3 & 1 & -5 & -7\
              0 & 10 & 8 & -12 & -14
              end{array}right]
              $$



              For this result matrix the checksum of row 4 is not valid, so an error must have occurred.





              Actually the checksum property means $(1,1,1,1)^T$ is a solution of the following system of equations that corresponds to the augmented matrix:



              $$begin{eqnarray}
              2x_1&+&5x_2&-&8x_3&+&8x_4&=&7\
              4x_1&+&3x_2&-&9x_3&+&9x_4&=&7\
              2x_1&+&3x_2&-&5x_3&+&7x_4&=&7\
              x_1&+&8x_2&-&7x_3&+&12x_4&=&14
              end{eqnarray}$$



              So this method is similar to the method in the accepted answer, where the solution given by the textbook is used.



              What if this method does not uncover the errors. This can happen if you made more then one error in a step- Then you can use a weighted checksum to try to uncover the error.






              share|cite|improve this answer



























                up vote
                2
                down vote













                Usually you don't know the solution of your equation. So you have to use other methods for check your calculation. One way is to add checksums to your calculation.



                Don't start your calculations with the original matrix



                $$left[begin{array}{ccc|c}
                2 & 5 & -8 & 8 \
                4 & 3 & -9 & 9 \
                2 & 3 & -5 & 7 \
                1 & 8 & -7 & 12
                end{array}right]
                $$



                but start with the matrix augmented by a column at the right. An element of this column is equal to the sum of the other elements of this row. So the augmented matrix is



                $$left[begin{array}{ccc|c|c}
                2 & 5 & -8 & 8 & 7 \
                4 & 3 & -9 & 9 & 7 \
                2 & 3 & -5 & 7 & 7 \
                1 & 8 & -7 & 12 & 14
                end{array}right]
                $$



                You do the same row operations as with the original matrices.



                So instead of



                $$left[begin{array}{ccc|c}
                2 & 5 & -8 & 8 \
                4 & 3 & -9 & 9 \
                2 & 3 & -5 & 7 \
                1 & 8 & -7 & 12
                end{array}right]
                overset{overset{large{R_1to R_1-R_3}}{{R_2to R_2-2R_3}}}{overset{R_3to R_3-2R_4}{largelongrightarrow}}
                left[begin{array}{ccc|c}
                0 & 2 & -3 & 1 \
                0 & -3 & 1 & -5 \
                0 & -13 & 9 & -17 \
                1 & 8 & -7 & 12
                end{array}right]$$



                you do



                $$left[begin{array}{ccc|c|c}
                2 & 5 & -8 & 8 & 7 \
                4 & 3 & -9 & 9 & 7 \
                2 & 3 & -5 & 7 & 7 \
                1 & 8 & -7 & 12 & 14
                end{array}right]
                overset{overset{large{R_1to R_1-R_3}}{{R_2to R_2-2R_3}}}{overset{R_3to R_3-2R_4}{largelongrightarrow}}
                left[begin{array}{ccc|c}
                0 & 2 & -3 & 1 & 0 \
                0 & -3 & 1 & -5 & -7 \
                0 & -13 & 9 & -17 & -21 \
                1 & 8 & -7 & 12 & 14
                end{array}right]$$



                Then you check if for the matrix you calculated the checksum is still valid.
                Here the checksum is correct.



                Now let's investigate the step where the error occurs. We get



                $$left[begin{array}{ccc|c|c}
                1 & 8 & -7 & 12 & 14 \
                0 & 2 & -3 & 1 & 0 \
                0 & -3 & 1 & -5 & -7 \
                0 & -13 & 9 & -17 & -21
                end{array}right]
                overset{R_4to R_4-R_3}{largelongrightarrow}
                left[begin{array}{ccc|c | c}
                1 & 8 & -7 & 12 & 14\
                0 & 2 & -3 & 1 & 0\
                0 & -3 & 1 & -5 & -7\
                0 & 10 & 8 & -12 & -14
                end{array}right]
                $$



                For this result matrix the checksum of row 4 is not valid, so an error must have occurred.





                Actually the checksum property means $(1,1,1,1)^T$ is a solution of the following system of equations that corresponds to the augmented matrix:



                $$begin{eqnarray}
                2x_1&+&5x_2&-&8x_3&+&8x_4&=&7\
                4x_1&+&3x_2&-&9x_3&+&9x_4&=&7\
                2x_1&+&3x_2&-&5x_3&+&7x_4&=&7\
                x_1&+&8x_2&-&7x_3&+&12x_4&=&14
                end{eqnarray}$$



                So this method is similar to the method in the accepted answer, where the solution given by the textbook is used.



                What if this method does not uncover the errors. This can happen if you made more then one error in a step- Then you can use a weighted checksum to try to uncover the error.






                share|cite|improve this answer

























                  up vote
                  2
                  down vote










                  up vote
                  2
                  down vote









                  Usually you don't know the solution of your equation. So you have to use other methods for check your calculation. One way is to add checksums to your calculation.



                  Don't start your calculations with the original matrix



                  $$left[begin{array}{ccc|c}
                  2 & 5 & -8 & 8 \
                  4 & 3 & -9 & 9 \
                  2 & 3 & -5 & 7 \
                  1 & 8 & -7 & 12
                  end{array}right]
                  $$



                  but start with the matrix augmented by a column at the right. An element of this column is equal to the sum of the other elements of this row. So the augmented matrix is



                  $$left[begin{array}{ccc|c|c}
                  2 & 5 & -8 & 8 & 7 \
                  4 & 3 & -9 & 9 & 7 \
                  2 & 3 & -5 & 7 & 7 \
                  1 & 8 & -7 & 12 & 14
                  end{array}right]
                  $$



                  You do the same row operations as with the original matrices.



                  So instead of



                  $$left[begin{array}{ccc|c}
                  2 & 5 & -8 & 8 \
                  4 & 3 & -9 & 9 \
                  2 & 3 & -5 & 7 \
                  1 & 8 & -7 & 12
                  end{array}right]
                  overset{overset{large{R_1to R_1-R_3}}{{R_2to R_2-2R_3}}}{overset{R_3to R_3-2R_4}{largelongrightarrow}}
                  left[begin{array}{ccc|c}
                  0 & 2 & -3 & 1 \
                  0 & -3 & 1 & -5 \
                  0 & -13 & 9 & -17 \
                  1 & 8 & -7 & 12
                  end{array}right]$$



                  you do



                  $$left[begin{array}{ccc|c|c}
                  2 & 5 & -8 & 8 & 7 \
                  4 & 3 & -9 & 9 & 7 \
                  2 & 3 & -5 & 7 & 7 \
                  1 & 8 & -7 & 12 & 14
                  end{array}right]
                  overset{overset{large{R_1to R_1-R_3}}{{R_2to R_2-2R_3}}}{overset{R_3to R_3-2R_4}{largelongrightarrow}}
                  left[begin{array}{ccc|c}
                  0 & 2 & -3 & 1 & 0 \
                  0 & -3 & 1 & -5 & -7 \
                  0 & -13 & 9 & -17 & -21 \
                  1 & 8 & -7 & 12 & 14
                  end{array}right]$$



                  Then you check if for the matrix you calculated the checksum is still valid.
                  Here the checksum is correct.



                  Now let's investigate the step where the error occurs. We get



                  $$left[begin{array}{ccc|c|c}
                  1 & 8 & -7 & 12 & 14 \
                  0 & 2 & -3 & 1 & 0 \
                  0 & -3 & 1 & -5 & -7 \
                  0 & -13 & 9 & -17 & -21
                  end{array}right]
                  overset{R_4to R_4-R_3}{largelongrightarrow}
                  left[begin{array}{ccc|c | c}
                  1 & 8 & -7 & 12 & 14\
                  0 & 2 & -3 & 1 & 0\
                  0 & -3 & 1 & -5 & -7\
                  0 & 10 & 8 & -12 & -14
                  end{array}right]
                  $$



                  For this result matrix the checksum of row 4 is not valid, so an error must have occurred.





                  Actually the checksum property means $(1,1,1,1)^T$ is a solution of the following system of equations that corresponds to the augmented matrix:



                  $$begin{eqnarray}
                  2x_1&+&5x_2&-&8x_3&+&8x_4&=&7\
                  4x_1&+&3x_2&-&9x_3&+&9x_4&=&7\
                  2x_1&+&3x_2&-&5x_3&+&7x_4&=&7\
                  x_1&+&8x_2&-&7x_3&+&12x_4&=&14
                  end{eqnarray}$$



                  So this method is similar to the method in the accepted answer, where the solution given by the textbook is used.



                  What if this method does not uncover the errors. This can happen if you made more then one error in a step- Then you can use a weighted checksum to try to uncover the error.






                  share|cite|improve this answer














                  Usually you don't know the solution of your equation. So you have to use other methods for check your calculation. One way is to add checksums to your calculation.



                  Don't start your calculations with the original matrix



                  $$left[begin{array}{ccc|c}
                  2 & 5 & -8 & 8 \
                  4 & 3 & -9 & 9 \
                  2 & 3 & -5 & 7 \
                  1 & 8 & -7 & 12
                  end{array}right]
                  $$



                  but start with the matrix augmented by a column at the right. An element of this column is equal to the sum of the other elements of this row. So the augmented matrix is



                  $$left[begin{array}{ccc|c|c}
                  2 & 5 & -8 & 8 & 7 \
                  4 & 3 & -9 & 9 & 7 \
                  2 & 3 & -5 & 7 & 7 \
                  1 & 8 & -7 & 12 & 14
                  end{array}right]
                  $$



                  You do the same row operations as with the original matrices.



                  So instead of



                  $$left[begin{array}{ccc|c}
                  2 & 5 & -8 & 8 \
                  4 & 3 & -9 & 9 \
                  2 & 3 & -5 & 7 \
                  1 & 8 & -7 & 12
                  end{array}right]
                  overset{overset{large{R_1to R_1-R_3}}{{R_2to R_2-2R_3}}}{overset{R_3to R_3-2R_4}{largelongrightarrow}}
                  left[begin{array}{ccc|c}
                  0 & 2 & -3 & 1 \
                  0 & -3 & 1 & -5 \
                  0 & -13 & 9 & -17 \
                  1 & 8 & -7 & 12
                  end{array}right]$$



                  you do



                  $$left[begin{array}{ccc|c|c}
                  2 & 5 & -8 & 8 & 7 \
                  4 & 3 & -9 & 9 & 7 \
                  2 & 3 & -5 & 7 & 7 \
                  1 & 8 & -7 & 12 & 14
                  end{array}right]
                  overset{overset{large{R_1to R_1-R_3}}{{R_2to R_2-2R_3}}}{overset{R_3to R_3-2R_4}{largelongrightarrow}}
                  left[begin{array}{ccc|c}
                  0 & 2 & -3 & 1 & 0 \
                  0 & -3 & 1 & -5 & -7 \
                  0 & -13 & 9 & -17 & -21 \
                  1 & 8 & -7 & 12 & 14
                  end{array}right]$$



                  Then you check if for the matrix you calculated the checksum is still valid.
                  Here the checksum is correct.



                  Now let's investigate the step where the error occurs. We get



                  $$left[begin{array}{ccc|c|c}
                  1 & 8 & -7 & 12 & 14 \
                  0 & 2 & -3 & 1 & 0 \
                  0 & -3 & 1 & -5 & -7 \
                  0 & -13 & 9 & -17 & -21
                  end{array}right]
                  overset{R_4to R_4-R_3}{largelongrightarrow}
                  left[begin{array}{ccc|c | c}
                  1 & 8 & -7 & 12 & 14\
                  0 & 2 & -3 & 1 & 0\
                  0 & -3 & 1 & -5 & -7\
                  0 & 10 & 8 & -12 & -14
                  end{array}right]
                  $$



                  For this result matrix the checksum of row 4 is not valid, so an error must have occurred.





                  Actually the checksum property means $(1,1,1,1)^T$ is a solution of the following system of equations that corresponds to the augmented matrix:



                  $$begin{eqnarray}
                  2x_1&+&5x_2&-&8x_3&+&8x_4&=&7\
                  4x_1&+&3x_2&-&9x_3&+&9x_4&=&7\
                  2x_1&+&3x_2&-&5x_3&+&7x_4&=&7\
                  x_1&+&8x_2&-&7x_3&+&12x_4&=&14
                  end{eqnarray}$$



                  So this method is similar to the method in the accepted answer, where the solution given by the textbook is used.



                  What if this method does not uncover the errors. This can happen if you made more then one error in a step- Then you can use a weighted checksum to try to uncover the error.







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited yesterday

























                  answered yesterday









                  miracle173

                  7,26822247




                  7,26822247






























                       

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