Find the solutions for the linear equation system: { x + 5 y + 11 z = − 5 2 x + 3 y + 8 z = 4 − x + 2 y + 3 z = − 9 [ 1 5 11 − 5 2 3 8 4 − 1 2 3 − 9 ] 2 f 3 + f 2 [ 1 5 11 − 5 0 6 14 − 14 − 1 2 3 − 9 ] f 1 + f 3 [ 1 5 11 − 5 0 6 14 − 14 0 7 14 − 14 ] − f 2 + f 3 [ 1 5 11 − 5 0 6 14 − 14 0 1 0 0 ] f 2 f 3 [ 1 5 11 − 5 0 1 0 0 0 6 14 − 14 ] − 6 f 2 + f 3 [ 1 5 11 − 5 0 1 0 0 0 0 14 − 14 ] 1 14 f 3 [ 1 5 11 − 5 0 1 0 0 0 0 1 − 1 ] − 5 f 2 + f 1 [ 1 0 11 − 5 0 1 0 0 0 0 1 − 1 ] − 11 f 3 + f 1 [ 1 0 0 6 0 1 0 0 0 0 1 − 1 ] 6 f 3 + f 1 [ 1 0 0 0 0 1 0 0 0 0 1 − 1 ] So the solution would be ( 0 , 0 , − 1 ) which is NOT right. Just looking at the first equation we'd have 0 + 5 ( 0 ) + 11 ( − 1 ) = − 11 ≠ − 5What did I do wrong?

Question

Find the solutions for the linear equation system:      {                            x          +          5          y          +          11          z          =          −          5                                      2          x          +          3          y          +          8          z          =          4                                      −          x          +          2          y          +          3          z          =          −          9                              [                            1                          5                          11                          −          5                                      2                          3                          8                          4                                      −          1                          2                          3                          −          9                      ]    2      f    3    +      f    2        [                            1                          5                          11                          −          5                                      0                          6                          14                          −          14                                      −          1                          2                          3                          −          9                      ]        f    1    +      f    3        [                            1                          5                          11                          −          5                                      0                          6                          14                          −          14                                      0                          7                          14                          −          14                      ]    −      f    2    +      f    3        [                            1                          5                          11                          −          5                                      0                          6                          14                          −          14                                      0                          1                          0                          0                      ]        f    2           f    3        [                            1                          5                          11                          −          5                                      0                          1                          0                          0                                      0                          6                          14                          −          14                      ]    −  6      f    2    +      f    3        [                            1                          5                          11                          −          5                                      0                          1                          0                          0                                      0                          0                          14                          −          14                      ]        1    14        f    3        [                            1                          5                          11                          −          5                                      0                          1                          0                          0                                      0                          0                          1                          −          1                      ]    −  5      f    2    +      f    1        [                            1                          0                          11                          −          5                                      0                          1                          0                          0                                      0                          0                          1                          −          1                      ]    −  11      f    3    +      f    1        [                            1                          0                          0                          6                                      0                          1                          0                          0                                      0                          0                          1                          −          1                      ]    6      f    3    +      f    1        [                            1                          0                          0                          0                                      0                          1                          0                          0                                      0                          0                          1                          −          1                      ]  So the solution would be   (  0  ,  0  ,  −  1  ) which is NOT right. Just looking at the first equation we&#039;d have  0  +  5  (  0  )  +  11  (  −  1  )  =  −  11  ≠  −  5What did I do wrong?

Kayley Jackson · Accepted Answer

The very first rowop: 2*2+3=7, not six

pablos28spainzd · Accepted Answer

One way to quickly find arithmetical errors in such row operations is to use modular arithmetic checks. For example, your first row operation is          r    2    ′    ←      r    2    +      2          r    3    ,  , so          r    2    ′    ≡      r    2      (  mod        2    )  .  . But this parity check fails since          r          2      ,      2        ′    =  6   is even but         r          2      ,      2        =  3   is odd. Hence the first row operation is in error at the second component. Such modular checks work very well in practice (just like casting nines (mod   9) works well to check integer arithmetic).

Find the solutions for the linear equation system: { <mtable columnalign="left left" ro

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