we have a system of linear equations as such: x + 2 y + ( a − 1 ) z = 1 − x − y + z = 0 − a x − ( a + 3 ) y − a z = − 3 − a x − ( a + 2 ) y + 0 ⋅ z = a 2 − 5 a − 2and i have to find the solution in R and Z 5 so i have no problem for R i get the matrix ( 1 2 a − 1 1 0 1 a 1 0 0 a 0 0 0 a a 2 − 5 ⋅ a ) but the questions i have are as follows:1. Can i use what i found for the augmented matrix and the discussion by parameter a in R to deduce Z 5 ?2. Or is there some other way i must reduce to row echelon form for Z 5 and then have the discussion for parameter a?3. If i had an 3x3 or 4x4 system to solve over a low prime Z p 1 and Z p 2 (eg 5 and 7) how would i go about doing it with the matrix gauss elimination?could i use the same augmented matrix and reduce it to row echelon over R and then use that augmented matrix for the rest like above or not?4. If i recall correctly there was a theorem about the rank of the original matrix and augmented that says something about the number of solutions but i do not recall how that would help me find solutions just eliminate the a's where there is none?

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we have a system of linear equations as such:  x  +  2  y  +  (  a  −  1  )  z  =  1    −  x  −  y  +  z  =  0    −  a  x  −  (  a  +  3  )  y  −  a  z  =  −  3    −  a  x  −  (  a  +  2  )  y  +  0  ⋅  z  =      a    2    −  5  a  −  2and i have to find the solution in       R   and             Z              5             so i have no problem for       R   i get the matrix      (                            1                          2                          a          −          1                          1                                      0                          1                          a                          1                                      0                          0                          a                          0                                      0                          0                          a                                      a            2                    −          5          ⋅          a                      )  but the questions i have are as follows:1. Can i use what i found for the augmented matrix and the discussion by parameter a in       R   to deduce             Z      5      ?2. Or is there some other way i must reduce to row echelon form for             Z      5       and then have the discussion for parameter a?3. If i had an 3x3 or 4x4 system to solve over a low prime             Z                        p                      1                               and             Z                        p                      2                               (eg 5 and 7) how would i go about doing it with the matrix gauss elimination?could i use the same augmented matrix and reduce it to row echelon over       R   and then use that augmented matrix for the rest like above or not?4. If i recall correctly there was a theorem about the rank of the original matrix and augmented that says something about the number of solutions but i do not recall how that would help me find solutions just eliminate the a&#039;s where there is none?

Zackery Harvey · Accepted Answer

If in producing the echelon form you only used multiplication or division by integers, provided you never divided by a multiple of 5, the same steps would produce the echelon form over             Z        5  .Assuming this is the case, the system over             Z        5   has solution if and only if the last column is not a pivot one. We need to distinguish the cases   a  ≠  0 and   a  =  0. If   a  ≠  0, you can go on with Gaussian elimination to      (                            1                          2                          a          −          1                          1                                      0                          1                          a                          1                                      0                          0                          1                          0                                      0                          0                          0                          1                      )  and the last column is a pivot column.If   a  =  0, the matrix is      (                            1                          2                          −          1                          1                                      0                          1                          0                          1                                      0                          0                          0                          0                                      0                          0                          0                          0                      )  and the system has solutions, with   z being a free variable (so five distinct solutions).

we have a system of linear equations as such: x + 2 y + ( a − 1

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