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

Montenovofe

Montenovofe

Answered

2022-07-10

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 2
and 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?

Answer & Explanation

Zackery Harvey

Zackery Harvey

Expert

2022-07-11Added 21 answers

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).

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