Montenovofe

2022-07-10

we have a system of linear equations as such:
$x+2y+\left(a-1\right)z=1\phantom{\rule{0ex}{0ex}}-x-y+z=0\phantom{\rule{0ex}{0ex}}-ax-\left(a+3\right)y-az=-3\phantom{\rule{0ex}{0ex}}-ax-\left(a+2\right)y+0\cdot z={a}^{2}-5a-2$
and i have to find the solution in $\mathbb{R}$ and ${\mathbb{Z}}_{\mathbb{5}}$ so i have no problem for $\mathbb{R}$ i get the matrix
$\left(\begin{array}{cccc}1& 2& a-1& 1\\ 0& 1& a& 1\\ 0& 0& a& 0\\ 0& 0& a& {a}^{2}-5\cdot a\end{array}\right)$
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 $\mathbb{R}$ to deduce ${\mathbb{Z}}_{\mathbb{5}}$?
2. Or is there some other way i must reduce to row echelon form for ${\mathbb{Z}}_{\mathbb{5}}$ and then have the discussion for parameter a?
3. If i had an 3x3 or 4x4 system to solve over a low prime ${\mathbb{Z}}_{{\mathbb{p}}_{\mathbb{1}}}$ and ${\mathbb{Z}}_{{\mathbb{p}}_{\mathbb{2}}}$ (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 $\mathbb{R}$ 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?

Zackery Harvey

Expert

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 ${\mathbb{Z}}_{5}$.
Assuming this is the case, the system over ${\mathbb{Z}}_{5}$ has solution if and only if the last column is not a pivot one. We need to distinguish the cases $a\ne 0$ and $a=0$. If $a\ne 0$, you can go on with Gaussian elimination to
$\left(\begin{array}{cccc}1& 2& a-1& 1\\ 0& 1& a& 1\\ 0& 0& 1& 0\\ 0& 0& 0& 1\end{array}\right)$
and the last column is a pivot column.
If $a=0$, the matrix is
$\left(\begin{array}{cccc}1& 2& -1& 1\\ 0& 1& 0& 1\\ 0& 0& 0& 0\\ 0& 0& 0& 0\end{array}\right)$
and the system has solutions, with $z$ being a free variable (so five distinct solutions).

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