Montenovofe

Answered

2022-07-10

we have a system of linear equations as such:

$x+2y+(a-1)z=1\phantom{\rule{0ex}{0ex}}-x-y+z=0\phantom{\rule{0ex}{0ex}}-ax-(a+3)y-az=-3\phantom{\rule{0ex}{0ex}}-ax-(a+2)y+0\cdot z={a}^{2}-5a-2$

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?

$x+2y+(a-1)z=1\phantom{\rule{0ex}{0ex}}-x-y+z=0\phantom{\rule{0ex}{0ex}}-ax-(a+3)y-az=-3\phantom{\rule{0ex}{0ex}}-ax-(a+2)y+0\cdot z={a}^{2}-5a-2$

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?

Answer & Explanation

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 ${\mathbb{Z}}_{5}$.

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).

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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