nonhomogenous and homogenous linear equations: non-specific solution

Question

Given the linear equations, over the field of $5$, so only numbers $0,1,2,3,4$ "exists" :

$$\left(\begin{matrix} 2 & 1 & 3 \\ 3 & 2 & 2 \\ 3 & 0 & 2 \\ \end{matrix} \right| \left. \begin{matrix} 1\\3\\2 \end{matrix} \right)$$

I have found a specific solution after forming the above into:

$$\left(\begin{matrix} 2 & 1 & 3 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \\ \end{matrix} \right| \left. \begin{matrix} 1\\3\\0 \end{matrix} \right)$$

with $z=0, y=3, x=4$.

What I struggle with is finding the non-specific solution. I have thought about just putting in the zero vector on the right hand site of my equations, but then I struggle to find a usable solution.

Answer 1

Same method as in $\mathbf R$, with computations done in $\mathbf F_5$: transform to the reduced row echelon form: \begin{align} \begin{bmatrix}\hskip-0.5em\begin{array}{ccc|c} 2&1&3&1\newline 3&2&2&3\newline 3&0&2&2 \end{array}\hskip-0.5em\end{bmatrix}\xrightarrow[R_i\to 2R_i\ (i=2,3)]{R_1\to3R_1} & \begin{bmatrix}\hskip-0.5em\begin{array}{ccc|c} 1&3&4&3\newline 1&4&4&1\newline 1&0&4&4 \end{array}\hskip-0.5em\end{bmatrix}\xrightarrow[R_3-R_1\to R_3]{R_2-R_1\to R_2} \begin{bmatrix}\hskip-0.5em\begin{array}{ccc|c} 1&3&4&3\newline 0&1&0&3\newline 0&2&0&1 \end{array}\hskip-0.5em\end{bmatrix}\xrightarrow[R_3+3R_2\to R_3]{R_1+2R_2\to R_1} %\\ \begin{bmatrix}\hskip-0.5em\begin{array}{ccc|c} 1&0&4&4\newline 0&1&0&3\newline 0&0&0&0 \end{array}\hskip-0.5em\end{bmatrix} \end{align} so the solutions are $$\begin{bmatrix}4\\3\\0\end{bmatrix}-z\begin{bmatrix}4\\0\\1\end{bmatrix}=\begin{bmatrix}4\\3\\0\end{bmatrix}+z\begin{bmatrix}1\\0\\1\end{bmatrix}.$$

Answer 2

So for the non-specific solution, I guess I have to first make the right hand side of my equation to the zero vector, so from

$$ \begin{matrix} 2 & 1 & 3 &|&1\\ 0 & 1 & 0 &|&3\\ 0 & 0 & 0 &|&0\\ \end{matrix} $$

I have

$$ \begin{matrix} 2 & 1 & 3 &|&0\\ 0 & 1 & 0 &|&0\\ 0 & 0 & 0 &|&0\\ \end{matrix} $$

Then I set the one free variable (rank of 2 and 3 variables: 3-2=1), and say z=1

$$ \begin{matrix} 2 & 1 & 3*1 &|&0\\ 0 & 1 & 0*1 &|&0\\ 0 & 0 & 0*1 &|&0\\ \end{matrix} $$

and move z to the right hand side of my equation:

$$ \begin{matrix} 2 & 1 &|&2*\\ 0 & 1 &|&0\\ \end{matrix} $$

*=2 because of -3 in F5 is 2.

and then I just solve it again for x and y. So I get y=0, x=1

and solution is:

$$\begin{bmatrix}4\\3\\0\end{bmatrix}+λ\begin{bmatrix}1\\0\\1\end{bmatrix}.$$

My remaining question (if what I did is correct) to this is: When do I know when to specifically set numbers instead of letters? As in, I set z=1. When am I supposed to set free variables to say α and β and calculate with that? Do I have to check solvability or something like that?