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Given the linear system in $\mathbb{Z_3}$:

$$ \left\{ \begin{array}{c} a+b+c+d=1 \\ b+c+e=2 \\ a+2e=0 \end{array} \right. $$

I used the row reduction with matrices and I got:

$$ \left\{ \begin{array}{c} a+b+c+d=1 \\ b+c+e=2 \\ d=2 \end{array} \right. $$

But now I don't know how to find the solutions.

user530437
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2 Answers2

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What you probably did is add the last two equations to get $$ 2=a+b+c, $$ since $3e=0$. Now the first equation gives you $d=1-2=2$. Going back to the system (and again using $2=-1$), $$ \left\{ \begin{array}{c} a+b+c=2 \\ b+c+e=2 \\ a-e=0 \end{array} \right. $$ and we already know $d=2$, $a=e$. Now theh first two equations are $a+b+c=2$, and so we are free to prescribe two of them. So, say, if you prescribe $b=t$, $c=s$, you have $$ a=2+2t+2s,\ \ d=2,\ \ e=a. $$

Martin Argerami
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I'd use the symbols $0,1,-1$ for the elements of $\mathbf Z/3\mathbf Z$. Here's how to put the augmented matrix in reduced row echelon form: \begin{align} &\begin{bmatrix} 1&1&1&1&0&\hspace{-0.6em}|\phantom{-}~1\\ 0&1&1&0&1&\hspace{-0.6em}|\:{-}1 \\1&0&0&0&-1&\hspace{-0.6em}|\phantom{-}~0 \end{bmatrix}\xrightarrow{R_1\leftarrow R_1-R_2} \begin{bmatrix} 1&0&0&1&-1&\hspace{-0.6em}|\:{-}1\\ 0&1&1&0&1&\hspace{-0.6em}|\,{-}1 \\1&0&0&0&-1&\hspace{-0.6em}|\phantom{-}~0 \end{bmatrix}\\[1ex] \xrightarrow[R_3\leftarrow R_1-R_3]{} &\begin{bmatrix} 1&0&0&1&-1&\hspace{-0.6em}|\:{-}1\\ 0&1&1&0&1&\hspace{-0.6em}|\,{-}1 \\0&0&0&1&0&\hspace{-0.6em}|\:{-}1 \end{bmatrix} \xrightarrow{R_1\leftarrow R_1-R_3} \begin{bmatrix} \color{red}1&\color{red}0&0&\color{red}0&-1&\hspace{-0.6em}|\:\phantom{-}0\\ \color{red}0&\color{red}1&1&\color{red}0&1&\hspace{-0.6em}|\,{-}1 \\\color{red}0&\color{red}0&0&\color{red}1&0&\hspace{-0.6em}|\:{-}1 \end{bmatrix} \end{align} Thus the solution are: $$\begin{cases} a=e\\b=-1-c-e\\d=-1 \end{cases}\enspace\text{or, in vector form:}\quad \begin{bmatrix}a\\b\\c\\d\\e\end{bmatrix}=\begin{bmatrix}\phantom{-}0\\-1\\\phantom{-}0\\-1\\\phantom{-}0 \end{bmatrix}-c\begin{bmatrix}\phantom{-}0\\\phantom{-}1\\-1\\\phantom{-}0\\\phantom{-}0\end{bmatrix} +e\begin{bmatrix}\phantom{-}1\\-1\\\phantom{-}0\\\phantom{-}0\\\phantom{-}1 \end{bmatrix}.$$

Bernard
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