My Problem is: to find all Solutions for the following given System of linear equations over the Field $K = \mathbb{Z}_{/7}$
The System is given with:
$$\begin{equation}
\begin{split}
x_1\quad\quad\quad + 3x_3 &= 5 \\
5x_1 + 3x_2 + 6x_3 &= 3 \\
6x_1 + 2x_2 + 5x_3 &= 6 \\
\end{split}
\end{equation}$$
My Approach was: i can see this is inhomogenous System of linear Equations. The given Field is $$\mathbb{Z}_{/7}=\{ 0,1,2,3,4,5,6\}$$ so i have to calculate in Modulo 7, i think. My task seems to be to find the general solution of this System.
I tried the Cramer's rule.
$$(A\ \ b)=\left(\begin{array}{ccc|c}
1 & 0 & 3 & 5 \\
5 & 3 & 6 & 3 \\
6 & 2 & 5 & 6
\end{array}\right)$$
Since this is not Zero:
$$\det A = \left|\begin{array}{ccc}
1 & 0 & 3 \\
5 & 3 & 6 \\
6 & 2 & 5
\end{array}\right|=-8 \neq 0$$
there must be a solution. i think.
$$x_1 = \frac{\det A_1}{\det A}=\frac{ \left|\begin{array}{ccc}
5 & 0 & 3 \\
3 & 3 & 6 \\
6 & 2 & 5
\end{array}\right| }{ \left|\begin{array}{ccc}
1 & 0 & 3 \\
5 & 3 & 6 \\
6 & 2 & 5
\end{array}\right| } =\frac{0}{-8}=0$$
$$x_2 = \frac{\det A_2}{\det A}=\frac{ \left|\begin{array}{ccc}
1 & 5 & 3 \\
5 & 3 & 6 \\
6 & 6 & 5
\end{array}\right| }{ \left|\begin{array}{ccc}
1 & 0 & 3 \\
5 & 3 & 6 \\
6 & 2 & 5
\end{array}\right| }=\frac{0}{-8}=0$$
$$x_3 = \frac{\det A_3}{\det A}=\frac{ \left|\begin{array}{ccc}
1 & 0 & 5 \\
5 & 3 & 3 \\
6 & 2 & 6
\end{array}\right| }{ \left|\begin{array}{ccc}
1 & 0 & 3 \\
5 & 3 & 6 \\
6 & 2 & 5
\end{array}\right| }=\frac{-7}{-8}=\frac{7}{8}$$
Okay now i have a solution.
$$\left(\begin{array}{c}
x_1 \\
x_2 \\
x_3
\end{array}\right) = \left(\begin{array}{c}
0 \\
0 \\
\frac{7}{8}
\end{array}\right)$$That Looks cool. But my Task is to find ALL Solutions. But the given System has as many unknowns as equations. So i learned there should be only one solution.
My conclusion is: there is either just one solution (and this one is "all") or i made one (or more) mistakes.
My question is: am i right? And in case i am wrong, what are my mistakes? How can i solve such a Task?
Best Answer
\begin{align} (A\ \ b)= \left(\begin{array}{ccc|c} 1 & 0 & 3 & 5 \\ 5 & 3 & 6 & 3 \\ 6 & 2 & 5 & 6 \end{array}\right) &\to \left(\begin{array}{ccc|c} 1 & 0 & 3 & 5 \\ 0 & 3 & 5 & 6 \\ 0 & 2 & 1 & 4 \end{array}\right)\quad R_2-5R1, R_3-6R_1 \\ &\to \left(\begin{array}{ccc|c} 1 & 0 & 3 & 5 \\ 0 & 1 & 4 & 2 \\ 0 & 2 & 1 & 4 \end{array}\right)\quad 5R_2 \\ &\to \left(\begin{array}{ccc|c} 1 & 0 & 3 & 5 \\ 0 & 1 & 3 & 2 \\ 0 & 0 & 0 & 0 \end{array}\right)\quad R_3-2R_2 \end{align} So the solution set is $$ \begin{pmatrix}5-3h\\2-3h\\h\end{pmatrix}\quad (h\in\mathbb{Z}/7\mathbb{Z}) $$
Let's look at the determinant, if you don't trust Gauss Elimination. I'll develop with respect to the first row: \begin{align} \det A&=(-1)^{1+1}\det\begin{pmatrix}3 & 6 \\ 2 & 5\end{pmatrix} +0+(-1)^{1+3}\cdot 3\det\begin{pmatrix}5 & 3 \\ 6 & 2\end{pmatrix}\\ &=(15-12)+3(10-18)\\ &=3+3(-8)=3-24=-21=0 \end{align} so you can't apply Cramer's rule.