Verwende Gauss-Algorithmus
\(\left(\begin{array}{llc}{[1]_{13}} & {[1]_{13}} & {[-1]_{13}} & {[3]_{13}} \\ {[2]_{13}} & {[0]_{13}} & {[1]_{13}}&{[5]_{13}} \\ {[1]_{13}} & {[1]_{13}} & {[3]_{13}} &{[1]_{13}}\end{array}\right)  \)
2.Zeile minus 2* 1. Zeile     und      3. Zeile minus 1. Zeile
\(\left(\begin{array}{llc}{[1]_{13}} & {[1]_{13}} & {[-1]_{13}} & {[3]_{13}}\\  {[0]_{13}} & {[11]_{13}} & {[3]_{13}}&{[12]_{13}} \\ {[0]_{13}} & {[0]_{13}} & {[4]_{13}} &{[11]_{13}}\end{array}\right)  \)
Also \(  x_3 = [6]_{13}\) denn 4*6 =24 ≡ 11 mod(13)
Dann die zweite Gleichung auswerten
\(    {[11]_{13}} \cdot x_2 + {[3]_{13}} \cdot [6]_{13}= {[12]_{13}}  \)
<=> \(    {[11]_{13}} \cdot x_2 +  [5]_{13}= {[12]_{13}}  \)
<=> \(    {[11]_{13}} \cdot x_2 = {[7]_{13}}  \)  |*6
<=> \(   x_2 = {[3]_{13}}  \) 
Und mit der ersten Gleichung x1 ausrechnen. Gibt  \(  x_3 = {[6]_{13}}  \)