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 \( x_3 = {[6]_{13}} \)