Let \(A=\left[\begin{array}{ccc}3 & 4 & 5 \\ 2 & -1 & 8 \\ 5 & 2 & 7\end{array}\right] B=\left[\begin{array}{c}18 \\ 13 \\ 20\end{array}\right]\)
The augmented matrix is
\({\left[\begin{array}{cccc}
3 & 4 & 5 & 18 \\
2 & -1 & 8 & 13 \\
5 & -2 & 7 & 20
\end{array}\right] \sim R_1 \rightarrow R_1-R_2} \)
\({\left[\begin{array}{cccc}
1 & 5 & -3 & 5 \\
2 & -1 & 8 & 13 \\
5 & -2 & 7 & 20
\end{array}\right]}\)
\(\begin{aligned}
\sim R_2& \rightarrow R_2-2R_1 \\
R_3 & \rightarrow R_3-5 R_1
\end{aligned}\left[\begin{array}{cccc}
1 & 5 & -3 & 5 \\
0 & -11 & 14 & 3 \\
0 & -27 & 22 & -5
\end{array}\right] \)
\(\sim R_2 \rightarrow-5 R_2+2 R_3\left[\begin{array}{cccc}
1 & 5 & -3 & 5 \\
0 & 1 & -26 & 25 \\
0 & -27 & 22 & -5
\end{array}\right]\)
\(\begin{aligned}
\sim R_1 & \rightarrow R_1-5 R_2 \\
R_3 & \rightarrow R_3+27 R_2
\end{aligned}\left[\begin{array}{cccc}
1 & 0 & 127 & 130 \\
0 & 1 & -26 & -25 \\
0 & 0 & -680 & -680
\end{array}\right]\)
\(\sim R_3 \rightarrow R_3+(-680)\left[\begin{array}{cccc}
1 & 0 & 127 & 135 \\
0 & 1 & -26 & -25 \\
0 & 0 & 1 & 1
\end{array}\right]\)
\(\begin{aligned}\sim R_1 & \rightarrow R_1-127 R_3 \\
R_2 & \rightarrow R_2+26 R_3
\end{aligned}\left[\begin{array}{llll}
1 & 0 & 0 & 3 \\
0 & 1 & 0 & 1 \\
0 & 0 & 1 & 1
\end{array}\right]\)
Hence the solution is \(x=3, y=1, z=1\).
