From the system of equation, we construct the matrix of coefficients:
[tex]\begin{bmatrix}{5} & {4} & {-4} & {|13} \\ {1} & {1} & {5} & {|2} \\ {1} & {-4} & {0} & {|17} \\ & & & \end{bmatrix}[/tex]
Now, we subtract the second row from the third row:
[tex]\begin{bmatrix}{5} & {4} & {-4} & {|13} \\ {1} & {1} & {5} & {|2} \\ {0} & {-5} & {-5} & {|15} \\ & & & \end{bmatrix}[/tex]
We exchange the first and the second row:
[tex]\begin{bmatrix}{1} & {1} & {5} & {|2} \\ {5} & {4} & {-4} & {|13} \\ {0} & {-5} & {-5} & {|15} \\ & & & \end{bmatrix}[/tex]
Subtracting 5 times the first row from the second row:
[tex]\begin{bmatrix}{1} & {1} & {5} & {|2} \\ {0} & {-1} & {-29} & {|3} \\ {0} & {-5} & {-5} & {|15} \\ & & & \end{bmatrix}[/tex]
Finally, we subtract 5 times the second row from the third row:
[tex]\begin{bmatrix}{1} & {1} & {5} & {|2} \\ {0} & {-1} & {-29} & {|3} \\ {0} & {0} & {140} & {|0} \\ & & & \end{bmatrix}[/tex]
Now, using back substitution:
[tex]\begin{gathered} 140z=0 \\ \Rightarrow z=0 \end{gathered}[/tex][tex]\begin{gathered} -y-29z=3 \\ -y=3 \\ \Rightarrow y=-3 \end{gathered}[/tex][tex]\begin{gathered} x+y+5z=2 \\ x-3=2 \\ \Rightarrow x=5 \end{gathered}[/tex]
The solution set is:
[tex]\lbrace(5,-3,0)\rbrace[/tex]
