In augmented matrix form, the system is
[tex]\begin{bmatrix}1&5&0&0\\1&6&1&1\\2&-1&-1&-21\end{bmatrix}[/tex]
Subtract row 1 from row 2, and subtract 2 times row 1 from row 3:
[tex]\begin{bmatrix}1&5&0&0\\0&1&1&1\\0&-11&-1&-21\end{bmatrix}[/tex]
Add 11 times row 1 to row 3:
[tex]\begin{bmatrix}1&5&0&0\\0&1&1&1\\0&0&10&-10\end{bmatrix}[/tex]
Divide through row 3 by 10:
[tex]\begin{bmatrix}1&5&0&0\\0&1&1&1\\0&0&1&-1\end{bmatrix}[/tex]
Row 3 says [tex]z=-1[/tex]. Substituting into row 2, we get [tex]y+z=1\implies y=2[/tex]. Substituting both into row 1, we get [tex]x+5y=0\implies x=-10[/tex]. So the solution is the single ordered triplet (-10, 2, -1).
