Answer
$$ \left[\begin{array}{l}{1} \\ {1}\\{2} \end{array}\right]+ 2\left[\begin{array}{l}{1} \\ {0} \\{-1} \end{array}\right]+0\left[\begin{array}{l}{-5} \\ {-1}\\{-1} \end{array}\right]=\left[\begin{array}{l}{3} \\ {1}\\{0} \end{array}\right].$$
Work Step by Step
Assume that
$$x_1\left[\begin{array}{l}{1} \\ {1}\\{2} \end{array}\right]+x_2\left[\begin{array}{l}{1} \\ {0} \\{-1} \end{array}\right]+x_3\left[\begin{array}{l}{-5} \\ {-1}\\{-1} \end{array}\right]=\left[\begin{array}{l}{3} \\ {1}\\{0} \end{array}\right],$$
then we have the system
$$
\begin{aligned}
x_{1}+x_{2}-5 x_{3} &=3 \\
x_{1} -1x_3 &=1\\
2x_1-x_{2}- x_{3} &=0
\end{aligned}.
$$
The augmented matrix is given by
$$
\left[ \begin {array}{cccc} 1&1&-5&3\\ 1&0&-1&1
\\ 2&-1&-1&0\end {array} \right]
.
$$
Using Gauss-Jordan elimination, we get the row-reduced echelon form as follows
$$\left[ \begin {array}{cccc} 1&0&0&1\\ 0&1&0&2
\\ 0&0&1&0\end {array} \right]
.$$
From which the solution is
$$x_1=1, \quad x_2=2 , \quad x_3=0.$$
Hence, we have
$$ \left[\begin{array}{l}{1} \\ {1}\\{2} \end{array}\right]+ 2\left[\begin{array}{l}{1} \\ {0} \\{-1} \end{array}\right]+0\left[\begin{array}{l}{-5} \\ {-1}\\{-1} \end{array}\right]=\left[\begin{array}{l}{3} \\ {1}\\{0} \end{array}\right].$$
