Elementary Linear Algebra 7th Edition

$$\left[\begin{array}{l}{1} \\ {1}\\{2} \end{array}\right]+ 2\left[\begin{array}{l}{1} \\ {0} \\{-1} \end{array}\right]+0\left[\begin{array}{l}{-5} \\ {-1}\\{-1} \end{array}\right]=\left[\begin{array}{l}{3} \\ {1}\\{0} \end{array}\right].$$
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 $$\left[\begin{array}{l}{1} \\ {1}\\{2} \end{array}\right]+ 2\left[\begin{array}{l}{1} \\ {0} \\{-1} \end{array}\right]+0\left[\begin{array}{l}{-5} \\ {-1}\\{-1} \end{array}\right]=\left[\begin{array}{l}{3} \\ {1}\\{0} \end{array}\right].$$