## Elementary Linear Algebra 7th Edition

$$3\left[\begin{array}{l}{1} \\ {3} \end{array}\right]+0\left[\begin{array}{l}{-1} \\ {-3} \end{array}\right]-2\left[\begin{array}{l}{2} \\ {1} \end{array}\right]=\left[\begin{array}{l}{-1} \\ {7} \end{array}\right].$$
Assume that $$x_1\left[\begin{array}{l}{1} \\ {3} \end{array}\right]+x_2\left[\begin{array}{l}{-1} \\ {-3} \end{array}\right]+x_3\left[\begin{array}{l}{2} \\ {1} \end{array}\right]=\left[\begin{array}{l}{-1} \\ {7} \end{array}\right],$$ then we have the system \begin{aligned} x_{1}- x_{2}+2 x_{3} &=-1 \\ 3x_{1}- 3x_{2} +x_3 &=7 \end{aligned}. The augmented matrix is given by $$\left[ \begin {array}{cccc} 1&-1&2&-1\\ 3&-3&1&7 \end {array} \right] .$$ Using Gauss-Jordan elimination, we get the row-reduced echelon form as follows $$\left[ \begin {array}{cccc} 1&-1&0&3\\ 0&0&1&-2 \end {array} \right] .$$ From which the solution is $$x_1=3+t, \quad x_2=t, \quad x_3=-2.$$ For example, take $t=0$, then $$3\left[\begin{array}{l}{1} \\ {3} \end{array}\right]+0\left[\begin{array}{l}{-1} \\ {-3} \end{array}\right]-2\left[\begin{array}{l}{2} \\ {1} \end{array}\right]=\left[\begin{array}{l}{-1} \\ {7} \end{array}\right].$$