## Elementary Linear Algebra 7th Edition

$$-3\left[\begin{array}{l}{1} \\ {-1}\\{0} \end{array}\right]+ 2\left[\begin{array}{l}{2} \\ {0} \\{1} \end{array}\right]+0\left[\begin{array}{l}{4} \\ {2}\\{3} \end{array}\right]=\left[\begin{array}{l}{1} \\ {3}\\{2} \end{array}\right].$$
Assume that $$x_1\left[\begin{array}{l}{1} \\ {-1}\\{0} \end{array}\right]+x_2\left[\begin{array}{l}{2} \\ {0} \\{1} \end{array}\right]+x_3\left[\begin{array}{l}{4} \\ {2}\\{3} \end{array}\right]=\left[\begin{array}{l}{1} \\ {3}\\{2} \end{array}\right],$$ then we have the system \begin{aligned} x_{1}+2x_{2}+4 x_{3} &=1 \\ -x_{1} +2x_3 &=3\\ x_{2}+3 x_{3} &=2 \end{aligned}. The augmented matrix is given by $$\left[ \begin {array}{cccc} 1&2&4&1\\ -1&0&2&3 \\ 0&1&3&2\end {array} \right] .$$ Using Gauss-Jordan elimination, we get the row-reduced echelon form as follows $$\left[ \begin {array}{cccc} 1&0&-2&-3\\ 0&1&3&2 \\ 0&0&0&0\end {array} \right] .$$ From which the solution is $$x_1=-3+2t, \quad x_2=2-3t, \quad x_3=t.$$ For example, take $t=0$, then $$-3\left[\begin{array}{l}{1} \\ {-1}\\{0} \end{array}\right]+ 2\left[\begin{array}{l}{2} \\ {0} \\{1} \end{array}\right]+0\left[\begin{array}{l}{4} \\ {2}\\{3} \end{array}\right]=\left[\begin{array}{l}{1} \\ {3}\\{2} \end{array}\right].$$