Answer
$$3\left[\begin{array}{l}{1} \\ {3} \end{array}\right]+0\left[\begin{array}{l}{-1} \\ {-3} \end{array}\right]-2\left[\begin{array}{l}{2} \\ {1} \end{array}\right]=\left[\begin{array}{l}{-1} \\ {7} \end{array}\right].$$
Work Step by Step
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
$$3\left[\begin{array}{l}{1} \\ {3} \end{array}\right]+0\left[\begin{array}{l}{-1} \\ {-3} \end{array}\right]-2\left[\begin{array}{l}{2} \\ {1} \end{array}\right]=\left[\begin{array}{l}{-1} \\ {7} \end{array}\right].$$
