Answer
$$4\left[\begin{array}{l}{-3} \\ {3}\\{4} \end{array}\right]-2\left[\begin{array}{l}{5} \\ {4} \\{-8} \end{array}\right]=\left[\begin{array}{l}{-22} \\ {4}\\{32} \end{array}\right],$$
Work Step by Step
Assume that
$$x_1\left[\begin{array}{l}{-3} \\ {3}\\{4} \end{array}\right]+x_2\left[\begin{array}{l}{5} \\ {4} \\{-8} \end{array}\right]=\left[\begin{array}{l}{-22} \\ {4}\\{32} \end{array}\right],$$
then we have the system
$$
\begin{aligned}
-3x_{1}+5x_{2}&=-22 \\
3x_{1} +4x_2 &=4\\
4x_1-8x_{2} &=32
\end{aligned}.
$$
The augmented matrix is given by
$$
\left[ \begin {array}{ccc} -3&5&-22\\ 3&4&4
\\ 4&-8&32\end {array} \right]
.
$$
Using Gauss-Jordan elimination, we get the row-reduced echelon form as follows
$$\left[ \begin {array}{ccc} 1&0&4\\ 0&1&-2
\\ 0&0&0\end {array} \right]
.$$
From which the solution is
$$x_1=4, \quad x_2=-2 .$$
Hence, we have
$$4\left[\begin{array}{l}{-3} \\ {3}\\{4} \end{array}\right]-2\left[\begin{array}{l}{5} \\ {4} \\{-8} \end{array}\right]=\left[\begin{array}{l}{-22} \\ {4}\\{32} \end{array}\right],$$
