Answer
$$x_1=2,\quad x_2= -3, \quad x_3=3.$$
Work Step by Step
We have the system
$$\left[ \begin {array}{ccc} -1&1&2\\ 2&3&1
\\ 5&4&2\end {array} \right]
\left[\begin{array}{rrr}{x_1}\\ {x_2} \\{x_3} \end{array}\right]=\left[\begin{array}{rrr}{1} \\ {-2}\\{4} \end{array}\right].$$
The coefficients matrix $A= \left[ \begin {array}{ccc} -1&1&2\\ 2&3&1
\\ 5&4&2\end {array} \right]
$ has the inverse
$$A^{-1}= \left[ \begin {array}{ccc} -\frac{2}{15}&-\frac{2}{5}&\frac{1}{3}\\ -\frac{1}{15}&\frac{4}{5}&-\frac{1}{3}\\ {\frac {7}{15}}&-\frac{3}{5}&\frac{1}{3}\end {array}
\right]
.
$$
The system has the solution
$$\left[\begin{array}{rrr}{x_1}\\ {x_2} \\{x_3} \end{array}\right]=\left[ \begin {array}{ccc} -\frac{2}{15}&-\frac{2}{5}&\frac{1}{3}\\ -\frac{1}{15}&\frac{4}{5}&-\frac{1}{3}\\ {\frac {7}{15}}&-\frac{3}{5}&\frac{1}{3}\end {array}
\right]\left[\begin{array}{rrr}{1} \\ {-2}\\{4} \end{array}\right]=\left[ \begin {array}{c} 2\\ -3\\ 3\end {array} \right]
.$$
That is
$$x_1=2,\quad x_2= -3, \quad x_3=3.$$
