$AX=B$ solve for $X$ ....... in MATRIX $$ 2x - 3y + 4z  = -19\\
6x + 4y - 2z  =8   \\
x  + 5y + 4z  =   23
$$
what I have done so far is I put the nubmer and $x, y $ and $ z$ in matrix form:
$$
\begin{bmatrix}
    2 & -3 & 4\\
    6 & 4 &-2\\
    1 & 5 & 4
\end{bmatrix}
\begin{bmatrix}
x\\
y\\
z
\end{bmatrix}=
\begin{bmatrix}
-19\\         
8\\
23
\end{bmatrix}
$$
step 2: I don't know where to go from here
 A: You can use Gaussian Elimination on the augmented coefficient matrix to solve for $x, y, z$ by expressing the matrix in reduced row echelon form. 
$$\begin{bmatrix}
    2 & -3 & 4 &\mid&19\\
    6 & 4 &-2&\mid & 8\\
    1 & 5 & 4& \mid &23
\end{bmatrix}$$
If you do this correctly, you should obtain the following:
$$\begin{bmatrix}
    1 & 0 & 0 &\mid&20/9\\
    0& 1 &0&\mid & 7/9\\
    0 & 0 & 1& \mid &38/9
\end{bmatrix}$$
 This means that $$\begin{bmatrix} x\\y\\z\end{bmatrix} = \begin{bmatrix} 20/9 \\ 7/9\\38/9\end{bmatrix} $$
A: You can also use Cramer's Rule:
$$
x=\frac{\left|\begin{array}{r}\color{#C00000}{-19}&-3&4\\\color{#C00000}{8}&4&-2\\\color{#C00000}{23}&5&4\end{array}\right|}{\left|\begin{array}{r}2&-3&4\\6&4&-2\\1&5&4\end{array}\right|}=-2
$$
$$
y=\frac{\left|\begin{array}{r}2&\color{#C00000}{-19}&4\\6&\color{#C00000}{8}&-2\\1&\color{#C00000}{23}&4\end{array}\right|}{\left|\begin{array}{r}2&-3&4\\6&4&-2\\1&5&4\end{array}\right|}=5
$$
$$
z=\frac{\left|\begin{array}{r}2&-3&\color{#C00000}{-19}\\6&4&\color{#C00000}{8}\\1&5&\color{#C00000}{23}\end{array}\right|}{\left|\begin{array}{r}2&-3&4\\6&4&-2\\1&5&4\end{array}\right|}=0
$$
In the numerator, replace the column in the matrix corresponding to the given variable by the column of results. Note that the bars denote the determinant of the matrix.
A: You are trying to solve a system of equation in the form
$$
A\cdot x = B
$$
where A is a $3$x$3$ matrix, $x$ is your $3$ elements vector and $B$ is your constant vector. You get your $x$ doing
$$
x=A^{-1}\cdot B
$$
Since I am lazy I used the computer to solve it. Your result is
$$
20/9, 7/9, 38/9
$$
