<s>[INST] $AX=B$ 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
 [/INST] 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.
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