Answer
$\{(2,-3,4)\}$.
Work Step by Step
The given system of equations is
$\left\{\begin{matrix}
4x& -5y &-6z&=&-1 \\
x& -2y & -5z&=&-12\\
2x& -y &+0z &=&7
\end{matrix}\right.$
The formula to determine the determinant is
$D=\begin{vmatrix}
a& b &c \\
d& e &f \\
g &h &i
\end{vmatrix}=a\begin{vmatrix}
e& f \\
h&i
\end{vmatrix}-b\begin{vmatrix}
d& f \\
g&i
\end{vmatrix}+c\begin{vmatrix}
d& e \\
g&h
\end{vmatrix}$
Determinant $D$ consists of the $x,y$ and $z$ coefficients.
$D=\begin{vmatrix}
4& -5 &-6 \\
1& -2 &-5 \\
2 &-1 &0
\end{vmatrix}=12$
For determinant $D_x$ replace the $x−$ coefficients with the constants.
$D_x=\begin{vmatrix}
-1& -5 &-6 \\
-12& -2 &-5 \\
7&-1 &0
\end{vmatrix}=24$
For determinant $D_y$ replace the $y−$ coefficients with the constants.
$D_y=\begin{vmatrix}
4& -1 &-6 \\
1& -12 &-5 \\
2 &7 &0
\end{vmatrix}=-36$
For determinant $D_z$ replace the $z−$ coefficients with the constants.
$D_z=\begin{vmatrix}
4& -5 &-1 \\
1& -2 &-12 \\
2 &-1 &7
\end{vmatrix}=48$
By using Cramer's rule we have.
$x=\frac{D_x}{D}=\frac{24}{12}=2$
and
$y=\frac{D_y}{D}=\frac{-36}{12}=-3$
and
$x=\frac{D_z}{D}=\frac{48}{12}=4$
Hence, the solution set is $\{(x,y,z)\} =\{(2,-3,4)\}$.
