Answer
$(x,y,z) =(1,3,-2)$
Work Step by Step
The given system of equations is
$\left\{\begin{matrix}
x& -y &+z&=&-4\\
2x& -3y & +4z&=&-15\\
5x& +y &-2z &=&12
\end{matrix}\right.$
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.
$D=\begin{vmatrix}
1& -1 &1 \\
2& -3 &4 \\
5&1 &-2
\end{vmatrix}=-5$
For determinant $D_x$ replace the $x−$ coefficients with the constants.
$D_x=\begin{vmatrix}
-4& -1 &1 \\
-15& -3 &4 \\
12&1 &-2
\end{vmatrix}=-5$
For determinant $D_y$ replace the $y−$ coefficients with the constants.
$D_y=\begin{vmatrix}
1& -4 &1 \\
2& -15 &4 \\
5&12 &-2
\end{vmatrix}=-15$
For determinant $D_z$ replace the $z−$ coefficients with the constants.
$D_z=\begin{vmatrix}
1& -1 &-4 \\
2& -3 &-15 \\
5&1 &12
\end{vmatrix}=10$
By using Cramer's rule we have.
$x=\dfrac{D_x}{D}=\dfrac{-5}{-5}=1$
and
$y=\dfrac{D_y}{D}=\dfrac{-15}{-5}=3$
and
$z=\dfrac{D_z}{D}=\dfrac{10}{-5}=-2$
Hence, the solution set is $(x,y,z) =(1,3,-2)$.
