Answer
$\{(23,-12,3)\}$.
Work Step by Step
The given system of equations is
$\left\{\begin{matrix}
x& +2y &+2z&=&5\\
2x& +4y & +7z&=&19\\
-2x& -5y &-2z &=&8
\end{matrix}\right.$
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.
$D=\begin{vmatrix}
1& 2 &2 \\
2& 4 &7 \\
-2 &-5 &-2
\end{vmatrix}=3$
For determinant $D_x$ replace the $x−$ coefficients with the constants.
$D_x=\begin{vmatrix}
5& 2 &2 \\
19& 4 &7 \\
8 &-5 &-2
\end{vmatrix}=69$
For determinant $D_y$ replace the $y−$ coefficients with the constants.
$D_y=\begin{vmatrix}
1& 5 &2 \\
2& 19 &7 \\
-2 &8 &-2
\end{vmatrix}=-36$
For determinant $D_z$ replace the $z−$ coefficients with the constants.
$D_z=\begin{vmatrix}
1& 2 &5 \\
2& 4 &19 \\
-2 &-5 &8
\end{vmatrix}=9$
By using Cramer's rule we have.
$x=\frac{D_x}{D}=\frac{69}{3}=23$
and
$y=\frac{D_y}{D}=\frac{-36}{3}=-12$
and
$z=\frac{D_z}{D}=\frac{9}{3}=3$
Hence, the solution set is $\{(x,y,z)\} =\{(23,-12,3)\}$.
