Answer
$(x,y,z) =\left(3,-\frac{8}{3},\frac{1}{9}\right)$
Work Step by Step
The given system of equations is
$\left\{\begin{matrix}
x& +4y &-3z&=&-8\\
3x& -y & +3z&=&12\\
x& +y &+6z &=&1
\end{matrix}\right.$
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.
$D=\begin{vmatrix}
1&4 &-3 \\
3& -1 &3 \\
1&1 &6
\end{vmatrix}=-81$
For determinant $D_x$ replace the $x−$ coefficients with the constants.
$D_x=\begin{vmatrix}
-8&4 &-3 \\
12& -1 &3 \\
1&1 &6
\end{vmatrix}=-243$
For determinant $D_y$ replace the $y−$ coefficients with the constants.
$D_y=\begin{vmatrix}
1&-8 &-3 \\
3& 12 &3 \\
1&1 &6
\end{vmatrix}=216$
For determinant $D_z$ replace the $z−$ coefficients with the constants.
$D_z=\begin{vmatrix}
1&4 &-8 \\
3& -1 &12 \\
1&1 &1
\end{vmatrix}=-9$
By using Cramer's rule we have.
$x=\dfrac{D_x}{D}=\dfrac{-243}{-81}=3$
and
$y=\dfrac{D_y}{D}=\dfrac{216}{-81}=-\dfrac{8}{3}$
and
$z=\dfrac{D_z}{D}=\dfrac{-9}{-81}=\dfrac{1}{9}$
Hence, the solution set is $(x,y,z) =\left(3,-\frac{8}{3},\frac{1}{9}\right)$.
