Answer
$(x,y,z) =(0,0,0)$
Work Step by Step
The given system of equations is
$\left\{\begin{matrix}
x& +4y &-3z&=&0\\
3x& -y & +3z&=&0\\
x& +y &+6z &=&0
\end{matrix}\right.$
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.
$D_x=\begin{vmatrix}
0& 4 &-3 \\
0& -1 &3 \\
0&1 &6
\end{vmatrix}=0$
For determinant $D_y$ replace the $y−$ coefficients with the constants.
$D_y=\begin{vmatrix}
1& 0 &-3 \\
3& 0 &3 \\
1 &0&6
\end{vmatrix}=0$
For determinant $D_z$ replace the $z−$ coefficients with the constants.
$D_z=\begin{vmatrix}
1& 4 &0 \\
3& -1 &0 \\
1 &1 &0
\end{vmatrix}=0$
By using Cramer's rule we have.
$x=\dfrac{D_x}{D}=\dfrac{0}{-81}=0$
and
$y=\dfrac{D_y}{D}=\dfrac{0}{-81}=0$
and
$z=\dfrac{D_z}{D}=\dfrac{0}{-81}=0$
Hence, the solution set is $(x,y,z) =(0,0,0)$.
