Answer
No solution or $\varnothing$.
Work Step by Step
The given system of equations is
$3x-y+2z=4$
$-6x+2y-4z=1$
$5x-3y+8z=0$
The augmented matrix is
$\Rightarrow \left[\begin{array}{ccc|c}
3 & -1 & 2& 4\\
-6 & 2 & -4& 1 \\
5&-3&8&0
\end{array}\right]$
Perform $R_1\rightarrow R_1/3$.
$\Rightarrow \left[\begin{array}{ccc|c}
3/3 & -1/3 & 2/3& 4/3\\
-6 & 2 & -4& 1 \\
5&-3&8&0
\end{array}\right]$
Simplify.
$\Rightarrow \left[\begin{array}{ccc|c}
1 & -1/3 & 2/3& 4/3\\
-6 & 2 & -4& 1 \\
5&-3&8&0
\end{array}\right]$
Perform $R_2\rightarrow R_2+6R_1$ and $R_3\rightarrow R_3-5 R_1$.
$\Rightarrow \left[\begin{array}{ccc|c}
1 & -1/3 & 2/3& 4/3\\
-6+6(1) & 2+6(-1/3) & -4+6(2/3)& 1+6(4/3) \\
5-5(1)&-3-5(-1/3)&8-5(2/3)&0-5(4/3)
\end{array}\right]$
Simplify.
$\Rightarrow \left[\begin{array}{ccc|c}
1 & -1/3 & 2/3& 4/3\\
0 & 0 & 0& 9 \\
0&-4/3&14/3&-20/3
\end{array}\right]$
Use back substitution for the second row.
$x(0)+y(0)+z(0)=9$
There are no values of $x,y$ and $z$ for which the above equation satisfy.
Hence, the system is inconsistent and has no solution.
