Answer
No.
Work Step by Step
The given system of equations is
$\left\{\begin{matrix}
x& +y &+z&=&0 \\
2x& -3y & +z&=&5\\
4x& +2y &+4z &=&3
\end{matrix}\right.$
The augmented matrix is
$\Rightarrow \left[\begin{array}{ccc|c}
1 & 1 & 1& 0\\
2 & -3 & 1& 5 \\
4&2&4&3
\end{array}\right]$
Perform $R_2\rightarrow R_2-2\times R_1$ and $R_3\rightarrow R_3-4 R_1$.
$\Rightarrow \left[\begin{array}{ccc|c}
1 & 1 & 1& 0\\
2-2(1) & -3-2(1) & 1-2(1)& 5-2(0) \\
4-4(1)&2-4(1)&4-4(1)&3-4(0)
\end{array}\right]$
Simplify.
$\Rightarrow \left[\begin{array}{ccc|c}
1 & 1 & 1& 0\\
0 & -5 & -1& 5 \\
0&-2&0&3
\end{array}\right]$
Perform $R_2\rightarrow \frac{R_2}{-5}$.
$\Rightarrow \left[\begin{array}{ccc|c}
1 & 1 & 1& 0\\
0/(-5) & -5/(-5) & -1/(-5)& 5/(-5) \\
0&-2&0&3
\end{array}\right]$
Simplify.
$\Rightarrow \left[\begin{array}{ccc|c}
1 & 1 & 1& 0\\
0 & 1 & 1/5& -1 \\
0&-2&0&3
\end{array}\right]$
Perform $R_1\rightarrow R_1- R_2$ and $R_3\rightarrow R_3+2 R_2$.
$\Rightarrow \left[\begin{array}{ccc|c}
1-0 & 1-1 & 1-1/5& 0-(-1)\\
0 & 1 & 1/5& -1 \\
0+2(0)&-2+2(1)&0+2(1/5)&3+2(-1)
\end{array}\right]$
Simplify.
$\Rightarrow \left[\begin{array}{ccc|c}
1 & 0 & 4/5& 1\\
0 & 1 & 1/5& -1 \\
0&0&2/5&1
\end{array}\right]$
Perform $R_3\rightarrow R_3(5/2)$.
$\Rightarrow \left[\begin{array}{ccc|c}
1 & 0 & 4/5& 1\\
0 & 1 & 1/5& -1 \\
0(5/2)&0(5/2)&2/5(5/2)&1(5/2)
\end{array}\right]$
Simplify.
$\Rightarrow \left[\begin{array}{ccc|c}
1 & 0 & 4/5& 1\\
0 & 1 & 1/5& -1 \\
0&0&1&5/2
\end{array}\right]$
Perform $R_1\rightarrow R_1-R_3(4/5)$ and $R_2\rightarrow R_2- R_3(1/5)$.
$\Rightarrow \left[\begin{array}{ccc|c}
1-0(4/5) & 0-0(4/5) & 4/5-1(4/5)& 1-(5/2)(4/5)\\
0-0(1/5) & 1-0(1/5) & 1/5-1(1/5)& -1-(5/2)(1/5) \\
0&0&1&5/2
\end{array}\right]$
Simplify.
$\Rightarrow \left[\begin{array}{ccc|c}
1 & 0 & 0& -1\\
0 & 1 & 0& -3/2 \\
0&0&1&5/2
\end{array}\right]$
Use back substitution to solve the linear system.
$\Rightarrow x=-1$
and
$\Rightarrow y=-3/2$.
and
$\Rightarrow z=5/2$.
The solution set is $\{(x,y,z)\}=\{(-1,-3/2,5/2)\}$.
