Answer
The solution set for the provided system of equations is $\left\{ \left( -3,\ 4,\ -2 \right) \right\}$.
Work Step by Step
The given system of equations can be written in matrix form as below: $\left[ \left. \begin{matrix}
3 & 4 & 2 \\
4 & -2 & -8 \\
1 & 1 & -1 \\
\end{matrix} \right|\begin{matrix}
3 \\
-4 \\
3 \\
\end{matrix} \right]$
Solve this matrix as below to get, $\left[ \left. \begin{matrix}
3 & 4 & 2 \\
0 & 6 & 4 \\
1 & 1 & -1 \\
\end{matrix} \right|\begin{matrix}
3 \\
16 \\
3 \\
\end{matrix} \right]$ $ By,\ 4{{R}_{3}}-{{R}_{2}}\to {{R}_{2}}$
$\left[ \left. \begin{matrix}
3 & 4 & 2 \\
0 & 6 & 4 \\
0 & 1 & 5 \\
\end{matrix} \right|\begin{matrix}
3 \\
16 \\
-6 \\
\end{matrix} \right]$ $ By,\ {{R}_{1}}-3{{R}_{3}}\to {{R}_{3}}$
$\left[ \left. \begin{matrix}
3 & 4 & 2 \\
0 & 6 & 4 \\
0 & 0 & 26 \\
\end{matrix} \right|\begin{matrix}
3 \\
16 \\
-52 \\
\end{matrix} \right]$ $ By,\ 6{{R}_{3}}-{{R}_{2}}\to {{R}_{3}}$
Convert the matrix into equation form, $\begin{align}
3x+4y+2z=3 & \\
6y+4z=16 & \\
26z=-52 & \\
\end{align}$
Solve the last equation to obtain the value of $ z $:
$\begin{align}
& 26z=-52 \\
& z=-\frac{52}{26} \\
& z=-2
\end{align}$
Substitute the value of $ z $ in $6y+4z=16$ to obtain the value of $ y $:
$\begin{align}
& 6y+4\times \left( -2 \right)=16 \\
& 6y-8=16 \\
& 6y=24 \\
& y=4
\end{align}$
Substitute the value of $ y,z $ in $3x+4y+2z=3$ to obtain the value of $ x $:
$\begin{align}
& 3x+4\times 4+2\times -2=3 \\
& 3x+16-4=3 \\
& 3x=-9 \\
& x=-3
\end{align}$
So, the solution is $\left\{ \left( -3,\ 4,\ -2 \right) \right\}$ .
