Answer
The values of w, x, y, z are $w=1,x=2,y=3,z=-2$.
Work Step by Step
Consider the system of equations:
$\begin{align}
& w+x+y+z=4 \\
& 2w+x-2y-z=0 \\
& w-2x-y-2z=-2 \\
& 3w+2x+y+3z=4
\end{align}$
The matrix corresponding to the system of equation is as follows:
$\left[ \begin{matrix}
1 & 1 & 1 & 1 & 4 \\
2 & 1 & -2 & -1 & 0 \\
1 & -2 & -1 & -2 & -2 \\
3 & 2 & 1 & 3 & 4 \\
\end{matrix} \right]$
The matrix will be converted into the echelon form by using the elementary row transformation.
${{R}_{2}}\to {{R}_{2}}+\left( -2 \right){{R}_{1}},{{R}_{3}}\to {{R}_{3}}+\left( -1 \right){{R}_{1}},{{R}_{4}}\to {{R}_{4}}+\left( -3 \right){{R}_{1}}$ give,
$\left[ \begin{matrix}
1 & 1 & 1 & 1 & 4 \\
0 & -1 & -4 & -3 & -8 \\
0 & -3 & -2 & -3 & -6 \\
0 & -1 & -2 & 0 & -8 \\
\end{matrix} \right]$
${{R}_{2}}\to \left( -1 \right){{R}_{2}}$ gives,
$\left[ \begin{matrix}
1 & 1 & 1 & 1 & 4 \\
0 & 1 & 4 & 3 & 8 \\
0 & -3 & -2 & -3 & -6 \\
0 & -1 & -2 & 0 & -8 \\
\end{matrix} \right]$
${{R}_{3}}\to {{R}_{3}}+\left( 3 \right){{R}_{2}},{{R}_{4}}\to {{R}_{4}}+\left( 1 \right){{R}_{2}}$ gives,
$\left[ \begin{matrix}
1 & 1 & 1 & 1 & 4 \\
0 & 1 & 4 & 3 & 8 \\
0 & 0 & 10 & 6 & 18 \\
0 & 0 & 2 & 3 & 0 \\
\end{matrix} \right]$
${{R}_{3}}\to \frac{1}{10}{{R}_{3}}$ gives,
$\left[ \begin{matrix}
1 & 1 & 1 & 1 & 4 \\
0 & 1 & 4 & 3 & 8 \\
0 & 0 & 1 & \frac{3}{5} & \frac{9}{5} \\
0 & 0 & 2 & 3 & 0 \\
\end{matrix} \right]$
${{R}_{4}}\to {{R}_{4}}+\left( -2 \right){{R}_{3}}$ gives,
$\left[ \begin{matrix}
1 & 1 & 1 & 1 & 4 \\
0 & 1 & 4 & 3 & 8 \\
0 & 0 & 1 & \frac{3}{5} & \frac{9}{5} \\
0 & 0 & 0 & \frac{9}{5} & -\frac{18}{5} \\
\end{matrix} \right]$
${{R}_{4}}\to \frac{5}{9}{{R}_{4}}$ gives,
$\left[ \begin{matrix}
1 & 1 & 1 & 1 & 4 \\
0 & 1 & 4 & 3 & 8 \\
0 & 0 & 1 & \frac{3}{5} & \frac{9}{5} \\
0 & 0 & 0 & 1 & -2 \\
\end{matrix} \right]$
Since the desired matrix is now in row echelon form, so, express the system of linear equations corresponding to the echelon form of the matrix as follows:
$w+x+y+z=4$ (I)
$x+4y+3z=8$ (II)
$y+\frac{3}{5}z=\frac{9}{5}$ (III)
$z=-2$ (IV)
Apply the back-substitution method:
Equation (IV) gives,
$z=-2$.
Substitute the value of z in the equation (III) as follows:
$\begin{align}
& y+\frac{3}{5}\left( -2 \right)=\frac{9}{5} \\
& y=\frac{9}{5}+\frac{6}{5} \\
& y=3
\end{align}$
Substitute the values of y and z in the equation (II) as follows:
$\begin{align}
& x+4\left( 3 \right)+3\left( -2 \right)=8 \\
& x=2
\end{align}$
Substitute the values of x, y, and z in the equation (I) as follows:
$\begin{align}
& w+\left( 2 \right)+\left( 3 \right)+\left( -2 \right)=4 \\
& w=1
\end{align}$
Therefore, the values of w, x, y, z are $w=1,x=2,y=3,z=-2$.
