Answer
The values of w, x, y, z are $w=2,x=1,y=-1,z=3$.
Work Step by Step
Consider the system of the equations:
$\begin{align}
& w+x+y+z=5 \\
& w+2x-y-2z=-1 \\
& w-3x-3y-z=-1 \\
& 2w-x+2y-z=-2
\end{align}$
The matrix corresponding to the system of equation is as follows:
$\left[ \begin{matrix}
1 & 1 & 1 & 1 & 5 \\
1 & 2 & -1 & -2 & -1 \\
1 & -3 & -3 & -1 & -1 \\
2 & -1 & 2 & -1 & -2 \\
\end{matrix} \right]$
The matrix will be converted into the echelon form by using the elementary row transformation.
${{R}_{2}}\to {{R}_{2}}+\left( -1 \right){{R}_{1}},{{R}_{3}}\to {{R}_{3}}+\left( -1 \right){{R}_{1}},{{R}_{4}}\to {{R}_{4}}+\left( -2 \right){{R}_{1}}$ give,
$\left[ \begin{matrix}
1 & 1 & 1 & 1 & 5 \\
0 & 1 & -2 & -3 & -6 \\
0 & -4 & -4 & -2 & -6 \\
0 & -3 & 0 & -3 & -12 \\
\end{matrix} \right]$
${{R}_{3}}\to {{R}_{3}}+\left( 4 \right){{R}_{2}},{{R}_{4}}\to {{R}_{4}}+\left( 3 \right){{R}_{2}}$ gives,
$\left[ \begin{matrix}
1 & 1 & 1 & 1 & 5 \\
0 & 1 & -2 & -3 & -6 \\
0 & 0 & -12 & -14 & -30 \\
0 & 0 & -6 & -12 & -30 \\
\end{matrix} \right]$
${{R}_{3}}\to -\frac{1}{12}{{R}_{3}}$ gives,
$\left[ \begin{matrix}
1 & 1 & 1 & 1 & 5 \\
0 & 1 & -2 & -3 & -6 \\
0 & 0 & 1 & \frac{7}{6} & \frac{5}{2} \\
0 & 0 & -6 & -12 & -30 \\
\end{matrix} \right]$
${{R}_{4}}\to {{R}_{4}}+\left( 6 \right){{R}_{3}}$ gives,
$\left[ \begin{matrix}
1 & 1 & 1 & 1 & 5 \\
0 & 1 & -2 & -3 & -6 \\
0 & 0 & 1 & \frac{7}{6} & \frac{5}{2} \\
0 & 0 & 0 & -5 & -15 \\
\end{matrix} \right]$
${{R}_{4}}\to -\frac{1}{5}{{R}_{4}}$ gives,
$\left[ \begin{matrix}
1 & 1 & 1 & 1 & 5 \\
0 & 1 & -2 & -3 & -6 \\
0 & 0 & 1 & \frac{7}{6} & \frac{5}{2} \\
0 & 0 & 0 & 1 & 3 \\
\end{matrix} \right]$
Since the desired matrix is in row echelon form, so, express the system of linear equations corresponding to the echelon form of matrix as follows:
$w+x+y+z=5$ (I)
$x-2y-3z=-6$ (II)
$y+\frac{7}{6}z=\frac{5}{2}$ (III)
$z=3$ (IV)
Apply the back-substitution method:
Equation (IV) gives,
$z=3$.
Substitute the value of z in the equation (III) as follows:
$\begin{align}
& y+\frac{7}{6}\left( 3 \right)=\frac{5}{2} \\
& y=\frac{5}{2}-\frac{7}{2} \\
& y=-1
\end{align}$
Substitute the values of y and z in the equation (II)as follows:
$\begin{align}
& x-2\left( -1 \right)-3\left( 3 \right)=-6 \\
& x=1
\end{align}$
Substitute the values of x, y, and z in the equation (I)as follows:
$\begin{align}
& w+\left( 1 \right)+\left( -1 \right)+\left( 3 \right)=5 \\
& w=2
\end{align}$
Therefore, the values of w, x, y, z are $w=2,x=1,y=-1,z=3$.
