Answer
The values of w, x, y, z are $w=1,x=3,y=0,z=-2$.
Work Step by Step
Consider the system of the equations:
$\begin{align}
& 2w+y-3z=8 \\
& w-x+4z=-10 \\
& 3w+5x-y-z=20 \\
& w+x-y-z=6
\end{align}$
The matrix corresponding to the system of equations is as follows:
$\left[ \begin{matrix}
2 & 0 & 1 & -3 & 8 \\
1 & -1 & 0 & 4 & -10 \\
3 & 5 & -1 & -1 & 20 \\
1 & 1 & -1 & -1 & 6 \\
\end{matrix} \right]$
The matrix will be converted into the echelon form by using the elementary row transformation.
${{R}_{1}}\to \frac{1}{2}{{R}_{1}}$ gives,
$\left[ \begin{matrix}
1 & 0 & \frac{1}{2} & -\frac{3}{2} & 4 \\
1 & -1 & 0 & 4 & -10 \\
3 & 5 & -1 & -1 & 20 \\
1 & 1 & -1 & -1 & 6 \\
\end{matrix} \right]$
${{R}_{2}}\to {{R}_{2}}+\left( -1 \right){{R}_{1}},{{R}_{3}}\to {{R}_{3}}+\left( -3 \right){{R}_{1}},{{R}_{4}}\to {{R}_{4}}+\left( -1 \right){{R}_{1}}$ give,
$\left[ \begin{matrix}
1 & 0 & \frac{1}{2} & -\frac{3}{2} & 4 \\
0 & -1 & -\frac{1}{2} & \frac{11}{2} & -14 \\
0 & 5 & -\frac{5}{2} & \frac{7}{2} & 8 \\
0 & 1 & -\frac{3}{2} & \frac{1}{2} & 2 \\
\end{matrix} \right]$
${{R}_{2}}\to -1{{R}_{2}}$ gives,
$\left[ \begin{matrix}
1 & 0 & \frac{1}{2} & -\frac{3}{2} & 4 \\
0 & 1 & \frac{1}{2} & -\frac{11}{2} & 14 \\
0 & 5 & -\frac{5}{2} & \frac{7}{2} & 8 \\
0 & 1 & -\frac{3}{2} & \frac{1}{2} & 2 \\
\end{matrix} \right]$
${{R}_{3}}\to {{R}_{3}}+\left( -5 \right){{R}_{2}},{{R}_{4}}\to {{R}_{4}}+\left( -1 \right){{R}_{2}}$ gives,
$\left[ \begin{matrix}
1 & 0 & \frac{1}{2} & -\frac{3}{2} & 4 \\
0 & 1 & \frac{1}{2} & -\frac{11}{2} & 14 \\
0 & 0 & -5 & 31 & -62 \\
0 & 0 & -2 & 6 & -12 \\
\end{matrix} \right]$
${{R}_{3}}\to -\frac{1}{5}{{R}_{3}}$ gives,
$\left[ \begin{matrix}
1 & 0 & \frac{1}{2} & -\frac{3}{2} & 4 \\
0 & 1 & \frac{1}{2} & -\frac{11}{2} & 14 \\
0 & 0 & 1 & -\frac{31}{5} & \frac{62}{5} \\
0 & 0 & -2 & 6 & -12 \\
\end{matrix} \right]$
${{R}_{4}}\to {{R}_{4}}+\left( 2 \right){{R}_{3}}$ gives,
$\left[ \begin{matrix}
1 & 0 & \frac{1}{2} & -\frac{3}{2} & 4 \\
0 & 1 & \frac{1}{2} & -\frac{11}{2} & 14 \\
0 & 0 & 1 & -\frac{31}{5} & \frac{62}{5} \\
0 & 0 & 0 & -\frac{32}{5} & \frac{64}{5} \\
\end{matrix} \right]$
${{R}_{4}}\to -\frac{5}{32}{{R}_{4}}$ gives,
$\left[ \begin{matrix}
1 & 0 & \frac{1}{2} & -\frac{3}{2} & 4 \\
0 & 1 & \frac{1}{2} & -\frac{11}{2} & 14 \\
0 & 0 & 1 & -\frac{31}{5} & \frac{62}{5} \\
0 & 0 & 0 & 1 & -2 \\
\end{matrix} \right]$
As, the desired matrix is in row echelon form, so, the system of the linear equations corresponding to the echelon form of the matrix will be as below:
$w+\frac{1}{2}y-\frac{3}{2}z=4$ (I)
$x+\frac{1}{2}y-\frac{11}{2}z=14$ (II)
$y-\frac{31}{5}z=\frac{62}{5}$ (III)
$z=-2$ (IV)
Apply the back-substitution method:
Equation (IV) gives:
$z=-2$.
Put the value of z in the equation (III) as follows:
$\begin{align}
& y-\frac{31}{5}\left( -2 \right)=\frac{62}{5} \\
& y=\frac{62}{5}-\frac{62}{5} \\
& y=0
\end{align}$
Put the values of y and z in the equation (II) as follows:
$\begin{align}
& x+\frac{1}{2}\left( 0 \right)-\frac{11}{2}\left( -2 \right)=14 \\
& x=14-11 \\
& x=3
\end{align}$
Put the values of x, y, and z in the equation (I) as follows:
$\begin{align}
& w+\frac{1}{2}\left( 0 \right)-\frac{3}{2}\left( -2 \right)=4 \\
& w=4-3 \\
& w=1
\end{align}$
Therefore, the values of w, x, y, z are $w=1,x=3,y=0,z=-2$.
