Answer
The values of w, x, y, z are $w=0,x=-3,y=0,z=-3$.
Work Step by Step
Consider the system of the equations:
$\begin{align}
& 3w-4x+y+z=9 \\
& w+x-y-z=0 \\
& 2w+x+4y-2z=3 \\
& -w+2x+y-3z=3
\end{align}$
The matrix corresponding to the system of equation is as follows:
$\left[ \begin{matrix}
3 & -4 & 1 & 1 & 9 \\
1 & 1 & -1 & -1 & 0 \\
2 & 1 & 4 & -2 & 3 \\
-1 & 2 & 1 & -3 & 3 \\
\end{matrix} \right]$
The matrix will be converted into the echelon form by using the elementary row transformation.
${{R}_{1}}\to \frac{1}{3}{{R}_{1}}$ gives,
$\left[ \begin{matrix}
1 & -\frac{4}{3} & \frac{1}{3} & \frac{1}{3} & 3 \\
1 & 1 & -1 & -1 & 0 \\
2 & 1 & 4 & -2 & 3 \\
-1 & 2 & 1 & -3 & 3 \\
\end{matrix} \right]$
${{R}_{2}}\to {{R}_{2}}+\left( -1 \right){{R}_{1}},{{R}_{3}}\to {{R}_{3}}+\left( -2 \right){{R}_{1}},{{R}_{4}}\to {{R}_{4}}+\left( 1 \right){{R}_{1}}$ give,
$\left[ \begin{matrix}
1 & -\frac{4}{3} & \frac{1}{3} & \frac{1}{3} & 3 \\
0 & \frac{7}{3} & -\frac{4}{3} & -\frac{4}{3} & -3 \\
0 & \frac{11}{3} & \frac{10}{3} & -\frac{8}{3} & -3 \\
0 & \frac{2}{3} & \frac{4}{3} & -\frac{8}{3} & 6 \\
\end{matrix} \right]$
${{R}_{2}}\to \frac{3}{7}{{R}_{2}}$ gives,
$\left[ \begin{matrix}
1 & -\frac{4}{3} & \frac{1}{3} & \frac{1}{3} & 3 \\
0 & 1 & -\frac{4}{7} & -\frac{4}{7} & -\frac{9}{7} \\
0 & \frac{11}{3} & \frac{10}{3} & -\frac{8}{3} & -3 \\
0 & \frac{2}{3} & \frac{4}{3} & -\frac{8}{3} & 6 \\
\end{matrix} \right]$
${{R}_{3}}\to {{R}_{3}}+\left( -\frac{11}{3} \right){{R}_{2}},{{R}_{4}}\to {{R}_{4}}+\left( -\frac{2}{3} \right){{R}_{2}}$ gives,
$\left[ \begin{matrix}
1 & -\frac{4}{3} & \frac{1}{3} & \frac{1}{3} & 3 \\
0 & 1 & -\frac{4}{7} & -\frac{4}{7} & -\frac{9}{7} \\
0 & 0 & \frac{38}{7} & -\frac{4}{7} & \frac{12}{7} \\
0 & 0 & \frac{12}{7} & -\frac{16}{7} & \frac{48}{7} \\
\end{matrix} \right]$
${{R}_{3}}\to \frac{7}{38}{{R}_{3}}$ gives,
$\left[ \begin{matrix}
1 & -\frac{4}{3} & \frac{1}{3} & \frac{1}{3} & 3 \\
0 & 1 & -\frac{4}{7} & -\frac{4}{7} & -\frac{9}{7} \\
0 & 0 & 1 & -\frac{2}{19} & \frac{6}{19} \\
0 & 0 & \frac{12}{7} & -\frac{16}{7} & \frac{48}{7} \\
\end{matrix} \right]$
${{R}_{4}}\to {{R}_{4}}+\left( -\frac{12}{7} \right){{R}_{3}}$ gives,
$\left[ \begin{matrix}
1 & -\frac{4}{3} & \frac{1}{3} & \frac{1}{3} & 3 \\
0 & 1 & -\frac{4}{7} & -\frac{4}{7} & -\frac{9}{7} \\
0 & 0 & 1 & -\frac{2}{19} & \frac{6}{19} \\
0 & 0 & 0 & -\frac{40}{19} & \frac{120}{19} \\
\end{matrix} \right]$
${{R}_{4}}\to -\frac{19}{40}{{R}_{4}}$ gives,
$\left[ \begin{matrix}
1 & -\frac{4}{3} & \frac{1}{3} & \frac{1}{3} & 3 \\
0 & 1 & -\frac{4}{7} & -\frac{4}{7} & -\frac{9}{7} \\
0 & 0 & 1 & -\frac{2}{19} & \frac{6}{19} \\
0 & 0 & 0 & 1 & -3 \\
\end{matrix} \right]$
Since the desired matrix is now in row echelon form, so express, the system of linear equations corresponding to the echelon form of matrix as follows:
$w-\frac{4}{3}x+\frac{1}{3}y+\frac{1}{3}z=3$ (I)
$x-\frac{4}{7}y-\frac{4}{7}z=-\frac{9}{7}$ (II)
$y-\frac{2}{19}z=\frac{6}{19}$ (III)
$z=-3$ (IV)
Apply the back-substitution method:
Equation (4) gives,
$z=-3$.
Substitute the value of z in the equation (3) as follows:
$\begin{align}
& y-\frac{2}{19}\left( -3 \right)=\frac{6}{19} \\
& y=0
\end{align}$
Substitute the values of y and z in the equation (2)as follows:
$\begin{align}
& x-\frac{4}{7}\left( 0 \right)-\frac{4}{7}\left( -3 \right)=-\frac{9}{7} \\
& x=-3
\end{align}$
Substitute the values of x, y, and z in the equation (1)as follows:
$\begin{align}
& w-\frac{4}{3}\left( -3 \right)+\frac{1}{3}\left( 0 \right)+\frac{1}{3}\left( -3 \right)=3 \\
& w=0
\end{align}$
Therefore, the values of w, x, y, z are $w=0,x=-3,y=0,z=-3$.
