Answer
Yes, the statement makes sense.
Work Step by Step
Consider the system of equations,
$\begin{align}
& x+y+z+w=-1 \\
& -x+4y+z-w=0 \\
& x-2y+z-2w=11 \\
& -x-2y+z+2w=-3
\end{align}$
First we will write the augmented matrix for the given system of equations:
Augmented matrix is:
$\left[ \left. \begin{matrix}
1 & 1 & 1 & 1 \\
-1 & 4 & 1 & -1 \\
1 & -2 & 1 & -2 \\
-1 & -2 & 1 & 2 \\
\end{matrix} \right|\begin{matrix}
-1 \\
0 \\
11 \\
-3 \\
\end{matrix} \right]$
Now, using the row operation, we will reduce the matrix to row echelon form.
${{R}_{2}}\to {{R}_{2}}+{{R}_{1}}$ ,
$\left[ \left. \begin{matrix}
1 & 1 & 1 & 1 \\
0 & 5 & 2 & 0 \\
1 & -2 & 1 & -2 \\
-1 & -2 & 1 & 2 \\
\end{matrix} \right|\begin{matrix}
-1 \\
-1 \\
11 \\
-3 \\
\end{matrix} \right]$
${{R}_{3}}\to {{R}_{3}}-{{R}_{1}}$ ,
$\left[ \left. \begin{matrix}
1 & 1 & 1 & 1 \\
0 & 5 & 2 & 0 \\
0 & -3 & 0 & -3 \\
-1 & -2 & 1 & 2 \\
\end{matrix} \right|\begin{matrix}
-1 \\
-1 \\
12 \\
-3 \\
\end{matrix} \right]$
${{R}_{4}}\to {{R}_{4}}+{{R}_{1}}$ ,
$\left[ \left. \begin{matrix}
1 & 1 & 1 & 1 \\
0 & 5 & 2 & 0 \\
0 & -3 & 0 & -3 \\
0 & -1 & 2 & 3 \\
\end{matrix} \right|\begin{matrix}
-1 \\
-1 \\
12 \\
-4 \\
\end{matrix} \right]$
${{R}_{2}}\to \frac{1}{5}{{R}_{2}}$ ,
$\left[ \left. \begin{matrix}
1 & 1 & 1 & 1 \\
0 & 1 & \frac{2}{5} & 0 \\
0 & -3 & 0 & -3 \\
0 & -1 & 2 & 3 \\
\end{matrix} \right|\begin{matrix}
-1 \\
\frac{-1}{5} \\
12 \\
-4 \\
\end{matrix} \right]$
${{R}_{3}}\to {{R}_{3}}+3{{R}_{2}}$ ,
$\left[ \left. \begin{matrix}
1 & 1 & 1 & 1 \\
0 & 1 & \frac{2}{5} & 0 \\
0 & 0 & \frac{6}{5} & -3 \\
0 & -1 & 2 & 3 \\
\end{matrix} \right|\begin{matrix}
-1 \\
\frac{-1}{5} \\
\frac{57}{5} \\
-4 \\
\end{matrix} \right]$
${{R}_{4}}\to {{R}_{4}}+{{R}_{2}}$ ,
$\left[ \left. \begin{matrix}
1 & 1 & 1 & 1 \\
0 & 1 & \frac{2}{5} & 0 \\
0 & 0 & \frac{6}{5} & -3 \\
0 & 0 & \frac{12}{5} & 3 \\
\end{matrix} \right|\begin{matrix}
-1 \\
\frac{-1}{5} \\
\frac{57}{5} \\
\frac{-21}{5} \\
\end{matrix} \right]$
${{R}_{3}}\to \frac{5}{6}{{R}_{3}}$ ,
$\left[ \left. \begin{matrix}
1 & 1 & 1 & 1 \\
0 & 1 & \frac{2}{5} & 0 \\
0 & 0 & 1 & \frac{-5}{2} \\
0 & 0 & \frac{12}{5} & 3 \\
\end{matrix} \right|\begin{matrix}
-1 \\
\frac{-1}{5} \\
\frac{19}{2} \\
\frac{-21}{5} \\
\end{matrix} \right]$
${{R}_{4}}\to {{R}_{4}}-\frac{12}{5}{{R}_{3}}$ ,
$\left[ \left. \begin{matrix}
1 & 1 & 1 & 1 \\
0 & 1 & \frac{2}{5} & 0 \\
0 & 0 & 1 & \frac{-5}{2} \\
0 & 0 & 0 & 9 \\
\end{matrix} \right|\begin{matrix}
-1 \\
\frac{-1}{5} \\
\frac{19}{2} \\
-27 \\
\end{matrix} \right]$
${{R}_{4}}\to \frac{1}{9}{{R}_{4}}$ ,
$\left[ \left. \begin{matrix}
1 & 1 & 1 & 1 \\
0 & 1 & \frac{2}{5} & 0 \\
0 & 0 & 1 & \frac{-5}{2} \\
0 & 0 & 0 & 1 \\
\end{matrix} \right|\begin{matrix}
-1 \\
\frac{-1}{5} \\
\frac{19}{2} \\
-3 \\
\end{matrix} \right]$
Now, the reduced system of equations is given by:
$x+y+z+w=-1$ …… (I)
$y+\frac{2}{5}z=\frac{-1}{5}$ …… (II)
$z-\frac{5}{2}w=\frac{19}{2}$ …… (III)
$w=-3$ …… (IV)
Substitute the value of $w$ in equation (III),
Thus,
$\begin{align}
& z-\frac{5}{2}\times -3=\frac{19}{2} \\
& z+\frac{15}{2}=\frac{19}{2} \\
& z=\frac{19}{2}-\frac{15}{2} \\
& z=\frac{4}{2}
\end{align}$
After solving it further we get,
Then, $z=2$
Now substitute the values of $z$ in equation (II) to get,
$\begin{align}
& y+\frac{2}{5}\times 2=\frac{-1}{5} \\
& y+\frac{4}{5}=\frac{-1}{5} \\
& y=\frac{-1}{5}-\frac{4}{5} \\
& y=\frac{-5}{5} \\
\end{align}$
After solving it further we get,
Then, $y=-1$
Now substitute the values of $y,z,w$ in equation (I) to find the value of x,
Thus,
$\begin{align}
& x-1+2-3=-1 \\
& x-2=-1 \\
& x=1
\end{align}$
Therefore,
$x=1,y=-1,z=2,w=-3$
Hence, the provided statement makes sense.
