Chapter 8 - Section 8.1 - Matrix Solutions to Linear Systems - Exercise Set - Page 895: 60

Yes, the statement makes sense.

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 write the augmented matrix for the given system of equations: Augmented matrix is: $\left[ \left. \begin{matrix} 1 & 1 & 1 & 1 \\ -1 & 4 & 1 & -1 \\ 1 & -2 & 1 & -2 \\ -1 & -2 & 1 & 2 \\ \end{matrix} \right|\begin{matrix} -1 \\ 0 \\ 11 \\ -3 \\ \end{matrix} \right]$ Apply Gauss Jordan Elimination Method. Use row operation to reduce the matrix to row echelon form ${{R}_{2}}\to {{R}_{2}}+{{R}_{1}}$, gives $\left[ \begin{matrix} 1 & 1 & 1 & 1 & -1 \\ 0 & 5 & 2 & 0 & -1 \\ 1 & -2 & 1 & -2 & 11 \\ -1 & -2 & 1 & 2 & -3 \\ \end{matrix} \right]$ ${{R}_{3}}\to {{R}_{3}}-{{R}_{1}}$, gives $\left[ \begin{matrix} 1 & 1 & 1 & 1 & -1 \\ 0 & 5 & 2 & 0 & -1 \\ 0 & -3 & 0 & -3 & 12 \\ -1 & -2 & 1 & 2 & -3 \\ \end{matrix} \right]$ ${{R}_{4}}\to {{R}_{4}}+{{R}_{1}}$, gives $\left[ \begin{matrix} 1 & 1 & 1 & 1 & -1 \\ 0 & 5 & 2 & 0 & -1 \\ 0 & -3 & 0 & -3 & 12 \\ 0 & -1 & 2 & 3 & -4 \\ \end{matrix} \right]$ ${{R}_{2}}\to \frac{1}{5}{{R}_{2}}$, gives $\left[ \begin{matrix} 1 & 1 & 1 & 1 & -1 \\ 0 & 1 & \frac{2}{5} & 0 & -\frac{1}{5} \\ 0 & -3 & 0 & -3 & 12 \\ 0 & -1 & 2 & 3 & -4 \\ \end{matrix} \right]$ ${{R}_{3}}\to {{R}_{3}}+3{{R}_{2}}$, gives $\left[ \begin{matrix} 1 & 1 & 1 & 1 & -1 \\ 0 & 1 & \frac{2}{5} & 0 & -\frac{1}{5} \\ 0 & 0 & \frac{6}{5} & -3 & \frac{57}{5} \\ 0 & -1 & 2 & 3 & -4 \\ \end{matrix} \right]$ ${{R}_{4}}\to {{R}_{4}}+{{R}_{2}}$, gives $\left[ \begin{matrix} 1 & 1 & 1 & 1 & -1 \\ 0 & 1 & \frac{2}{5} & 0 & -\frac{1}{5} \\ 0 & 0 & \frac{6}{5} & -3 & \frac{57}{5} \\ 0 & 0 & \frac{12}{5} & 3 & -\frac{21}{5} \\ \end{matrix} \right]$ ${{R}_{3}}\to \frac{5}{6}{{R}_{3}}$, gives $\left[ \begin{matrix} 1 & 1 & 1 & 1 & -1 \\ 0 & 1 & \frac{2}{5} & 0 & -\frac{1}{5} \\ 0 & 0 & 1 & -\frac{5}{2} & \frac{19}{2} \\ 0 & 0 & \frac{12}{5} & 3 & -\frac{21}{5} \\ \end{matrix} \right]$ ${{R}_{4}}\to {{R}_{4}}-\frac{12}{5}{{R}_{3}}$, gives $\left[ \begin{matrix} 1 & 1 & 1 & 1 & -1 \\ 0 & 1 & \frac{2}{5} & 0 & -\frac{1}{5} \\ 0 & 0 & 1 & -\frac{5}{2} & \frac{19}{2} \\ 0 & 0 & 0 & 9 & -27 \\ \end{matrix} \right]$ ${{R}_{4}}\to \frac{1}{9}{{R}_{4}}$, gives $\left[ \begin{matrix} 1 & 1 & 1 & 1 & -1 \\ 0 & 1 & \frac{2}{5} & 0 & -\frac{1}{5} \\ 0 & 0 & 1 & -\frac{5}{2} & \frac{19}{2} \\ 0 & 0 & 0 & 1 & -3 \\ \end{matrix} \right]$ ${{R}_{3}}\to {{R}_{3}}+\frac{5}{2}{{R}_{4}}$, gives $\left[ \begin{matrix} 1 & 1 & 1 & 1 & -1 \\ 0 & 1 & \frac{2}{5} & 0 & -\frac{1}{5} \\ 0 & 0 & 1 & 0 & 2 \\ 0 & 0 & 0 & 1 & -3 \\ \end{matrix} \right]$ ${{R}_{1}}\to {{R}_{1}}-{{R}_{4}}$, gives $\left[ \begin{matrix} 1 & 1 & 1 & 0 & 2 \\ 0 & 1 & \frac{2}{5} & 0 & -\frac{1}{5} \\ 0 & 0 & 1 & 0 & 2 \\ 0 & 0 & 0 & 1 & -3 \\ \end{matrix} \right]$ ${{R}_{2}}\to {{R}_{2}}-\frac{2}{5}{{R}_{1}}$, gives $\left[ \begin{matrix} 1 & 1 & 1 & 0 & 2 \\ 0 & 1 & 0 & 0 & -1 \\ 0 & 0 & 1 & 0 & 2 \\ 0 & 0 & 0 & 1 & -3 \\ \end{matrix} \right]$ ${{R}_{1}}\to {{R}_{1}}-{{R}_{3}}$, gives $\left[ \begin{matrix} 1 & 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & -1 \\ 0 & 0 & 1 & 0 & 2 \\ 0 & 0 & 0 & 1 & -3 \\ \end{matrix} \right]$ ${{R}_{1}}\to {{R}_{1}}-{{R}_{2}}$, gives $\left[ \begin{matrix} 1 & 0 & 0 & 0 & 1 \\ 0 & 1 & 0 & 0 & -1 \\ 0 & 0 & 1 & 0 & 2 \\ 0 & 0 & 0 & 1 & -3 \\ \end{matrix} \right]$ Thus, the solution is: $x=1,y=-1,z=2,w=-3$ Hence, the provided statement makes sense.