Chapter 8 - Section 8.1 - Matrix Solutions to Linear Systems - Exercise Set - Page 894: 44

The solution set of the given system of equations is $\left( {{e}^{1}},{{e}^{-1}},{{e}^{2}}\text{ and }{{e}^{-3}} \right)$.

The solution set of the given system of equations is $\left( {{e}^{1}},{{e}^{-1}},{{e}^{2}}\text{ and }{{e}^{-3}} \right)$. It is provided that, $\ln w=A,\ln x=B,\ln y=C\text{ and }\ln z=D$ Now, the resulting system of equations is: $\begin{align} & A+B+C+D=-1 \\ & -A+4B+C-D=0 \\ & A-2B+C-2D=11 \\ & -A-2B+C+2D=-3 \end{align}$ First write the augmented matrix for the given system of equations. Therefore, the 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, reduce the matrix to row echelon form by using the row operation First apply ${{R}_{2}}\to {{R}_{2}}+{{R}_{1}}$, to get $\left[ \left. \begin{matrix} 1 & 1 & 1 & 1 \\ 0 & 5 & 2 & 0 \\ 1 & -2 & 1 & -2 \\ -1 & -2 & 1 & 2 \\ \end{matrix} \right|\begin{matrix} -1 \\ -1 \\ 11 \\ -3 \\ \end{matrix} \right]$ ${{R}_{3}}\to {{R}_{3}}-{{R}_{1}}$, to get $\left[ \left. \begin{matrix} 1 & 1 & 1 & 1 \\ 0 & 5 & 2 & 0 \\ 0 & -3 & 0 & -3 \\ -1 & -2 & 1 & 2 \\ \end{matrix} \right|\begin{matrix} -1 \\ -1 \\ 12 \\ -3 \\ \end{matrix} \right]$ ${{R}_{4}}\to {{R}_{4}}+{{R}_{1}}$, to get $\left[ \left. \begin{matrix} 1 & 1 & 1 & 1 \\ 0 & 5 & 2 & 0 \\ 0 & -3 & 0 & -3 \\ 0 & -1 & 2 & 3 \\ \end{matrix} \right|\begin{matrix} -1 \\ -1 \\ 12 \\ -4 \\ \end{matrix} \right]$ ${{R}_{2}}\to \frac{1}{5}{{R}_{2}}$, to get $\left[ \left. \begin{matrix} 1 & 1 & 1 & 1 \\ 0 & 1 & \frac{2}{5} & 0 \\ 0 & -3 & 0 & -3 \\ 0 & -1 & 2 & 3 \\ \end{matrix} \right|\begin{matrix} -1 \\ \frac{-1}{5} \\ 12 \\ -4 \\ \end{matrix} \right]$ ${{R}_{3}}\to {{R}_{3}}+3{{R}_{2}}$, to get $\left[ \left. \begin{matrix} 1 & 1 & 1 & 1 \\ 0 & 1 & \frac{2}{5} & 0 \\ 0 & 0 & \frac{6}{5} & -3 \\ 0 & -1 & 2 & 3 \\ \end{matrix} \right|\begin{matrix} -1 \\ \frac{-1}{5} \\ \frac{57}{5} \\ -4 \\ \end{matrix} \right]$ ${{R}_{4}}\to {{R}_{4}}+{{R}_{2}}$, to get $\left[ \left. \begin{matrix} 1 & 1 & 1 & 1 \\ 0 & 1 & \frac{2}{5} & 0 \\ 0 & 0 & \frac{6}{5} & -3 \\ 0 & 0 & \frac{12}{5} & 3 \\ \end{matrix} \right|\begin{matrix} -1 \\ \frac{-1}{5} \\ \frac{57}{5} \\ \frac{-21}{5} \\ \end{matrix} \right]$ ${{R}_{3}}\to \frac{5}{6}{{R}_{3}}$, to get $\left[ \left. \begin{matrix} 1 & 1 & 1 & 1 \\ 0 & 1 & \frac{2}{5} & 0 \\ 0 & 0 & 1 & \frac{-5}{2} \\ 0 & 0 & \frac{12}{5} & 3 \\ \end{matrix} \right|\begin{matrix} -1 \\ \frac{-1}{5} \\ \frac{19}{2} \\ \frac{-21}{5} \\ \end{matrix} \right]$ ${{R}_{4}}\to {{R}_{4}}-\frac{12}{5}{{R}_{3}}$, to get $\left[ \left. \begin{matrix} 1 & 1 & 1 & 1 \\ 0 & 1 & \frac{2}{5} & 0 \\ 0 & 0 & 1 & \frac{-5}{2} \\ 0 & 0 & 0 & 9 \\ \end{matrix} \right|\begin{matrix} -1 \\ \frac{-1}{5} \\ \frac{19}{2} \\ -27 \\ \end{matrix} \right]$ ${{R}_{4}}\to \frac{1}{9}{{R}_{4}}$, to get $\left[ \left. \begin{matrix} 1 & 1 & 1 & 1 \\ 0 & 1 & \frac{2}{5} & 0 \\ 0 & 0 & 1 & \frac{-5}{2} \\ 0 & 0 & 0 & 1 \\ \end{matrix} \right|\begin{matrix} -1 \\ \frac{-1}{5} \\ \frac{19}{2} \\ -3 \\ \end{matrix} \right]$ ${{R}_{3}}\to {{R}_{3}}+\frac{5}{2}{{R}_{4}}$, to get $\left[ \left. \begin{matrix} 1 & 1 & 1 & 1 \\ 0 & 1 & \frac{2}{5} & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{matrix} \right|\begin{matrix} -1 \\ \frac{-1}{5} \\ 2 \\ -3 \\ \end{matrix} \right]$ ${{R}_{1}}\to {{R}_{1}}-{{R}_{4}}$, to get $\left[ \left. \begin{matrix} 1 & 1 & 1 & 0 \\ 0 & 1 & \frac{2}{5} & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{matrix} \right|\begin{matrix} 2 \\ \frac{-1}{5} \\ 2 \\ -3 \\ \end{matrix} \right]$ ${{R}_{2}}\to {{R}_{2}}-\frac{2}{5}{{R}_{1}}$, to get $\left[ \left. \begin{matrix} 1 & 1 & 1 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{matrix} \right|\begin{matrix} 2 \\ -1 \\ 2 \\ -3 \\ \end{matrix} \right]$ ${{R}_{1}}\to {{R}_{1}}-{{R}_{3}}$, to get $\left[ \left. \begin{matrix} 1 & 1 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{matrix} \right|\begin{matrix} 0 \\ -1 \\ 2 \\ -3 \\ \end{matrix} \right]$ ${{R}_{1}}\to {{R}_{1}}-{{R}_{2}}$, to get $\left[ \left. \begin{matrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{matrix} \right|\begin{matrix} 1 \\ -1 \\ 2 \\ -3 \\ \end{matrix} \right]$ By the Gauss Jordan Elimination Method, $A=1,B=-1,C=2,D=-3$ Now, $\begin{align} & \ln w=A \\ & w={{e}^{A}} \\ & w={{e}^{1}} \end{align}$ And $\begin{align} & \ln x=B \\ & x={{e}^{B}} \\ & x={{e}^{-1}} \end{align}$ And $\begin{align} & \ln y=C \\ & y={{e}^{C}} \\ & y={{e}^{2}} \end{align}$ And $\begin{align} & \ln z=D \\ & z={{e}^{D}} \\ & z={{e}^{-3}} \end{align}$ Hence, the solution set of the given system of equations is $\left( {{e}^{1}},{{e}^{-1}},{{e}^{2}}\text{ and }{{e}^{-3}} \right)$.