Chapter 8 - Review Exercises - Page 949: 3

The solution to the system of equations is $1,3,-4$.

The matrix corresponding to the given system of equations is as follows: $\left[ \begin{array}{*{35}{l}} 1 & 2 & 3 & -5 \\ 2 & 1 & 1 & 1 \\ 1 & 1 & -1 & 8 \\ \end{array} \right]$ Using the elementary row transformation we will find the echelon form of the matrix. In the first step, we will make zeros in column 1 (except for the entry at row 1 and column 1). Multiply row 1 by 2 and subtract it from row 2, ${{R}_{2}}={{R}_{2}}-2{{R}_{1}}$, $\left[ \begin{array}{*{35}{l}} 1 & 2 & 3 & -5 \\ 0 & -3 & -5 & 11 \\ 1 & 1 & -1 & 8 \\ \end{array} \right]$ Subtract row 1 from row 3, ${{R}_{3}}={{R}_{3}}-{{R}_{1}}$ $\left[ \begin{array}{*{35}{l}} 1 & 2 & 3 & -5 \\ 0 & -3 & -5 & 11 \\ 0 & -1 & -4 & 13 \\ \end{array} \right]$ The second step is to work on the entry at row 2 and column 2 and make it 1 by performing row division ${{R}_{2}}=\frac{{{R}_{2}}}{-3}$, $\left[ \begin{array}{*{35}{l}} 1 & 2 & 3 & -5 \\ 0 & 1 & \frac{5}{3} & \frac{-11}{3} \\ 0 & -1 & -4 & 13 \\ \end{array} \right]$ Then make all other entries in this column zero; multiply 2 in row 2 and subtract row 1 from row 2 ${{R}_{1}}={{R}_{1}}-2{{R}_{2}}$, $\left[ \begin{array}{*{35}{l}} 1 & 0 & \frac{-1}{3} & \frac{7}{3} \\ 0 & 1 & \frac{5}{3} & \frac{-11}{3} \\ 0 & -1 & -4 & 13 \\ \end{array} \right]$ Add row 2 to row 3 ${{R}_{3}}={{R}_{3}}+{{R}_{2}}$, $\left[ \begin{array}{*{35}{l}} 1 & 0 & \frac{-1}{3} & \frac{7}{3} \\ 0 & 1 & \frac{5}{3} & \frac{-11}{3} \\ 0 & 0 & \frac{-7}{3} & \frac{28}{3} \\ \end{array} \right]$ The third step is to work on row 3 and column 3 and make it 1. Divide row 3 by $\frac{-7}{3}$ ${{R}_{3}}=\frac{-3}{7}{{R}_{3}}$, $\left[ \begin{array}{*{35}{l}} 1 & 0 & \frac{-1}{3} & \frac{7}{3} \\ 0 & 1 & \frac{5}{3} & \frac{-11}{3} \\ 0 & 0 & 1 & -4 \\ \end{array} \right]$ Multiply row 3 by $\frac{1}{3}$ and add it to row 1 ${{R}_{1}}={{R}_{1}}+\frac{1}{3}{{R}_{3}}$, $\left[ \begin{array}{*{35}{l}} 1 & 0 & 0 & 1 \\ 0 & 1 & \frac{5}{3} & \frac{-11}{3} \\ 0 & 0 & 1 & -4 \\ \end{array} \right]$ Multiply row 3 by $\frac{5}{3}$ and subtract it from row 2 ${{R}_{2}}={{R}_{2}}-\frac{5}{3}{{R}_{3}}$, $\left[ \begin{array}{*{35}{l}} 1 & 0 & 0 & 1 \\ 0 & 1 & 0 & 3 \\ 0 & 0 & 1 & -4 \\ \end{array} \right]$ The above matrix represents, $\begin{align} & 1\cdot x+0\cdot y+0\cdot z=1 \\ & 0\cdot x+1\cdot y+0\cdot z=3 \\ & 0\cdot x+0\cdot y+1\cdot z=-4 \end{align}$ Hence, the solution to the system is $\left[ \left( 1,3,-4 \right) \right]$.