Question 3.2.3: To solve the system x1 + 2x2 + x3 = 4 3x1 + 8 x2 + 7 x3 = 20...
To solve the system
x_{1} + 2 x_{2} + x_{3} = 4
3 x_{1} + 8 x_{2} + 7 x_{3} = 20
2 x_{1} + 7 x_{2} + 9 x_{3} = 23 (11)
whose augmented coefficient matrix is exhibited in (3), we carry out the following sequence of elementary row operations, corresponding to the steps in the solution of Example 6 in Section 3.1:
\left [ \begin{matrix} 1 & 2 & 1 & 4 \\ 3 & 8 & 7 & 20 \\ 2 & 7 & 9 & 23 \end{matrix} \right ] (12)
\underrightarrow{(- 3) R_{1} + R_{2}} \left [ \begin{matrix} 1 & 2 & 1 & 4 \\ 0 & 2 & 4 & 8 \\ 2 & 7 & 9 & 23 \end{matrix} \right ]
\underrightarrow{(- 2) R_{1} + R_{3}} \left [ \begin{matrix} 1 & 2 & 1 & 4 \\ 0 & 2 & 4 & 8 \\ 0 & 3 & 7 & 15 \end{matrix} \right ]
\underrightarrow{(\frac{1}{2}) R_{2}} \left [ \begin{matrix} 1 & 2 & 1 & 4 \\ 0 & 1 & 2 & 4 \\ 0 & 3 & 7 & 15 \end{matrix} \right ]
\underrightarrow{(- 3) R_{2} + R_{3}} \left [ \begin{matrix} 1 & 2 & 1 & 4 \\ 0 & 1 & 2 & 4 \\ 0 & 0 & 1 & 3 \end{matrix} \right ] (13)
The final matrix here is the augmented coefficient matrix of the system
x_{1} + 2 x_{2} + x_{3} = 4
x_{2} + 2 x_{3} = 4
x_{3} = 3. (14)
whose unique solution (readily found by back substitution) is x_{1} = 5, x_{2} = – 2, x_{3} = 3.