Question 3.3.3: To use Gauss-Jordan elimination to solve the linear system x...
To use Gauss-Jordan elimination to solve the linear system
x_{1} + x_{2} + x_{3} + x_{4} = 12
x_{1} + 2 x_{2} + 5 x_{4} = 17
3 x_{1} + 2 x_{2} + 4 x_{3} – x_{4} = 31, (2)
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we transform its augmented coefficient matrix into reduced echelon form, as follows:
\left [ \begin{matrix} 1 & 1 & 1 & 1 & 12 \\ 1 & 2 & 0 & 5 & 17 \\ 3 & 2 & 4 & – 1 & 31 \end{matrix} \right ] \underrightarrow{(- 1) R_{1} + R_{2}} \left [ \begin{matrix} 1 & 1 & 1 & 1 & 12 \\ 0 & 1 & – 1 & 4 & 5\\ 3 & 2 & 4 & – 1 & 31 \end{matrix} \right ]\underrightarrow{(- 3) R_{1} + R_{3}} \left [ \begin{matrix} 1 & 1 & 1 & 1 & 12 \\ 0 & 1 & – 1 & 4 & 5 \\ 0 & – 1 & 1 & – 4 & – 5 \end{matrix} \right ]
\underrightarrow{(1) R_{2} + R_{3}} \left [ \begin{matrix} 1 & 1 & 1 & 1 & 12 \\ 0 & 1 & – 1 & 4 & 5 \\ 0 & 0 & 0 & 0 & 0 \end{matrix} \right ]
\underrightarrow{(- 1) R_{2} + R_{1}} \left [ \begin{matrix} 1 & 0 & 2 & – 3 & 7 \\ 0 & 1 & – 1 & 4 & 5 \\ 0 & 0 & 0 & 0 & 0 \end{matrix} \right ]
Thus the reduced echelon form of the system in (2) is
x_{1} + 2 x_{3} – 3 x_{4} = 7
x_{2} – x_{3} + 4 x_{4} = 5
0 = 0. (3)
The leading variables are x_{1} and x_{2} ; the free variables are x_{3} and x_{4} . If we set
x_{3} = s and x_{4} = t,
then (3) immediately yields
x_{1} = 7 – 2 s + 3 t,
x_{2} = 5 + s – 4 t.
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