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)

Question

To use Gauss-Jordan elimination to solve the linear system

[latex]x_{1} + x_{2} + x_{3} + x_{4} = 12[/latex]

[latex]x_{1} + 2 x_{2} + 5 x_{4} = 17[/latex]

[latex]3 x_{1} + 2 x_{2} + 4 x_{3} - x_{4} = 31[/latex], (2)

Accepted Answer

we transform its augmented coefficient matrix into reduced echelon form, as follows:

[latex]\left [ \begin{matrix} 1 & 1 & 1 & 1 & 12 \ 1 & 2 & 0 & 5 & 17 \ 3 & 2 & 4 & - 1 & 31 \end{matrix} ight ] \underrightarrow{(- 1) R_{1} + R_{2}}[/latex]

[latex]\left [ \begin{matrix} 1 & 1 & 1 & 1 & 12 \ 0 & 1 & - 1 & 4 & 5\ 3 & 2 & 4 & - 1 & 31 \end{matrix} ight ] [/latex]

[latex]\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} ight ][/latex]

[latex]\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} ight ][/latex]

[latex]\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} ight ][/latex]

Thus the reduced echelon form of the system in (2) is

[latex]x_{1} + 2 x_{3} - 3 x_{4} = 7[/latex]

[latex]x_{2} - x_{3} + 4 x_{4} = 5[/latex]

0 = 0. (3)

The leading variables are [latex]x_{1}[/latex] and [latex]x_{2}[/latex] ; the free variables are [latex]x_{3}[/latex] and [latex]x_{4}[/latex] . If we set

[latex]x_{3} = s and x_{4} = t,[/latex]

then (3) immediately yields

[latex]x_{1} = 7 - 2 s + 3 t,[/latex]

[latex]x_{2} = 5 + s - 4 t.[/latex]

Question 3.3.3: To use Gauss-Jordan elimination to solve the linear system x...

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