Gaussian Elimination Four Variables Lesson

Over the course of the last two lessons, we have learned how to solve a linear system in two and three variables using Gaussian Elimination and Gauss-Jordan Elimination. In this lesson, we will look at an example of how to solve a linear system in four variables using Gauss-Jordan Elimination.

Gauss-Jordan Four-Variable System

• Obtain a 1 as the first element in the first column
• Use the first row to transform the remaining elements in the first column into zeros
• Obtain a 1 as the second element in the second column
• Use the second row to transform the remaining elements in the second column into zeros
• Obtain a 1 as the third element in the third column
• Use the third row to transform the remaining elements in the third column into zeros
• Obtain a 1 as the fourth element in the fourth column
• Use the fourth row to transform the remaining elements in the fourth column into zeros

Example #1: Solve each linear system. $$2x_1+x_2 -x_3 + 2x_4=-6$$ $$3x_1+4x_2 + x_4=1$$ $$x_1 + 5x_2 + 2x_3 + 6x_4=-3$$ $$5x_1 + 2x_2 - x_3 - x_4=3$$ Let's set up our augmented matrix: $$\left[ \begin{array}{cccc|c}2&1&-1&2&-6\\ 3&4&0&1&1\\ 1&5&2&6&-3 \\5&2&-1&-1&3 \end{array}\right] $$ Our main goal is to get 1's down the main diagonal, with zeros above and below. Our end result: $$\left[ \begin{array}{cccc|c}1&0&0&0&1\\ 0&1&0&0&0\\ 0&0&1&0&4 \\0&0&0&1&-2 \end{array}\right] $$ $$(1, 0, 4, -2)$$

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