How do I choose a free variable?

Question

I have a question regarding the Gaussian method for solving linear equations. I had to solve 2 equations with 3 unknowns and naturally with the elimination process I had 2 variables left. I thought that it didn't matter which variable I chose as a free variable, but apparently after inputting random values after choosing the last variable as the free one, the values didn't match. So is there any method of how to choose a free variable? \begin{cases} & 4x_{1} + 2x_{2} + 3x_{3} = -2 \\ & 2x_{1} + 8x_{2} - x_{3} = -10 \end{cases}

\begin{pmatrix} 4 & 2 & 3 & | & -2 \\ 2 & 8 & -1 & | & -10 \end{pmatrix}

\begin{pmatrix} 2 & 8 & -1 & | & -10 \\ 4 & 2 & 3 & | & -2 \end{pmatrix}

\begin{pmatrix} 2 & 8 & -1 & | & -10 \\ 0 & -14 & 5 & | & 18 \end{pmatrix}

$$ -14x_{2} + 5x_{3} = 18 $$

$$ x_{3} = C $$

$$ x_2 = \frac{18-5C}{-14} $$

$$ 4x_1 = -2 -3C - 2*\frac{18-5C}{-14} $$

$$ 4x_1 = -2 -3C - \frac{18-5C}{-7} $$

$$ x_1 = \frac{-14-21C-18+5C}{28} $$

$$ x_1 = \frac{-16C-32}{28} = \frac{-8C-16}{14} $$

this is what I did, but using x3 as a free variable was not a correct choice, because after inputting a random value, the first equation didn't equal to -2. so I want to find out why do I necessarily have to choose x2 instead of x3.

Answer 1

You did it correctly. You just made a little arithmetic mistake near the end. You got to here $$4x_1 = -2 -3C - \frac{18-5C}{-7}$$ correctly but the line beneath that is not correct. It should actually be $$x_1 = \frac{-14-21C+18-5C}{28}$$ So just be careful of your signs, and keep doing what you're doing because other than that mistake everything else you wrote is correct.

You can verify for yourself that this expression for $x_1$, along with what you gave for $x_2$ and $x_3$, solve both of your equations.

Answer 2

As noted by Bye_World in the comments, in Reduced Row Echelon Form, the matrix specified becomes:

\begin{pmatrix} 1 & 0 & \frac{13}{14} & | & \frac{1}{7} \\ 0 & 1 & -\frac{5}{14} & | & -\frac{9}{7} \end{pmatrix}

You then note that you can use $x_3$ as the free variable. So you get:

$$x_1 = -\frac{13}{14}x_3 + \frac{1}{7}$$ $$x_2 = \frac{5}{14}x_3 + -\frac{9}{7}$$

Hope this helps!

Answer 3

Choosing free variables depends upon the pivot position means pivot columns and pivot rows where leading coefficient 1 is shown in row echelon form,,if we suppose need three pivot position but one is missing ,,like 2 equations with 3 unknown variables ,,so we got for example x1 and x2 but not x3,, so this implies ,that is nonpivot columns so here we are free to choose ..when ever we see nonpivot columns we can choose them as free variables...