I don't understand how a row of zeros gives a free variable when solving systems of linear equations. Here's an example matrix and let us say that we're trying to solve Ax=0:
$$\left[
\begin{matrix}
2 & -3 & 0 \\
3 & 5 & 0 \\
0 & 1 & 0 \\
\end{matrix}
\right]$$
This makes sense to me – you have a COLUMN of numbers corresponding to the number of times the variable $x_3$ is used for each equation and since they are all zeros, $x_3$ could have been any real number because we never get to manipulate our equations to determine a value for it. Hence, it is a free variable as it is not subject to the constraints of the equations.
$$\left[
\begin{matrix}
2 & -3 & 5 \\
0 & 5 & 1 \\
0 & 0 & 0 \\
\end{matrix}
\right]$$
Now for this second example, $x_3$ would still be a free variable. Why is this so? $x_3$ is being used in the other two equations where you could certainly come up with a finite set of answers for this variable rather than saying "It could've been anything!" right?
Also is it entirely arbitrary that $x_3$ is the free variable or could it be decided that $x_1$ or $x_2$ is free instead?
Could someone explain to me in a more layman or simplified form on why a row of zeros magically makes a free variable?
Please help me 🙁
Best Answer
Fewer equations than unknowns means you cannot solve the set in a unique way. If you have three variables and in effect two equations as you have, then any one of the three variables can be seen as free. However, once you choose a value for it, the value of the two other variables will be forced.
Studying interactions like this in general polynomial equations (i.e. the variables appear with natural number exponents, no roots or logarithms or anything) is the field of algebraic geometry. When working with matrices, row reducing and finding inverses and stuff, all of the variables only have the exponent $1$, and therefore it's called linear algebra.