How does a row of zeros make a free variable in linear systems of equations?

Question

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 :(

Answer 1

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.

Answer 2

It's because the row rank is equal to the column rank. If the matrix has more rows than columns, a row of all zeros does not imply a free variable. For example:

$$ \left[\matrix{1 & 2\cr 3 & 4\cr 0 & 0\cr}\right]$$

One way you find free variables is: put the matrix into row-echelon form, then any column that does not contain the first nonzero entry of any row corresponds to a free variable. In your example

$$\left[\matrix{2 & -3 & 5\cr 0 & 5 & 1\cr 0 & 0 & 0\cr}\right]$$ the first nonzero entries of the rows are in columns 1 and 2, so column 3 corresponds to a free variable. However, column 1 or 2 could also correspond to a free variable. In the case of column $2$, just consider changing the order of the columns so $3$ comes before $2$. In the case of column $1$, it's a bit less obvious: you'd have to do some more row operations.

Answer 3

How many $\langle x,y,z\rangle$ satisfy both $x + 5y + 3z = 0$ and $2x-3y+z =0$?

Infinitely many: $x + 5y + 3z = 0$ and $2x-3y+z =0$ are non-parallel planes; as such, it should be obvious that they intersect in a line, and there is no unique solution to this system.

What does this question have to do with $\begin{bmatrix}1&5&3 \\ 2&-3&1 \\ 0&0&0\end{bmatrix}$?

The above is the exact same problem as finding the solutions to $\begin{bmatrix}1&5&3 \\ 2&-3&1 \\ 0&0&0\end{bmatrix}\begin{bmatrix}x\\ y\\ z\end{bmatrix}=\begin{bmatrix}0\\ 0\\ 0\end{bmatrix}.$
The key to realize here is that if a system of equations is going to have finite solutions, you need a certain number of equations. Having a row of zeroes means you have one less equation than necessary to have finite solutions, and hence you have an infinite set of solutions.

It is indeed entirely arbitrary that $x_3$ is the free variable. Remember, we're just choosing variable names at our own convenience - if we let $\mathbf{x}=\begin{bmatrix}x_1\\ x_3 \\x_2\end{bmatrix}$, then $x_2$ would be our free variable. But because it makes sense to name the coordinates of $\mathbf{x}$ in numerical order, it's much easier to remember and to write $\mathbf{x}=\begin{bmatrix}x_1\\ x_2 \\x_3\end{bmatrix}$ instead.