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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 :(

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A row of zeroes (or, in general, the matrix's row rank being less than the number of columns) means that there are not enough equations to completely determine all the variables. In this case, you can pick any of the three variables as a free variable, and determine the other two in terms of it. Sorry if this isn't clear; maybe someone will post a more clear answer. – ShreevatsaR Jun 8 '14 at 5:27
The row of zeros itself does nothing: tack a row of zeros to any linear system, and you'll get an equivalent new system (same number of unknowns, same solutions). Therefore the row of zeros is only significant if the number of equations is given; in your question it seems as if you might only be considering systems with as many equations as unknowns (why would you restrict to that?), but if that is the case you should state so clearly. – Marc van Leeuwen Jun 8 '14 at 6:47
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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.

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I understand now! I should have remembered that about systems of equations: fewer equations than unknowns means it can't be uniquely solved. So we essentially assign some value A to one of the variables which then forces answers for the other two variables under the constraint that x3 equals A. Thanks a lot, now I've been able to determine and properly write the eigenvectors for a matrix without being stumped at what to do with rows of zeros. – Ursa Major Jun 8 '14 at 6:26

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.

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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.

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Ahh okay, so even with our variable matrix being: $$\left[ \begin{matrix} x \\ y \\ z \\ \end{matrix} \right]$$ Assuming there was a single row of zeros, regardless of row number, we could pick any one variable to be free regardless of position. – Ursa Major Jun 8 '14 at 6:36
Pretty much. It's just much simpler to conceptualize (and/or visualize) if we follow certain conventions to choose our free variables (and infinitely easier to communicate them as well!). – Pockets Jun 8 '14 at 6:43

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