Augmented matrix gives no solutions when augmented with variables, but will give solutions when augmented with some numbers

Question

I have a homework problem that boils down to this:

1. x1 = 1a + 1b
2. x2 = 1b + 1c
3. x3 = 1c + 1d
4. x4 = 1d + 1e
5. x5 = 1e + 1f
6. x6 = 1f + 1a

I write the matrix to solve for x.

I row-reduce the matrix.

The bottom row says a bunch of zeroes sum to a nonzero number, which is false. AFAIK, this means the system of equations has no solutions.

However, if I plug in actual values for x1-x6 (i.e. they all equal 1), solutions do exist.

AFAIK, this means that there are infinite solutions. How come I can't arrive to this conclusion by using variables in the original matrix?

EDIT I got the above answers using Wolfram Alpha

Answer 1

The system is overdetermined, in that you can use the equations to show that you must have $$x_1-x_2+x_3-x_4+x_5 = x_6. $$ (The point of this is that the coefficient matrix, $A$, for $a,b,c,d,e,f$ is singular, so that the above is required for the equation to have any solution. On the other hand, there is a vector $v$ so that $A \cdot v = 0$, which you can add onto any solution you can find while still satisfying the equations, so any solution is not unique.)

Wolfram|Alpha has probably assumed that this is not the case when it ran its row-reduction algorithm, and thus produced a result inconsistent with what you expect. If you try putting numbers in that do not satisfy the equality above, you should find a similar result.

Answer 2

You're not saying how you're doing row reduction, but if you're using mathematical software (like Maple), you can get answers like this.

If you enter your augmented matrix into Maple (version 17) and ask for the RREF, Maple will implicitly assume that any expression is nonzero, unless it's the number zero. So $\left[\begin{matrix}1&1&a\\ 1&1&b\end{matrix}\right]$ reduces to $\left[\begin{matrix}1&1&0\\ 0&0&1\end{matrix}\right]$, because it doesn't know that $b-a$ can sometimes be zero.

The command GaussianElimination(A, 'method'='FractionFree') gets it right, even though it only gives a row echelon form: $\left[\begin{matrix}1&1&a\\ 0&0&b-a\end{matrix}\right]$; it implies that there is a solution if $b-a=0$.

(NOTE: The OP mentioned WolframAlpha while I was writing this up. It's still the same basic principle, though.)