In practice reducing matrices to row echelon form, I've encountered scenarios in which I have a row like row 3 in the following matrix:

$$ \begin{bmatrix} 1 & -3 & 3\\ 0 & 1 & 3 \\ 0 & 0 & 3\\ \end{bmatrix} $$

This matrix consists of input $x$ and $y$, with the third column as the solution vector.

When trying to convert matrices to echelon form (which this is now in), I may sometimes have a row like the third. It is a funny thing, as when I try to convert the matrix to reduced echelon form, and performing the row operation, say:

$$Upon \ R2 \Rightarrow R2-R3$$

The matrix becomes

$$ \begin{bmatrix} 1 & -3 & 3\\ 0 & 1 & 0 \\ 0 & 0 & 3\\ \end{bmatrix} $$

And then, for row 1 (to put zeros above the third pivot):

$$Upon \ R1 \Rightarrow R1-R3$$

$$ \begin{bmatrix} 1 & -3 & 0\\ 0 & 1 & 0 \\ 0 & 0 & 3\\ \end{bmatrix} $$

And then to scale the third pivot to $1$ I scale the matrix $R3$ to $(1/3)R3$. This feels like cheating, allow I'm merely preforming row operations. I literally could've just drawn zeros above the third pivot and changing the $3$ to a $1$ instead of going through all that hassle. In addition, I haven't a clue what that row would mean in terms of its row picture. $0x + 0y = 3$? That seems preposterous. Is me arriving to such a row in my operations a mistake? And if not, what gives with this?

To recap, my question is this: is the condition where the third row looks like way it does still allow for its matrix to have an existing reduced echelon form for it, and why does the fact that the matrix has no solution looking the way it does explain why row operations on it to put zeros above the third pivot are so trivial?

  • $\begingroup$ If your calculations are correct, then the only interpretation is that the system doesn't have a solution, because there is no x and y that satisfy $0x+0y=3$ $\endgroup$ Sep 18 '17 at 19:42
  • $\begingroup$ This still makes it reducible, though, right? And how does that lend to explain how putting zeros above the third pivot is so trivial? $\endgroup$
    – sangstar
    Sep 18 '17 at 19:43
  • $\begingroup$ What's the original problem? I'm not sure I understand what your question is, maybe edit your original post with the original problem. $\endgroup$ Sep 18 '17 at 19:47
  • $\begingroup$ @MatheusFernandes Sure, I edited it. $\endgroup$
    – sangstar
    Sep 18 '17 at 19:52

I think I understand your question now.

Let's tackle this problem piece by piece, ok? First of all, you have a system of 2 variables $x$ and $y$, but you have 3 equations in the system.

Surely you remember that you only need two equations to reach a solution for two variables, so what would be the consequence of there being three equations, not two?

One consequence would be that the third line is unnecessary to solve the problem, that it's a multiple of the other lines, a linear combination of the other two lines, like this:

$$x+y=4$$ $$2x-3y=-7$$ $$3x-2y=-3$$

You see, adding the first two equations gives you the third equation.

If you applied the Guass-Jordan method to that system, you'd get the following matrix:

\begin{bmatrix} 1 & 0 & 1\\ 0 & 1 & 3\\ 0 & 0 & 0\\ \end{bmatrix}

Meaning, quite rightly, that $x=1$, $y=3$ and that $0x+0y=0$, which is not really that useful to us. Now, if the system only had two equations, not three, and in the end you had $0x+0y=0$, that could only mean that there infinite solutions to this problem. You see, there would be an infinite combinations of $x$ and $y$ that fit that solution.

In the example I gave you, however, there is only one solution, and the extra line is just that: extra.

In your problem, though, you seem to be confused in two parts. One is that you could just "replace" the number above the 3 in the third row with zeros. That's not really what you're doing, you are actually multiplying the third row by -1 and adding it to other rows. The thing is, the other numbers are 0, so they don't actually change anything, which I think is source of your confusion.

The second part is that pesky $0x+0y=3$. The matrix is indeed correct, that line is fine as it is, you don't have to change it. What it means here is something impossible, how can a number times 0 plus another number times 0 equal 3? The answer is, of course, that there are no numbers that can satisfy that equation, and the only conclusion is that the system has no solutions.

That's the reason people use the Gauss-Jordan method, it lets you know if a system has no solutions, one solution, or an infinite number of solutions.

  • $\begingroup$ I see. So I've discovered the system has no solution. I still could turn it into reduced row echelon form, though right? It just seems to be very easy to create $0$'s above that pivot in the quaint third row. $\endgroup$
    – sangstar
    Sep 19 '17 at 19:18
  • $\begingroup$ It's just what the method is, the only reason it's easy is because the other numbers are all $0$. Notice that that's what happens naturally whenever you use the Gauss-Jordan method, starting top to bottom, left to right, you never interfere with what you've already done, because of the zeroes. $\endgroup$ Sep 20 '17 at 6:44

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