ymar

@MattN Hi. I just read on Area51 that you've started to learn Polish. If you need help with that, I'd be happy to try. I think I should still get an email if you ping me here.

I've accepted your answer :)

Okay, now it just asks me if I want to save the changes, and when I do, the box remains in the state I've put it in. It didn't do that before the code change for some reason, and it always unticked itself. But maybe I don't have to understand why that happened. :) Thank you very much!

I didn't want to stay ticked.

The macro works now! My only question is whys that box was ticked after startup. That was my problem all that time earlier...

Okay, thank you. I'm sorry I'm at work, so I can't talk continuously, but I can keep this window open and post messages from time to time.

Will that numerical value be the same each time I run the macro? It's important for the macro to work wach time without midifications.

Okay, thank you. Now after startup (before running the macro), the box "Microsoft Scripting Runtime" is ticked. Is this what should happen?

I'm getting a "75 - Path/File access error" after replacing the lines...

The program stops after I press F8 with " MkDir (cTmpFld)" highlighted

@nixda Now there's no error message at all, and nothing seems to happen. Should I list the lines that actually do get highlighted? (Not all of them do.)

@nixda The error message appears after I press F8 with this line highlighted: "MsgBox Err.Number & " - " & Err.Description".

@nixda The solution in the second link is to a different problem. Rules alone are not enough to print attachments automatically. They only allow to print messages.

@nixda As for the second question, yes a couple of times, mostly in this version, and nothing seemed to happen. But I didn't add any rules here -- I just ran the script as Diane Poremsky advises in her post.

OK, I think I'll go or I'll yawn my guts out. I'm really tired. See you some time later!

I don't really remember. :) I remember something was definitely wrong with it.

Just a couple of algorithms, and you had to pretend you're a computer and the piece of paper is a screen.

Of course. But then you don't actually have to do it. You just think of it. But the course was closer to accountancy than to mathematics. Matrix after matrix, eliminate this, augment that...

I started to hate Gaussian elimination when I took a linear programming course. It was so monotonous I couldn't bear it.

Oh, that's really weird.

Oh, so you had no LA in your first semester at university?

I don't understand. You are in your second year, right?

How many semesters of LA do you have to do?

Yes.

Yes, that's how I read the proof, but it really doesn't matter at all.

Fair enough.

May I ask why you deleted your m.se account, kan?

Then yes, I think you're right. I still think he structures the proof differently, but it doesn't make any difference. That's it.

I don't know what an erring row is.

Yes. :) If these two equations are equivalent, then r'=s', which is obvious.

OK.

no. These are two equivalent systems.

...that is simply x=r' and x=s'. These are equivalent only if r'=s'.

You can just as well omit them. But then you get $I_nx=r'$ and $I_nx=s'$.

They translate to 0=0 as equations.

Do you see that the zero rows don't matter?

No! We had a much more complicated situation then. Now the equations are as simple as they get.

@kan this

But this is clear, isn't it?

That if the two systems are equivalent, then r'=s'.

@kan we still need to show what you objected to originally.

@kan The outcome is always two new equivalent matrices.

@egreg He says this: R'= something or something else. The something else is a matrix in which the last column is dominant. But the last column in R' is by definition the first column in R in which R differs from S.

Anyway, I think we agree now that the systems of equations represented by R' and S' are equivalent.

@egreg yes, that's true. I think it's a bit of a moot point. He still considers the case where the first column in which they differ is dominant (these are the matrices to the right), but it's clearly impossible (and so he says).

(Only he doesn't say he doesn't remove any dominant columns -- a priori, he may remove some that are to the right of the first column in which the two matrices differ.)

Well that's true.

In general, if I have any two equivalent matrices A and B, in a reduced echelon form or not, if I delete the same columns in A and B, then the resulting matrices will still be equivalent.

This is what I was saying.

Yes, but this is not the problem. [1 2 2] is equivalent to [2 4 4]. I have to delete the same columns in the first and in the second. If I delete the first ones, I'll get [2 2] and [4 4] which are still equivalent.