# System of linear equations - types of solutions.

• August 26th 2010, 06:48 AM
System of linear equations - types of solutions.
http://i33.tinypic.com/28qsntd.jpg

Will put up what i have done in a couple of mins
• August 26th 2010, 06:58 AM
• August 26th 2010, 08:09 AM
Have i done the echelon system reduction correct?
• August 26th 2010, 09:00 AM
Ackbeet

$\left[\begin{matrix}1 &2\alpha &1\\ 0 &1 &\alpha+1\\ 0 &\alpha-3 &\alpha-5\end{matrix}\left|\begin{matrix}1\\ 2\\ 3\alpha-4\end{matrix}\right].$

What row operation did you do next?
• August 26th 2010, 09:39 AM
Quote:

Originally Posted by Ackbeet

$\left[\begin{matrix}1 &2\alpha &1\\ 0 &1 &\alpha+1\\ 0 &\alpha-3 &\alpha-5\end{matrix}\left|\begin{matrix}1\\ 2\\ 3\alpha-4\end{matrix}\right].$

What row operation did you do next?

3rd - (2nd x ( $\alpha$ -3))
• August 26th 2010, 09:49 AM
Ackbeet

1. Your fourth line doesn't look like that at all.
2. I don't think that's the right ERO for this stage in the game. I would do 3rd - (2nd x ( $\alpha$-3)).

Come to think of it, that's what your fourth line looks like. I got your numeral 1's mixed up with your parentheses (maybe you should make your parentheses more curved, so they don't get confused with 1's). You actually did what I suggested in my second comment.

[EDIT]: I think you edited your post. Pick up my comments from here on down.

So I now follow you down to the fifth line. Your row reduction is correct, I believe.

Your logic is incorrect in your answer to i. In order for that entry in the matrix to be nonzero, you must have both $\alpha\not=2$ and $\alpha\not=1,$ (not or, as you wrote).

I would agree that for ii, it's impossible. And I agree with your answer for iii.

Does that help?
• August 26th 2010, 09:52 AM
Quote:

Originally Posted by Ackbeet

1. Your fourth line doesn't look like that at all.
2. I don't think that's the right ERO for this stage in the game. I would do 3rd - (2nd x ( $\alpha$-3)).

Come to think of it, that's what your fourth line looks like. I got your numeral 1's mixed up with your parentheses (maybe you should make your parentheses more curved, so they don't get confused with 1's). You actually did what I suggested in my second comment.

[EDIT]: I think you edited your post. Pick up my comments from here on down.

So I now follow you down to the fifth line. Your row reduction is correct, I believe.

Your logic is incorrect in your answer to i. In order for that entry in the matrix to be nonzero, you must have both $\alpha\not=2$ and $\alpha\not=1,$ (not or, as you wrote).

I would agree that for ii, it's impossible. And I agree with your answer for iii.

Does that help?

Im just worried because iv asks for a solution for ii, but i have got it to be impossible.
• August 26th 2010, 09:56 AM
Ackbeet
Hmm. Well, that is a bit puzzling. However, I think your row reduction is correct. Perhaps you could say the empty set?
• August 28th 2010, 05:45 AM
If instead of $3\alpha+1$ there were just $\alpha$ then for $\alpha =1$ parts ii) & iv) have answers.