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Urgent Matrices Help - LU

Right, so I have my linear algebra exam tomorrow, was quite confident, until this seemingly easy question had me stumped.

http://awesomescreenshot.com/0d97i9r05

That's the question. My problem is with the LU decomposition/factorisation.

This is what I can get to:

http://awesomescreenshot.com/0927i9u36

Now, the markscheme gives the following as the next step:

http://awesomescreenshot.com/0617i9vdf

However, I get a very different answer. They seem to have subtracted -1 lots of the second row from the bottom row. I however subtracted 1 times the 3rd row from the bottom row - but this gives a different final answer. Am I not understanding the factorisation properly?
(edited 11 years ago)
Not sure about this factorising technique, but all they've done is add row 2 to row 3 & 4 (it just looks like Gaussian elimination to me).
Reply 2
Avatar for ViralRiver
ViralRiver
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16
Original post by Brit_Miller
Not sure about this factorising technique, but all they've done is add row 2 to row 3 & 4 (it just looks like Gaussian elimination to me).


Yeah I know what they've done, but I used a different row (which I thought should still work), but my final answer is different to theirs.
Reply 3
Avatar for ViralRiver
ViralRiver
OP
16
I managed to replicate this problem in another question. Is there a rule as to which rows you can use?
Original post by ViralRiver
Yeah I know what they've done, but I used a different row (which I thought should still work), but my final answer is different to theirs.


I don't think you should use the third row. I think you should only use the row with the leading element in it. I haven't got my notes on hand but I kept messing up gaussian elimination doing something similar. Maybe you'll shoot me down and say it doesn't matter, I'm not totally confident.
Reply 5
Avatar for BabyMaths
BabyMaths
Original post by ViralRiver
I managed to replicate this problem in another question. Is there a rule as to which rows you can use?


I don't think so. Just use your common sense. Some things would be pointless.

Your row operations give you EA=U and then A = E^-1 U = L U unless I've forgotten more linear algebra than I think I have.
Original post by ViralRiver
I managed to replicate this problem in another question. Is there a rule as to which rows you can use?


The way I do it, I leave the top row (unless the row starts with a zero), then use that row the make the other rows' first element equal to zero. Then do the same process with the second row and so on until they cancel or an answer pops out.
Reply 7
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ViralRiver
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16
Hmm not sure what you mean?

I've uploaded an attachment of a similar question showing the problem.
IMG_2476[1].jpg507.2KB
Original post by ViralRiver
Hmm not sure what you mean?

I've uploaded an attachment of a similar question showing the problem.


The second method is the way I do it.
Reply 9
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ViralRiver
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16
So what is the correct order of elimination? Rows descending, columns ascending?
Reply 10
Avatar for BabyMaths
BabyMaths
Original post by ViralRiver
So what is the correct order of elimination? Rows descending, columns ascending?


I'm not sure what you mean there.

I've just made up an example and managed to get the LU factorisation so I do remember how to do it.. :u:
Reply 11
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ViralRiver
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16
Original post by BabyMaths
I'm not sure what you mean there.

I've just made up an example and managed to get the LU factorisation so I do remember how to do it.. :u:


Well my problem is choosing the wrong rows, so there must be some rule to say which row you can choose for elimination and when. If you see my image, I chose 2 different rows, but I got different answers depending on when I chose them.
I think they can be equivalent...
Original post by ViralRiver
Well my problem is choosing the wrong rows, so there must be some rule to say which row you can choose for elimination and when. If you see my image, I chose 2 different rows, but I got different answers depending on when I chose them.


i think they can be equivalent...
Reply 14
Avatar for ViralRiver
ViralRiver
OP
16
They're not equivalent.
Reply 15
Avatar for BabyMaths
BabyMaths
The LU factorisation is unique so any correct method will give the same result.

I would add multiples of row 1 to get zeros in the rest of the first column then add multiples of row 2 to rows 3 and 4 to get zeros under the 2,2 position etc..

It works.

Your question comes out as


L=
1 0 0 0
0 1 0 0
1 -1 1 0
2 -1 -3 1

and U=
2 3 1 2
0 3 1 2
0 0 2 3
0 0 0 2

Now I'll check the latex and post that properly in a minute..
Reply 16
Avatar for BabyMaths
BabyMaths
L=(100001001−1102−1−31)L=\begin{pmatrix} 1 & 0 & 0 & 0 \\0 & 1 & 0 & 0 \\1 & -1 & 1 & 0 \\2 & -1 & -3 & 1 \end{pmatrix} and U=(2312031200230002)U=\begin{pmatrix} 2 & 3 & 1 & 2 \\0 & 3 & 1 & 2 \\0 & 0 & 2 & 3 \\0 & 0 & 0 & 2 \end{pmatrix}
Reply 17
Avatar for ViralRiver
ViralRiver
OP
16
Original post by BabyMaths
L=(100001001−1102−1−31)L=\begin{pmatrix} 1 & 0 & 0 & 0 \\0 & 1 & 0 & 0 \\1 & -1 & 1 & 0 \\2 & -1 & -3 & 1 \end{pmatrix} and U=(2312031200230002)U=\begin{pmatrix} 2 & 3 & 1 & 2 \\0 & 3 & 1 & 2 \\0 & 0 & 2 & 3 \\0 & 0 & 0 & 2 \end{pmatrix}


Are you saying that both examples in my image lead to the same factorisation?
Reply 18
Avatar for BabyMaths
BabyMaths
Original post by ViralRiver
Are you saying that both examples in my image lead to the same factorisation?



It doesn't look like they do.

I get these steps..

(231203120−3110−3−7−9)\begin{pmatrix} 2 & 3 & 1 & 2 \\0 & 3 & 1 & 2 \\0 & -3 & 1 & 1 \\0 & -3 & -7 & -9 \end{pmatrix}

(23120312002300−6−7)\begin{pmatrix} 2 & 3 & 1 & 2 \\0 & 3 & 1 & 2 \\0 & 0 & 2 & 3 \\0 & 0 & -6 & -7 \end{pmatrix}

(2312031200230002)\begin{pmatrix} 2 & 3 & 1 & 2 \\0 & 3 & 1 & 2 \\0 & 0 & 2 & 3 \\0 & 0 & 0 & 2 \end{pmatrix}

Why worry about another method?

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