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I have the following matrix:

$ A = \begin{bmatrix} 1  & 2  & -3 \\ -5 & -4 & 13 \\ 2  & 1  & -5 \\ \end{bmatrix} $

If you attempt to find the LU using what my book calls "storage of multipliers" or I assume to really be pivoting you run into a problem, when using this technique. That is U cannot have a value that is 0 in the diagonal entries.

In which case the following should be the wrong answer?

$ L = \begin{bmatrix} 1  & 0    & 0 \\ -5 & 1    & 0 \\ 2  & -1/2 & 1 \\ \end{bmatrix} $

$ U = \begin{bmatrix} 1 & 2 & -3 \\ 0 & 6 & -2 \\ 0 & 0 & 0  \\ \end{bmatrix} $

In which case one would have to use a permutation matrix P? (This wasn't covered for us but I assume it's the only way to solve the problem correctly, since U is not of the correct form?)

I'm basically confused as to if my answer is wrong or it's wrong and there is a more appropriate way to solve the problem. Thanks.

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The use of a $P$ matrix is usually referred as partial pivoting and is needed to improve numerical conditioning. –  user13838 Oct 23 '11 at 18:58
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1 Answer 1

up vote 3 down vote accepted

Why do you say that's the "wrong answer"?

If you're worried about the 0 on the diagonal, there's nothing you can do about it: since $A$ is singular, $\det A=0$ and so $\det L \det U = 0$ since determinants are multiplicative. Since the determinant of a triangular matrix is just the product of the diagonal entries, this means at least one of $L$ and $U$ must have a zero on the diagonal. $L$ is constrained to have all 1s on the diagonal, so it must be $U$.

What you may be remembering is that not every singular matrix has a LU decomposition. But when they do (such as this A) it is expected for some of the entries of $U$ to be zero.

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Ah okay thanks for the explanation that was where I was confused was the 0 in the diagonal. –  eWizardII Oct 23 '11 at 20:32
    
Shouldn't the following statement be the other way around? " since $A$ is singular, $\det A=0$ ". I mean, $A$ is singular since $\det A=0$, am I right? –  Gigili Dec 3 '13 at 16:34
    
The two are equivalent, so if you know $A$ has determinant zero then you know it is singular, sure. I myself noticed that the second row of $A$ is $-1$ times the first row plus $-2$ times the third. –  user7530 Dec 3 '13 at 16:41
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