Efficiently Calculating Determinants Using Cofactors and Expansion

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lo2
Can this be written.

[tex] \[<br /> \det{\textbf{A}}=a_{11}a'_{11}+a_{12}a'_{12}+\ldots+a_{1n}a'{1n}<br /> \]<br /> [/tex]

As. (Where I just use p instead of n for obvious reasons)

[tex] \[<br /> \sum^{n}_{p=1}{a_{1p}a'_{1p}}<br /> \][/tex]

Since.

[tex] \[<br /> a'_{ij}=(-1)^{i+j}(\det{\textbf{A}_{ij}})<br /> \][/tex]

And yes we are talking determinants.
 
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If you want help then write in a clear manner that people can understand, preferably in complete sentences - if people have to work hard at understanding what the question is they will tend not to bother working out what you intend.

What is a'_{rs}? What are you trying to prove?

I suspect that that doesn't matter. If all you're asking is does that summation expand to give the long hand version, then the answer is 'yes'. Just write down what the summation means.

Det has nothing to do with it as far as I can tell. But because you made a big point of saying it was abuot Det I have no idea if I've answered the question you intended to ask.
 
lo2 said:
[tex]a_{11}a'_{11}+a_{12}a'_{12}+\ldots+a_{1n}a'_{1n} = \sum^{n}_{p=1}{a_{1p}a'_{1p}}[/tex]

yes, these are equal.

Note, I cleaned up your latex. There is no need to include a \[ inside tex tags - I don't think it does anything.
 
I do think I made myself very clear, but well anyway thank you for the help!
 
You can delete it yourself.

If you think you were clear, can I ask what you think A_{ij} is? 'Cos we don't have a clue. And the 'since' is completely misleading. As is the sentence structure ('As.' is not a sentence, and is confusingly, given the post, like a plural form of A).
 
matt grime said:
You can delete it yourself.

How to do that?