- #1

- 190

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[tex]\sum_{j=1}^n \sum_{k=1}^n a_{ik}b_{kj} = \left(\sum_{j=1}^n b_{kj}\right) \left(\sum_{k=1}^n a_{ik}\right) = 1\cdot 1 = 1[/tex]

Thanks!

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- Thread starter IniquiTrance
- Start date

- #1

- 190

- 0

[tex]\sum_{j=1}^n \sum_{k=1}^n a_{ik}b_{kj} = \left(\sum_{j=1}^n b_{kj}\right) \left(\sum_{k=1}^n a_{ik}\right) = 1\cdot 1 = 1[/tex]

Thanks!

- #2

- 22,129

- 3,298

What you could do is:

[tex]\sum_{j=1}^n\sum_{k=1}^n a_{ik}b{kj}= \sum_{k=1}^n \sum_{j=1}^n a_{ik}b_{kj} = \sum_{k=1}^n \left(a_{ik} \sum_{j=1}^n b_{kj}\right)= \sum_{k=1}^n a_{ik}=1[/tex]

- #3

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Thank you very much. I knew I wasn't understanding something.

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