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I need to prove (or disprove) that if A(nxn) is a matrix so that the sum of elements in every row equals 1, then the sum of elements in every row of A

I tried a bunch of examples and they all worked, so I'm assuming this is true and I'm trying to prove it:

We know that:

[tex]A^{-1} = \frac{ajd(A)}{|A|}[/tex]

[tex]\sum_{i=1}^n a_{ki}^{-1} = \sum_{i=1}^n \frac{(adj(A))_{ki}}{|A|} = \sum_{i=1}^n \frac{(-1)^{k+i}M_{ik}}{|A|}[/tex]

But by defintion:

[tex]|A| = \sum_{i=1}^n (-1)^{k+i}a_{ki}M_{ki}[/tex]

So we get:

[tex]\sum_{i=1}^n a_{ki}^{-1} = \frac{\sum_{i=1}^n (-1)^{k+i}M_{ik}}{\sum_{i=1}^n (-1)^{k+i}a_{ki}M_{ki}}[/tex]

And from here I don't know what to do. All that's left is to show that this expression equals 1... somehow. I think I need to use the fact that |A| = 1/|A

Thanks.

^{-1}also equals 1.I tried a bunch of examples and they all worked, so I'm assuming this is true and I'm trying to prove it:

We know that:

[tex]A^{-1} = \frac{ajd(A)}{|A|}[/tex]

[tex]\sum_{i=1}^n a_{ki}^{-1} = \sum_{i=1}^n \frac{(adj(A))_{ki}}{|A|} = \sum_{i=1}^n \frac{(-1)^{k+i}M_{ik}}{|A|}[/tex]

But by defintion:

[tex]|A| = \sum_{i=1}^n (-1)^{k+i}a_{ki}M_{ki}[/tex]

So we get:

[tex]\sum_{i=1}^n a_{ki}^{-1} = \frac{\sum_{i=1}^n (-1)^{k+i}M_{ik}}{\sum_{i=1}^n (-1)^{k+i}a_{ki}M_{ki}}[/tex]

And from here I don't know what to do. All that's left is to show that this expression equals 1... somehow. I think I need to use the fact that |A| = 1/|A

^{-1}|, no?Thanks.

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