Find the determinant of \[\begin{pmatrix}a&b&b&\cdots&b\\b&a&b&\cdots&b\\b&b&a&\cdots&b\\\vdots&\vdots&\vdots&\quad&\vdots\\b&b&b&\cdots&a\end{pmatrix}_{n\times n}\]

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Find the determinant of \[\begin{pmatrix}a&b&b&\cdots&b\\b&a&b&\cdots&b\\b&b&a&\cdots&b\\\vdots&\vdots&\vdots&\quad&\vdots\\b&b&b&\cdots&a\end{pmatrix}_{n\times n}\]

Mathematics
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Book says use the Leibniz formula...but that formula is tough.
\( n a(a^2-b^2)? \)
@Anwar: If it is a \(2\times 2\) matrix, then the determinant is \(a^2-b^2\). But we don't get that if we plug in \(n=2\) to your answer.

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Other answers:

Oh you're right. This is right for n>2. It can be shown by mathematical induction.
I did it quickly, so it might not be right. I have to go now, I'll do it again when I am back.
Although my answer seems to be correct :D
What do you think?!
The answer should be true for \(n\geq 1\)
Sorry, I made a mistake.
It has be something like for \(n=3; 3a(a^2-b^2)\) and for \(n=4; 4a(3a(a^2-b^2))\) I'll write the full solution when I come back. I just have to go for like 30 minutes or something.
anwar got a sec?
Here the form I came up with for \(n≥2\): \[{n! \over 2} a^{n-2}(a^2-b^2)\]
What's up satellite?!
i was hoping you could take a second to look at this my attempted answer to the fib question. i think he/she said something bigger than 25 , but i got 25
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on the other hand i frequently make mistakes. and not just in my typesetting.
I could not find the mistake you made, if you did make one.
thanks another pair of eyes is always good. still it worries me that they said they got something larger than 25. i am not going to fret any more about it.

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