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Mathematics 22 Online
OpenStudy (watchmath):

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

OpenStudy (anonymous):

Book says use the Leibniz formula...but that formula is tough.

OpenStudy (anonymous):

\( n a(a^2-b^2)? \)

OpenStudy (watchmath):

@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.

OpenStudy (anonymous):

Oh you're right. This is right for n>2. It can be shown by mathematical induction.

OpenStudy (anonymous):

I did it quickly, so it might not be right. I have to go now, I'll do it again when I am back.

OpenStudy (anonymous):

Although my answer seems to be correct :D

OpenStudy (anonymous):

What do you think?!

OpenStudy (watchmath):

The answer should be true for \(n\geq 1\)

OpenStudy (anonymous):

Sorry, I made a mistake.

OpenStudy (anonymous):

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.

OpenStudy (anonymous):

anwar got a sec?

OpenStudy (anonymous):

Here the form I came up with for \(n≥2\): \[{n! \over 2} a^{n-2}(a^2-b^2)\]

OpenStudy (anonymous):

What's up satellite?!

OpenStudy (anonymous):

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

OpenStudy (anonymous):

on the other hand i frequently make mistakes. and not just in my typesetting.

OpenStudy (anonymous):

I could not find the mistake you made, if you did make one.

OpenStudy (anonymous):

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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