## watchmath 5 years ago Find the determinant of $\begin{pmatrix}a&b&b&\cdots&b\\b&a&b&\cdots&b\\b&b&a&\cdots&b\\\vdots&\vdots&\vdots&\quad&\vdots\\b&b&b&\cdots&a\end{pmatrix}_{n\times n}$

1. anonymous

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

2. anonymous

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

3. 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.

4. anonymous

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

5. 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.

6. anonymous

Although my answer seems to be correct :D

7. anonymous

What do you think?!

8. watchmath

The answer should be true for $$n\geq 1$$

9. anonymous

10. 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.

11. anonymous

anwar got a sec?

12. anonymous

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

13. anonymous

What's up satellite?!

14. 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

15. anonymous

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

16. anonymous

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

17. 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.