A size n determinant has all 1s on the main diagonal and the first superdiagonal. It has all -1s on the first subdiagonal. It has zeros in all other places.
Show that this determinant evaluates to
Fibonacci number F(n+1).
The determinant corresponding to n=5 is depicted below:
| 1 1 0 0 0 |
|-1 1 1 0 0 |
| 0 -1 1 1 0 | = 8 = F(6)
| 0 0 -1 1 1 |
| 0 0 0 -1 1 |
A graphic solution is easy as you see from the sample that
|n|=1*|a|- 1*|b|+0*|c|-0*|d|+...+0|w| (n size of the matrix and choosing the first row to calculate)
But |a|=|n-1|
So: |n|=|n-1|-|b|
But |b|=(-1)|n-2| (the other terms of the first column of this matrix are zeroes)
So: |n|=|n-1|+|n-2|
Then:
|1|=1=F(2)
|2|=| 1 1|
|-1 1| = 2= F(3)
So: |3|=|1|+|2| = F(2)+F(3) = F(4) and with recursion:
|n| = |n-1|+|n-2| = F(n)+F(n-1)= F(n+1)
Edited on April 19, 2016, 12:57 pm
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Posted by armando
on 2016-04-19 11:23:17 |
Visual solution
