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Fibonacci Determinant (Posted on 2016-04-19) Difficulty: 3 of 5
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 |

See The Solution Submitted by Brian Smith    
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Visual solution Comment 1 of 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
  Posted by armando on 2016-04-19 11:23:17

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