Fibonacci Determinant (Posted on 2016-04-19)

 | 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 |

	Submitted by Brian Smith
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Let D[x] denote the determinate of size x. The determinant for n=1 is trivial: | 1 | = 1 = F(2). The determinant for n=2: | 1 1 | = 2 = F(3) |-1 1 | To evaluate higher orders, use cofactor expansion along the bottom row. Only the last two terms need to be evaluated as they are the only nonzero entries in the bottom row. The minor for the last entry is D[x-1], so its cofactor is (-1)^(2n)*|D[n-1]| = |D[n-1]| The minor for the other entry has its rightmost column of n-2 0s and one 1. Then cofactor expansion along that column yields a subminor of D[n-2]. Then the second cofactor is (-1)^(2n-1)*(1*(-1)^(2n-2))*|D[n-2]| = -|D[n-2]| Then |D[n]| = 1*|D[n-1]| + (-1)*(-|D[n-2]|) = |D[n-1]| + |D[n-2]|. This is the same formula as the Finonacci sequence; since the first two terms are the second and third Fibonacci numbers then the sequence of determinants is the Fibonacci numbers shifted by 1 term. Hence |D[n]| = F(n+1).

	Subject	Author	Date
	Visual solution	armando	2016-04-19 11:23:17