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Matrixed Fibonacci (Posted on 2019-04-08) Difficulty: 3 of 5
Fn is the n-th Fibonacci number defined by the recurrence relation Fn = Fn-1 + Fn-2 with F1 = F2 = 1. If n is a perfect square and n > 4, then find the value of the determinant below

| F1          F2          ... F√n  |
| F√n+1     F√n+2    ... F2√n |
|    .           .                .    |
|    .           .                .    |
| Fn-√n+1   Fn-√n+2  ... Fn   |

No Solution Yet Submitted by Danish Ahmed Khan    
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Solution Analytic Solution Comment 2 of 2 |
If the matrix is size 1, its determinant is trivially 1.  If the matrix is size 2, then its determinant is easily evaluated to be 1.

For all larger matrices, pick any three adjacent column vectors.  The sum of the first two vectors equals the third vector.  This dependency means that the rank of the matrix is less than its size and therefore the determinant must be 0.  A stronger statement can be deduced by looking at all triplets of consecutive column vectors and finding that the rank of all the matrices equals 2.
  Posted by Brian Smith on 2019-04-09 11:49:43
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