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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 answer (computer exploration) | Comment 1 of 2
The following program tests matrices from 2x2 to 8x8. It finds the 2x2 (not included in the puzzle) has determinant 1, but the ones included in the puzzle (n>4) all have determinant zero.

The determinant evaluation routine was swiped from an old puzzle solution program I had written.  Hopefully it wasn't assuming anything about the matrices whose determinants were being evaluated; the determinants for the 3x3 and 4x4 cases were verified as zero by a TI-84 plus CE calculator, which has a built-in determinant function.

DefDbl A-Z
Dim crlf$, fib(100)


Private Sub Form_Load()
 Form1.Visible = True
 Text1.Text = ""
 crlf = Chr(13) + Chr(10)
 
 maxsize = 10
 a = 1: b = 1
 fib(1) = a: fib(2) = b
 
 For nxt = 3 To 100
   fib(nxt) = fib(nxt - 2) + fib(nxt - 1)
   Text1.Text = Text1.Text & Str(fib(nxt))
 Next
 For sz = 1 To 8
   ReDim triangle(sz, sz)
   For row = 1 To sz
    For col = 1 To sz
      triangle(row, col) = fib((row - 1) * sz + col)
    Next
   Next
   Text1.Text = Text1.Text & sz & Str(det(sz, triangle())) & crlf
   DoEvents
 Next
 
 
 
 Text1.Text = Text1.Text & crlf & " done"
  
End Sub

Function det(sz, triangle())
  If sz = 2 Then
    xx = xx
  End If
  If sz = 1 Then
    det = triangle(1, 1)
    Exit Function
  End If
  ReDim matrix(sz - 1, sz - 1)
  tot = 0
  For col = 1 To sz
    newrow = 0
    For r = 2 To sz
      newrow = newrow + 1
      newcol = 0
      For c = 1 To sz
        If c <> col Then
          newcol = newcol + 1
          matrix(newrow, newcol) = triangle(r, c)
        End If
      Next
    Next r
    term = triangle(1, col) * det(sz - 1, matrix())
    If col Mod 2 = 0 Then term = -term
    tot = tot + term
  Next col
  det = tot
End Function


  Posted by Charlie on 2019-04-08 16:41:42
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