The original matrix

1 1

1 0

can be considered to exhibit Fibonacci numbers 0, 1 and 2.

Successive powers of the matrix exhibit further Fibonacci numbers as noted:

2 1       F(3) F(2)

1 1       F(2) F(1)

3 2       F(4) F(3)

2 1       F(3) F(1)

5 3       F(5) F(4)

3 2       F(4) F(3)

8 5       F(6) F(5)

5 3       F(5) F(4)

13 8       F(7) F(6)

 8 5       F(6) F(5)

21 13       F(8) F(7)

13  8       F(7) F(6)

34 21       F(9) F(8)

21 13       F(8) F(7)

55 34       F(10) F(9)

34 21       F(9)  F(8)

89 55       F(11) F(10)

55 34       F(10) F(9)

DefDbl A-Z

Dim crlf$, matrix(2, 2)

Private Sub Form_Load()

 Form1.Visible = True

 

 Text1.Text = ""

 crlf = Chr$(13) + Chr$(10)

 matrix(1, 1) = 1

 matrix(1, 2) = 1

 matrix(2, 1) = 1

 matrix(2, 2) = 0

 

 For i = 1 To 9

   a = matrix(1, 1) + matrix(2, 1)

   b = matrix(1, 2) + matrix(2, 2)

   c = matrix(1, 1)

   d = matrix(1, 2)

   matrix(1, 1) = a

   matrix(1, 2) = b

   matrix(2, 1) = c

   matrix(2, 2) = d

   Text1.Text = Text1.Text & a & Str(b) & "       " & "F(" & whichFib(a) & ") F(" & whichFib(b) & ")" & crlf

   Text1.Text = Text1.Text & c & Str(d) & "       " & "F(" & whichFib(c) & ") F(" & whichFib(d) & ")" & crlf & crlf

 Next

 

 Text1.Text = Text1.Text & crlf & " done"

  

End Sub

Function whichFib(x)

  wf = Log(x) / Log((1 + Sqr(5)) / 2)

  whichFib = -Int(-wf) + 1 ' ceiling function, then plus 1

End Function

The whichFib() annotation was added, of course, after the pattern was observed.

The program calculates whichFib(1) as 1, which is a correct possibility, but to keep the sequence straight, in the output I overwrote with 2, the other correct possibility, in two places on the table.

Removing the matrix notation, in each iteration, and adding generation subscripts:

a(n) = a(n-1) + c(n-1) = a(n-1) + a(n-2)

b(n) = b(n-1) + d(n-1) = b(n-1) + b(n-2)

c(n) = a(n-1)

d(n) = b(n-1)

which explains the Fibonacci pattern.