Problem 11678
(American Mathematical Monthly, Vol.119, December 2012) Proposed by Farrukh Ataev Rakhimjanovich (Uzbekistan).
Let Fkbe the kth Fibonacci number, where F0= 0 and F1= 1. For n ≥ 1 let Anbe an (n+1)×(n+1) matrix with entries aj,k given by a0,k = ak,0 = Fk for 0 ≤ k ≤ n and by aj,k = aj−1,k+ aj,k−1 for j, k ≥ 1. Compute the determinant of An.
Solution proposed by Roberto Tauraso, Dipartimento di Matematica, Universit`a di Roma “Tor Vergata”, via della Ricerca Scientifica, 00133 Roma, Italy.
Let Bn be the (n + 1) × (n + 1) upper triangular matrix with entries
bj,k=
1 if i = j,
−1 if i = j − 1 or i = j − 2, 0 otherwise.
then, by the definition of the numbers aj,k, it is easy to verify that
Btn· An· Bn=
0 1 0 · · · 0 1 0 0 · · · 0 0 0
... ... 2Pn−1 0 0
where Pn−1is the the (n − 1) × (n − 1) Pascal matrix:
(Pn−1)j,k=i + j i
for i = 0, . . . , n − 2 and j = 0, . . . , n − 2.
Hence
det(An) = det(Btn· An· Bn) = − det(2Pn−1) = −2n−1
because det(Mn) = 1, det(Pn−1) = 1.
