We will use the determinant identity det  1 ai+ bj  n×n

Problem 11969

(American Mathematical Monthly, Vol.124, March 2017) Proposed by A. Dzhumadil’daev (Kazakhstan).

Letx1, . . . , xn be indeterminates, and letA be the n-by-n matrix with (i, j) entry 1/ cos(xi− x^j).

Prove

det(A) = (−1)(ⁿ²) Y

1≤i<j≤n

tan²(xi− x^j).

Solution proposed by Roberto Tauraso, Dipartimento di Matematica, Universit`a di Roma “Tor Vergata”, via della Ricerca Scientifica, 00133 Roma, Italy.

Solution. We will use the determinant identity

det

 1 ai+ bj



n×n

!

= Q

1≤i<j≤n(ai− aj)(bi− bj) Q

1≤i,j≤n(ai+ bj)

which appears in T. Muir’s book A Treatise on the Theory of Determinants, par. 353, p. 348 (see also Monthly problem 10387). The identity can be proved by induction by noting that if we subtract to the i-th row the n-th row multiplied by (an+ bn)/(ai+ bn) for i = 1, . . . , n, we get

det

 1 ai+ bj



n×n

!

= 1

an+ bn

det

 (ai− an)(bj− bn) (ai+ bj)(ai+ bn)(an+ bj)



n−1×n−1

!

=

Q

1≤i≤n−1(ai− aⁿ)(bi− bⁿ) (an+ bn)Q

1≤i≤n−1(ai+ bn)(an+ bi)det

 1 ai+ bj



n−1×n−1

! .

Let ai= bi= e^2ixi then

cos(xi− xj) =e^i(xi−xj)+ e^{−i(xi−xj)}

2 = ai+ aj

2e^ixie^ixj and

tan(xi− xj) = sin(xi− x^j)

cos(xi− xj)= e^i(xi−xj)− e^{−i(xi−xj)}

i(e^i(xi−xj)+ e^{−i(xi−xj)})= ai− a^j i(ai+ aj). Hence, by Muir’s identity,

det(A) = det 2e^ixie^ixj ai+ aj



= 2ⁿ

n

Y

i=1

e^ixi

n

Y

j=1

e^ixj · det

 1 ai+ aj



=

n

Y

i=1

(2ai) · Q

1≤i<j≤n(ai− a^j)² Q

1≤i,j≤n(ai+ aj) = Q

1≤i<j≤n(ai− a^j)² Q

1≤i<j≤n(ai+ aj)²

= Y

1≤i<j≤n

 ai− a^j ai+ aj

2

= Y

1≤i<j≤n

(i tan(xi− x^j))²

= (−1)(ⁿ²) Y

1≤i<j≤n

tan²(xi− xj).