428 Algebra
=
∣
∣∣
∣∣
∣∣
∣
−y− 10 0 − 1
0 −y 1 − 1
− 11 −y− 1 − 1
−10 0−y∣
∣∣
∣∣
∣∣
∣
,
which, after expanding with the rule of Laplace, becomes
−
∣
∣∣
∣
−y − 1
1 −y− 1∣
∣∣
∣·
∣
∣∣
∣
−y− 1 − 1
− 1 −y∣
∣∣
∣=−(y(^2) +y− 1 ) (^2).
Hence the original determinant is equal to(y− 2 )(y^2 +y− 1 )^2 .Ify=2, the space
of solutions is therefore one-dimensional, and it is easy to guess the solutionx 1 =x 2 =
x 3 =x 4 =x 5 =λ,λ∈R.
Ify=−^1 +
√
5
2 or ify=
− 1 −
√
5
2 , the space of solutions is two-dimensional. In both
cases, the minor
∣∣
∣∣
∣∣
−y 10
1 −y 1
01 −y
∣∣
∣∣
∣∣
is nonzero, hencex 3 ,x 4 , andx 5 can be computed in terms ofx 1 andx 2. In this case the
general solution is
(λ, μ,−λ+yμ,−y(λ+μ), yλ−μ), λ, μ∈R.Remark.The determinant of the system can also be computed using the formula for the
determinant of a circulant matrix.
(5th International Mathematical Olympiad, 1963, proposed by the Soviet Union)
233.Taking the logarithms of the four relations from the statement, we obtain the fol-
lowing linear system of equations in the unknowns lna,lnb,lnc,lnd:
−xlna+lnb+lnc+lnd= 0 ,
lna−ylnb+lnc+lnd= 0 ,
lna+lnb−zlnc+lnd= 0 ,
lna+lnb+lnc−tlnd= 0.We are given that this system has a nontrivial solution. Hence the determinant of the
coefficient matrix is zero, which is what had to be proved.
(Romanian mathematics competition, 2004)
234.First solution: Suppose there is a nontrivial solution(x 1 ,x 2 ,x 3 ). Without loss of
generality, we may assumex 1 ≤x 2 ≤x 3. Letx 2 =x 1 +m,x 3 =x 1 +m+n,m, n≥0.
The first and the last equations of the system become