Linear System with Parameter

Here is problem 3 from the Fifth International Internet Mathematical Olympiad for Students. This is an online competition run by the Ariel University Center of Samaria, Israel.

Find all values of the parameter y for which the following system has a solution

(1) x₅ + x₂ = yx₁

(2) x₁ + x₃ = yx₂

(3) x₂ + x₄ = yx₃

(4) x₃ + x₅ = yx₄

(5) x₄ + x₁ = yx₅

For each value of y, find all the solutions.

Solution

First of all observe that, for any value of y, the system has a trivial solution: x₁ = x₂ = x₃ = x₄ = x₅ = 0. The task is to find the values of y for which there is a nontrivial solution. Add up all five equations to obtain

(y - 2)(x₁ + x₂ + x₃ + x₄ + x₅) = 0.

So, either y = 2 or x₁ + x₂ + x₃ + x₄ + x₅ = 0. Clearly the value y = 2 is bound to play a prominent role in the solution. Curiously, there are other y's of no lesser importance.

But, to follow up, assume first that y = 2. The system becomes

(1')	x₅ + x₂	= 2x₁
(2')	x₁ + x₃	= 2x₂
(3')	x₂ + x₄	= 2x₃
(4')	x₃ + x₅	= 2x₄
(5')	x₄ + x₁	= 2x₅

Since the system is homogeneous and we are looking for a non-trivial solution, we may omit the last equation and solve the rest with the Gaussian elimination, getting subsequently,

- 2	1	0	0	1
1	- 2	1	0	0
0	1	- 2	1	0
0	0	1	- 2	1

⇒

- 2	1	0	0	1
0	- 3	2	0	1
0	0	- 4	3	1
0	0	0	- 5	5

By the back substitution, x₅ = x₄ = x₃ = x₂ = x₁, with arbitrary x₁.

Now we may assume y ≠ 2 and x₁ + x₂ + x₃ + x₄ + x₅ = 0. We'll replace (quite arbitrarily) the last equation with this one and employ again the Gaussian elimination.

1	1	1	1	1
-y	1	0	0	1
1	-y	1	0	0
0	1	-y	1	0
0	0	1	-y	1

⇒

1	1	1	1	1
0	y+1	y	y	y+1
0	-(y+1)	0	-1	-1
0	1	-y	1	0
0	0	1	-y	1

⇒

1	1	1	1	1
0	1	-y	1	0
0	y+1	y	y	y+1
0	-(y+1)	0	-1	-1
0	0	1	-y	1

⇒

1	1	1	1	1
0	1	-y	1	0
0	0	y	y-1	y
0	0	-y(y+1)	y	-1
0	0	1	-y	1

⇒

1	1	1	1	1
0	1	-y	1	0
0	0	-y(y+1)	y	-1
0	0	y	y-1	y
0	0	y+1	-1	y+1

⇒

1	1	1	1	1
0	1	-y	1	0
0	0	-y(y+1)	y	-1
0	0	0	y²+y-1	y²+y-1
0	0	0	0	y²+y-1

If y² + y - 1 ≠ 0, the last equation gives x₅ = 0 and the back substitution leads to a trivial solution which was already accounted for. So let us assume that y² + y - 1 = 0. The quadratic formula gives two distinct solutions

y_1,2 = (-1 ± √5) / 2.

With the parameter y equal to one of them, the last equation implies that x₅ can be chosen arbitrarily; and, from the fourth equation, the same is true of x₄.

Since, for the chosen y, -y(y + 1) = -1, the third equation gives

x₃ = yx₄ - x₅

From the second equation,

x₂ = -y(x₄ + x₅)

and, from the first equation,

x₁ = -x₄ + yx₅.

Thus the system has three sets of solutions:

A trivial solution for any y,
A 1-parameter solution for y = 2,
A 2-parameter solution for y = (-1 ± √5) / 2.

The latter family, perhaps surprisingly, includes the golden ratio.

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