1965 IMO Problems/Problem 2

Problem

Consider the system of equations $a_{11}x_1 + a_{12}x_2 + a_{13}x_3 = 0$ $a_{21}x_1 + a_{22}x_2 + a_{23}x_3 = 0$ $a_{31}x_1 + a_{32}x_2 + a_{33}x_3 = 0$ with unknowns $x_1$ , $x_2$ , $x_3$ . The coefficients satisfy the conditions:

(a) $a_{11}$ , $a_{22}$ , $a_{33}$ are positive numbers;

(b) the remaining coefficients are negative numbers;

(c) in each equation, the sum of the coefficients is positive.

Prove that the given system has only the solution $x_1 = x_2 = x_3 = 0$ .

Solution

Clearly if the $x_i$ are all equal, then they are equal to 0. Now let's assume WLOG that $x_1=0$ . If $x_2$ or $x_3$ is 0, then the other is clearly zero, so let's consider the case where neither are 0. $a_{12}$ and $a_{21}$ are negative, so exactly one of $x_2$ or $x_3$ is positive. Unfortunately this means that one of $a_{22}x_2 + a_{23}x_3$ or $a_{32}x_2 + a_{33}x_3 = 0$ is positive and the other is negative, so the equation couldn't possibly be satisfied if $x_2$ or $x_3$ isn't 0. We have covered the case where one of the $x_i$ is 0, now let's assume that none of them are 0.

If two are positive and one is negative, then when the negative $x_i$ is paired with one of the positive $a_i$ , the corresponding equation is negative. This is bad. If two are negative and one is positive, then when the positive $x_i$ is paired with one of the positive $a_i$ , the corresponding equation is positive. This is also bad. Therefore the $x_i$ all have the same sign.

Case 1: The $x_i$ are all positive. WLOG $x_1\leq x_2\leq x_3$ . Now consider the third equation, $a_{31}x_1 + a_{32}x_2 + a_{33}x_3 = 0$ . Therefore $x_2(a_{31} +a_{32}+a_{33})+ a_{31}(x_1-x_2)+a_{33}(x_3-x_2)= 0$ , but all of the terms on the LHS are non-negative and the first one is positive, so this is impossible.

Case 2: The $x_i$ are all negative. WLOG $x_1\geq x_2\geq x_3$ . Consider the third equation, $a_{31}x_1 + a_{32}x_2 + a_{33}x_3 = 0$ . Therefore $x_3(a_{31}+a_{32}+a_{33})+a_{31}(x_1-x_3)+a_{32}(x_2-x_3)=0$ , but all of the terms on the LHS are non-positive and the first one is negative, so this is impossible.

Therefore at least one of the $x_i$ is 0, which implies all of them are 0.

Solution 2

We will prove that the matrix

$a_{11}\ \ \ \ a_{12}\ \ \ \ a_{13}$

$a_{21}\ \ \ \ a_{22}\ \ \ \ a_{23}$

$a_{31}\ \ \ \ a_{32}\ \ \ \ a_{33}$

has its determinant $> 0.$

The determinant is

$a_{11}a_{22}a_{33} + a_{12}a_{23}a_{31} + a_{13}a_{32}a_{21} - a_{13}a_{22}a_{31} - a_{11}a_{23}a_{32} - a_{33}a_{12}a_{21}.$

After a little algebraic manipulation we can rewrite this as

$[a_{33}(a_{11} + a_{13}) - a_{13}(a_{31} + a_{33})](a_{21} + a_{22} + a_{23}) - (a_{21}a_{33} - a_{23}a_{31})(a_{11} + a_{12} + a_{13}) - (a_{11}a_{23} - a_{13}a_{21})(a_{31} + a_{32} + a_{33})$

(or we can just verify that this is true).

Note that $a_{11} + a_{12} + a_{13} > 0$ implies $a_{11} + a_{12} > 0$ and $a_{11} + a_{13} > 0.$ The expression above is clearly $> 0$ . To show this in a simple way, I will just write out the sign of each factor:

$[(+)(+) - (-)(+)]\ (+) - ((-)(+) - (-)(-))\ (+) - ((+)(-) - (-)(-))\ (+)$

so now we can see that the end expression is $(+).$

[Solution by pf02, November 2024]

1965 IMO (Problems) • Resources
Preceded by Problem 1	1 • 2 • 3 • 4 • 5 • 6	Followed by Problem 3
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1965 IMO Problems/Problem 2

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Solution 2