# Difference between revisions of "1965 IMO Problems/Problem 2"

(Created page with '== Problem == Consider the system of equations <cmath>a_{11}x_1 + a_{12}x_2 + a_{13}x_3 = 0</cmath> <cmath>a_{21}x_1 + a_{22}x_2 + a_{23}x_3 = 0</cmath> <cmath>a_{31}x_1 + a_{32}…') |
(I'm pretty sure this solution is good.) |
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== Solution == | == Solution == | ||

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

+ | |||

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

+ | |||

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

+ | |||

+ | Case 2: The <math>x_i</math> are all negative. WLOG <math>x_1\leq x_2\leq x_3</math>. Consider the third equation, <math>a_{31}x_1 + a_{32}x_2 + a_{33}x_3 = 0</math>. Therefore <math>x_3(a_{31}+a_{32}+a_{33})+a_{31}(x_1-x_3)+a_{32}(x_2-x_3)=0</math>, 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 <math>x_i</math> is 0, which implies all of them are 0. |

## Revision as of 01:04, 8 July 2010

## Problem

Consider the system of equations with unknowns , , . The coefficients satisfy the conditions:

(a) , , 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 .

## Solution

Clearly if the are all equal, then they are equal to 0. Now let's assume WLOG that . If or is 0, then the other is clearly zero, so let's consider the case where neither are 0. and are negative, so exactly one of or is positive. Unfortunately this means that one of or is positive and the other is negative, so the equation couldn't possibly be satisfied if or isn't 0. We have covered the case where one of the is 0, now let's assume that none of them are 0.

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

Case 1: The are all positive. WLOG . Now consider the third equation, . Therefore , but all of the terms on the LHS are non-negative and the first one is positive, so this is impossible.

Case 2: The are all negative. WLOG . Consider the third equation, . Therefore , 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 is 0, which implies all of them are 0.