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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 ==
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{{solution}}
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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.
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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.
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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.
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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.
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Therefore at least one of the <math>x_i</math> is 0, which implies all of them are 0.

Revision as of 00:04, 8 July 2010

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 $x_1\leq x_2\leq x_3$. 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.

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