Consider the sytem 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 ofthe coefficients is positive. Prove that the given system has only the solution $x_1=x_2=x_3=0$.
Problem
Source: IMO 1965, Day 1, Problem 2
Tags: linear algebra, matrix, function, algebra, system of equations, IMO, IMO 1965
boxedexe
16.08.2007 01:40
It seems that this problem needs a solution. Solution. Assume without loss of generality that $ |x_{3}|\geq|x_{2}|\geq |x_{1}|$. Assume to the contrary that $ |x_{3}|>0$. Then it follows, from the given conditions, that \begin{eqnarray*}0&=&\left| a_{31}\frac{x_{1}}{x_{3}}+a_{32}\frac{x_{2}}{x_{3}}+a_{33}\right|\cdot|x_{3}|\\ &\geq& |x_{3}|(a_{33}-|a_{32}|-|a_{31}|)\\ &=& |x_{3}|(a_{33}+a_{32}+a_{31})>0\end{eqnarray*} Contradiction! It follows that $ x_{1}=x_{2}=x_{3}=0,$ as required. $ \Box$
TaiPan~SP!
21.10.2007 05:00
This problem can also be solved using determinants . P.S. Sorry for reviving the old thread, I wanted to look at the original problem because this problem is printed wrong in my book (otherwise it make no sense).
Jorge Miranda
30.01.2010 20:05
We want to prove that the matrix $ A = [a_{ij}]$ has nonzero determinant. We'll prove that $ \det A > 0$.
We have that $ \det A$ is a linear function in each entry of $ A$. The coefficient of $ a_{11}$ is $ a_{22}a_{33} - a_{23}a_{32}$, but since $ a_{22} > - a_{21} - a_{23} > - a_{23} > 0$ and $ a_{33} > - a_{32} > 0$, $ a_{22}a_{33} - a_{23}a_{32} > 0$.
From this we conclude that $ \det A$ is decreased as $ a_{11}$ is decreased. Analogously, it decreases when $ a_{22},a_{33}$ decrease, so it's greater than the determinant when the sum of each row is $ 0$. But in that case the determinant is zero because the columns are linearly dependent (their sum is zero), so we are done.
Mathias_DK
30.01.2010 22:23
The system of equations has the unique solution $ x_1 = x_2 = x_3 = 0$ iff $ b_1 = (a_{11},a_{12},a_{13}), b_2 = (a_{21},a_{22},a_{23}), b_3 = (a_{31}, a_{32}, a_{33})$ is linear independent. Assume they are not. So there exists $ (x_1,x_2,x_3) \neq (0,0,0)$ s.t. $ x_1b_1 + x_2b_2 + x_3b_3 = 0$. c) implies that not all of $ x_1,x_2,x_3$ has the same sign. wlog asume that $ x_1$ has another sign than $ x_2,x_3$ and $ x_1 \neq 0$. Hence the first entry of $ x_1b_1 + x_2b_2 + x_3b_3$ has the same sign as $ x_1$ since $ a_{11} > 0$. Contradiction. So $ b_1,b_2,b_3$ are linear independent and the system of equations has the unique solution $ x_1 = x_2 = x_3 = 0$.
LOTRFan123
11.08.2016 07:20
Would this solution work? If not, could anyone explain a solution using this approach?
Instead of letting $a$ be the coefficients, let $x$ be the coefficients. Now we are trying to find three solutions to the equation $x_1a+x_2b+x_3c=0$, for coefficients $(x_1, x_2, x_3)$, such that
a/b)in each of the equations, one of $a$, $b$ or $c$ is positive while the other two are negative and each variable gets a chance to be positive.
c) When the coefficients are all 1, the sum is positive for all three solutions.
Condition a and b means that one solution must exist in each of the octants where $(x_1,x_2,x_3)$ are $(+, -, -), (-,+,-)$, and $(-,-,+)$. Now all we have to show is that each valid equation can only pass through two octants of those three octants maximum.
If any of the coefficients are negative, then it won't pass through all three octants.
Now, we only consider planes that pass through all three octants and prove that they aren't valid solutions.
Since $x_1a+x_2b+x_3c=0$, by condition 3, any solution must be "behind" the (1,1,1) plane in one of the octants in order to have a sum of less than 0. WLOG, in the $(-,-,+)$ octant, $-(a+b)>c$. However, going back to the original form, $a,b$ correspond to $a_{31}, a_{32}$ and $c$ corresponds to $a_{33}$, and this would mean the sum of the coefficients would be negative, which is a contradiction.
Since no positive or negative values work for $(x_1, x_2, x_3)$, the only thing they can be is all 0, and that works obviously. QED?
Thanks.
Mathematicsislovely
27.08.2022 02:06
Let $$A=\begin{pmatrix}a_{11}&a_{12}&a_{13}\\a_{21}&a_{22}&a_{23}\\a_{31}&a_{32}&a_{33}\end{pmatrix}$$In order to show that $\dim(\ker(A))=0$ we only need to show that $rank(A)=3$.(By rank nullity Theorem.) Suppose FTSoC the columns $C_i$ ($i=1,2,3$) of $A$ are linearly dependent.Then there are reals $b_1,b_2,b_3$ not all 0 such that $$b_1C_1+b_2C_2+b_3C_3=\textbf{0}$$ Let $i\in\{1,2,3\}$ be the integer such that $b_i$ has maximum absolute value among $b_1,b_2,b_3$. But then $$|a_{ii}|=|\sum_{j\ne i} \frac{b_j}{b_i} a_{ij}|\le \sum_{j\ne i} |\frac{b_j}{b_i} a_{ij}|\le \sum_{j\ne i} | a_{ij}|<|a_{ii}|$$But this is impossible.Hence $rank(A)=3$.$\blacksquare$