Prove that there exists exactly one solution

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In summary, the linear system of the form a_{1,1}x + a_{1,2}y = b_{1} and a_{2,1}x + a_{2,2}y = b_{2} has exactly one solution in x and y if a_{1,1}a_{2,2} \neq a_{1,2}a_{2,1}. This can be proven by showing that if the determinant of the coefficient matrix is nonzero, then there is a unique solution, and if the determinant is zero, then there are either infinitely many solutions (if the determinant of the augmented matrix is also zero) or no solutions (if the determinant of the augmented matrix is nonzero
  • #1
4Fun
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Homework Statement



Prove that the linear system of the form:

[tex]
a_{1,1}x + a_{1,2}y = b_{1}\\
a_{2,1}x + a_{2,2}y = b_{2}
[/tex]

has exactly one solution in x and y if [itex]a_{1,1}a_{2,2} \neq a_{1,2}a_{2,1}[/itex]



The Attempt at a Solution



I proved the contrapositive:

Assume that the given system does not have exactly one solution in x and y, then it either has no solution or it has infinitely many solutions.

Case 1: Assume that there is no solution to this system. Since it is a system of two equations, I can refer to a geometric explanation. Two equations do not intersect if they are parallel, i.e. their slopes are equal and b1 != b2. For their slopes to be equal it has to be the case that [itex]a_{1,1}a_{2,2} = a_{1,2}a_{2,1}[/itex].

Case 2: Assume there are infinitely many solutions, in that case the equations are the same and their slopes are equal, therefore again fulfilling [itex]a_{1,1}a_{2,2} = a_{1,2}a_{2,1}[/itex].

Now I think the proof is correct, but I don't like to rely on geometric intuition, if it would have been a system of 3 equations I wouldn't be able to prove this type of problem. I tried applying the Gauss algorithm to this system, but it did not yield anything useful.
Could anybody give me a hint for another approach please?
 
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  • #2
From matrix theory: your original condition is equivalent to "the determinant is non-zero"; this means that the coefficient matrix can be inverted. So AX=B can be rewritten as X=A^-1*B ... which is the only solution.

If det(A)=0 then A^-1 does not exist ... and then there is not a unique solution.
 
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  • #3
Thanks for your quick reply. We haven't covered matrices and determinants yet. Is there another way, without using determinants to prove this?
 
  • #4
In that case just draw the possible lines, using the given condition to control the slopes. Note that if you put both equations into y=mx+b form you get the slopes ... and from the condition given the slopes are not equal.

Hence the lines are not parallel. Now you have to remember a famous theorem from Euclid's geometry ...
 
  • #5
4Fun said:

Homework Statement



Prove that the linear system of the form:

[tex]
a_{1,1}x + a_{1,2}y = b_{1}\\
a_{2,1}x + a_{2,2}y = b_{2}
[/tex]

has exactly one solution in x and y if [itex]a_{1,1}a_{2,2} \neq a_{1,2}a_{2,1}[/itex]



The Attempt at a Solution



I proved the contrapositive:

Assume that the given system does not have exactly one solution in x and y, then it either has no solution or it has infinitely many solutions.

Case 1: Assume that there is no solution to this system. Since it is a system of two equations, I can refer to a geometric explanation. Two equations do not intersect if they are parallel, i.e. their slopes are equal and b1 != b2. For their slopes to be equal it has to be the case that [itex]a_{1,1}a_{2,2} = a_{1,2}a_{2,1}[/itex].

Case 2: Assume there are infinitely many solutions, in that case the equations are the same and their slopes are equal, therefore again fulfilling [itex]a_{1,1}a_{2,2} = a_{1,2}a_{2,1}[/itex].

Now I think the proof is correct, but I don't like to rely on geometric intuition, if it would have been a system of 3 equations I wouldn't be able to prove this type of problem. I tried applying the Gauss algorithm to this system, but it did not yield anything useful.
Could anybody give me a hint for another approach please?

Just start solving the equations by hand. Since ##a_{11}a_{22} \neq a_{12} a_{21}## we cannot have both ##a_{11}## and ##a_{12}## equal to zero, so at least one of them is nonzero. Say, for example, that ##a_{11} \neq 0##. That means that you can solve for ##x## in terms of ##y## from the first equation (since dividing by ##a_{11}## is OK). If you plug that expression for ##x## into the second equation, you will finally get an equation involving ##y## alone. This will have the form ##A y = B##. When does that equation have a unique solution? When does it have no solutions? When does it have infinitely many solutions?
 
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  • #6
Ok thanks a lot. I did as you said and when solving for y I got a term with a11a22 - a12a21 in the denominator. So if a11a22 - a12a21 != 0, then the term is defined and there are unique solutions for y and x. If a11a22 - a12a21 = 0, the system would have no solutions, and I'm not sure about the case of infinitely many solutions. For that to happen I would need A=0 and B=0 in the equation Ay=B, so when solving for y both the term in the numerator and denominator would have to be 0, I guess.
 
  • #7
If your expression is zero the lines are parallel ... how many lines satisfy this condition?
 
  • #8
4Fun said:
Ok thanks a lot. I did as you said and when solving for y I got a term with a11a22 - a12a21 in the denominator. So if a11a22 - a12a21 != 0, then the term is defined and there are unique solutions for y and x. If a11a22 - a12a21 = 0, the system would have no solutions, and I'm not sure about the case of infinitely many solutions. For that to happen I would need A=0 and B=0 in the equation Ay=B, so when solving for y both the term in the numerator and denominator would have to be 0, I guess.

You are confusing the issue by premature "division". You obtain and equation of the form
[tex] A y = B, \: A = a_{11} a_{22} - a_{21} a_{12}, \: B = a_{11} b_2 - a_{21}b_1.[/tex]
If ##A \neq 0## you get a unique solution because at that point you know you are allowed to divide by ##A##. Next: if ##A = 0## you get infinitely many solutions if ##B = 0## (that is, any y value will satisfy 0*y=0) and there are no solutions at all if ##B \neq 0## (that is, '0*y = nonzero' is impossible).
 
Last edited:

1. What does it mean to prove that there exists exactly one solution?

Proving that there exists exactly one solution means to demonstrate that there is only one possible answer or solution to a problem or equation. It involves showing that all other options have been eliminated and that only one option remains.

2. How can you prove that there exists exactly one solution?

To prove that there exists exactly one solution, one must use logical reasoning and evidence to show that all other possibilities have been ruled out. This can be done through various mathematical or scientific methods depending on the specific problem or question.

3. Why is it important to prove that there exists exactly one solution?

Proving that there exists exactly one solution is important because it ensures that there is a definitive answer to a problem or question. It eliminates any ambiguity or uncertainty and provides a clear understanding of the solution.

4. What are some common methods used to prove that there exists exactly one solution?

Some common methods used to prove that there exists exactly one solution include mathematical proofs, logical reasoning, and scientific experiments. These methods involve carefully analyzing the problem and providing evidence to support the existence of one unique solution.

5. Can there be more than one solution to a problem even if it is proven that there exists exactly one solution?

No, if it has been proven that there exists exactly one solution, then there can only be one possible answer or solution to the problem. This is the whole purpose of proving that there exists exactly one solution - to demonstrate that all other options have been eliminated.

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