Where is the Mistake? Solving a System of Equations

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Discussion Overview

The discussion revolves around solving a system of equations and identifying mistakes in the solution process. Participants explore conditions under which the system has a single solution, no solution, or infinitely many solutions, focusing on the parameter 'a' and its impact on the system's behavior.

Discussion Character

  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant poses a system of equations and asks for the values of 'a' that lead to different types of solutions.
  • Another participant suggests that setting a = -1 leads to an undefined system with infinitely many solutions.
  • It is proposed that a = 3 results in a situation indicating no solution.
  • A participant argues that for any other value of 'a', a unique solution is obtained.
  • Concerns are raised about the operations performed on the rows of the matrix when a = 2, leading to confusion about the nature of the solutions.
  • One participant identifies a specific operation as problematic, suggesting that it introduces proportional rows that do not reflect the original system's characteristics.
  • Another participant emphasizes the importance of treating the case where a = 2 separately due to the division by zero issue that arises in the calculations.

Areas of Agreement / Disagreement

Participants express differing views on the implications of specific values of 'a' and the correctness of the operations performed. The discussion remains unresolved regarding the exact nature of the mistakes and the conditions for different types of solutions.

Contextual Notes

Participants note limitations in the operations performed on the matrix, particularly concerning the treatment of the case when a = 2, which leads to division by zero and affects the interpretation of the solutions.

Yankel
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Hello

I have a system of equations with the question "for which values of a the system has: a single solution, no solution and infinite number of solutions". In addition, I have some solution, and I need to find the mistake in the solution, I need some help with it...

So, for which values of a, the next system:

2x+y=1
4x+ay+z=0
3y+az=2

has a single solution, no solution and infinite number of solutions ?

View attachment 434

if a=2, the matrix has proportional rows (r2 and r3), and we get infinite number of solutions. However, if we set a=2 in the original system, there is a single solution.

Where is the mistake ?

Thanks !
 

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You have it there in your graphic.

a = -1

This results in the third row disappearing, making an underfined system and infinitely many solutions.

a = 3

This results in an inappropriate result indicating no solution.

a = Anything Else

Unique Solution.
 
yes, but put a=2, you get

2 1 0 1
0 0 1 -2
0 0 -3 6

now multiply the 3rd row by (1/3) and add the 2nd row to it, you get:

2 1 0 1
0 0 1 -2
0 0 0 0

which is not a single solution...something is wrong here...
 
Yankel said:
yes, but put a=2, you get

2 1 0 1
0 0 1 -2
0 0 -3 6

now multiply the 3rd row by (1/3) and add the 2nd row to it, you get:

2 1 0 1
0 0 1 -2
0 0 0 0

which is not a single solution...something is wrong here...
The operation $R_3 \to (a-2)R_3 -3R_2$ is the culprit. When $a=2$, that has the effect of killing the third row and replacing it by a multiple of the second row. So it is hardly surprising that rows 2 and 3 are then proportional. This is not so in the original system, it is something that you have introduced by performing a dodgy operation.

In fact, operations of the form $R_i\to pR_i+qR_j$ are best avoided unless $p=1.$ What you should have done instead of $R_3 \to (a-2)R_3 -3R_2$ is $R_2\to \frac1{a-2}R_2.$ When $a=2$ that gives a division by zero, which should have sounded an alarm bell and warned you to treat that case separately.
 
Understood, thank you !
 

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