Linear Equations - Cramer's Rule

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Homework Help Overview

The discussion revolves around determining whether a given set of linear equations has a unique solution, specifically excluding trivial solutions where x=y=z=0. The equations presented are: x+2y-4z=8, 4x-6y+12z=19, and -6x+3y-6z=-20. The context involves the application of Cramer's Rule and matrix representation.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the implications of using Cramer's Rule without formally solving the equations. There is an exploration of rewriting the system as a matrix equation and questioning the conditions under which the matrix guarantees a unique solution. Some participants analyze the determinant of the matrix formed by the coefficients of the equations to infer properties about the solution set.

Discussion Status

The discussion is active, with participants providing insights into the implications of a zero determinant and the distinction between having no unique solution versus having no solution at all. There is a recognition of the potential for multiple interpretations of the terms used in the context of linear equations.

Contextual Notes

Participants are navigating the constraints of the homework prompt, which explicitly instructs not to formally solve the equations. The discussion also highlights the importance of precise language when describing the nature of the solutions to the system of equations.

ZedCar
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Homework Statement



Does the following set of linear equations have a unique solution?

(excluding any trivial solutions when x=y=z=0)

Do not attempt to formally solve the equations.

x+2y-4z=8
4x-6y+12z=19
-6x+3y-6z=-20



Homework Equations





The Attempt at a Solution



My immediate thought would be to solve the equations using Cramers Rule. However, it states not to formally solve the equations. So I assume I'm not allowed to do this.

So what does the question mean me to do?
 
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ZedCar said:

Homework Statement



Does the following set of linear equations have a unique solution?

(excluding any trivial solutions when x=y=z=0)

Do not attempt to formally solve the equations.

x+2y-4z=8
4x-6y+12z=19
-6x+3y-6z=-20



Homework Equations





The Attempt at a Solution



My immediate thought would be to solve the equations using Cramers Rule. However, it states not to formally solve the equations. So I assume I'm not allowed to do this.

So what does the question mean me to do?

Rewrite the system of three equations as a matrix equation in the form Ax = b. Here x represents the vector <x, y, z>T, and b represents the constant vector <8, 19, -20>T. What conditions on matrix A guarantee a unique solution to the equation?
 
By ^T do you mean transpose?

Just wanted to check before attempting to try and solve this.
 
If I take the matrix;

1, 2, -4
4, -6, 12
-6, 3, -6

The determinant = 0

Therefore it is not invertible.

Therefore this implies it has no unique solution.

Is this correct?
 
ZedCar said:
By ^T do you mean transpose?

Just wanted to check before attempting to try and solve this.
Yes.
ZedCar said:
If I take the matrix;

1, 2, -4
4, -6, 12
-6, 3, -6

The determinant = 0

Therefore it is not invertible.

Therefore this implies it has no unique solution.

Is this correct?
A better way is to say that it has no solution. A system of equations can have
1) a unique solution
2) multiple solutions (an infinite number of them)
3) no solutions

If you say "no unique solution" this might be interpreted as multiple solutions.
 
Okay. Thank you.
 
But isn't "no unique solution" what he wants to say? The question, after all, was "Does the following set of linear equations have a unique solution?" The fact that the determinant of this 3 by 3 set of equations is 0 means it maps R3 into a proper subspace of R3. If the right hand side doesn't happen to be in that subspace, there is no solution but if it does, an entire subspace will be mapped to it.
 
HallsofIvy said:
But isn't "no unique solution" what he wants to say? The question, after all, was "Does the following set of linear equations have a unique solution?" The fact that the determinant of this 3 by 3 set of equations is 0 means it maps R3 into a proper subspace of R3. If the right hand side doesn't happen to be in that subspace, there is no solution but if it does, an entire subspace will be mapped to it.
For this particular problem, there is no solution. I am distinguishing between "no unique solution" and "no solution" as the former term might be interpreted by some to mean that there are multiple solutions (i.e., not a unique solution). That was the distinction I was trying to make.
 

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