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Reverse-engineering a system of linear equations from solution, using matrices

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dark494
#1
Jan14-10, 10:28 PM
P: 1
1. The problem statement, all variables and given/known data

Find a system of linear equations, with 3 unknowns, given that the solutions are the points (1,1,1) and (3,5,0) on a line.

2. Relevant equations

None

3. The attempt at a solution

A solution that lies on a line tells me that I'm looking at the line of intersection between 2 planes. I'm supposed to be using matrices to solve this, but I've only ever done so in the other direction: taking a system of linear equations and reducing them to reduced-row-echelon-form to find the solution set.

Since there are infinitely many solutions, there must be at least 1 dependent variable. I figured that I would need to first find the equation for the line from the points of the solution, but I ended up with this:

(x-1)/2 = (y-1)/4 = (z-1)/-1

which only confused me more. Then I tried to formulate an augmented matrix to try and find the coefficients of the linear equations, but didn't get very far after reduction, and I realized that I was still missing the right hand side of the matrix:

[a 0b 2.5c ?]
[0a b -1.5c ?]

And this is where I stand. Any help is much appreciated.
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HallsofIvy
#2
Jan15-10, 04:52 AM
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P: 39,345
Quote Quote by dark494 View Post
1. The problem statement, all variables and given/known data

Find a system of linear equations, with 3 unknowns, given that the solutions are the points (1,1,1) and (3,5,0) on a line.

2. Relevant equations

None

3. The attempt at a solution

A solution that lies on a line tells me that I'm looking at the line of intersection between 2 planes. I'm supposed to be using matrices to solve this, but I've only ever done so in the other direction: taking a system of linear equations and reducing them to reduced-row-echelon-form to find the solution set.

Since there are infinitely many solutions, there must be at least 1 dependent variable. I figured that I would need to first find the equation for the line from the points of the solution, but I ended up with this:

(x-1)/2 = (y-1)/4 = (z-1)/-1
Showing that this line is the intersection of the planes given by (x-1)/2= (y-1)/4 and (x-1)/2= (z-1)/(-1) which are the same as 4(x-1)= 2(y-1) or 4x- 2y= 2 and -(x-1)= 2(z-1) or x+ 2z= 3. That is, those points satisfy 4x- 2y= 2 and x+ 2z= 3. That's your system of equations.

which only confused me more. Then I tried to formulate an augmented matrix to try and find the coefficients of the linear equations, but didn't get very far after reduction, and I realized that I was still missing the right hand side of the matrix:

[a 0b 2.5c ?]
[0a b -1.5c ?]

And this is where I stand. Any help is much appreciated.


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