How to Determine Values for a and b in a 3x4 System for Infinite Solutions?

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SUMMARY

The forum discussion centers on determining the values of a and b in a 3x4 system of equations to achieve infinite solutions. The consensus is that setting a = -1 and b = 6 allows the second equation to be derived from the first and third equations. The method involves manipulating the equations through subtraction and multiplication, demonstrating that one equation can be expressed as a linear combination of the others. This approach is validated by multiple contributors who emphasize the effectiveness of creative problem-solving techniques in linear algebra.

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Ewan_C
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[Solved] 3x4 system of equations

Homework Statement



Consider the following system of three equations in x, y and z.

2x + 4y + 5z = 17
4x + ay + 3z = b
8x + 7y + 13z = 40

Give values for a and b in the second equation that make this system consistent, but with an infinite set of solutions.

The Attempt at a Solution



I found the answers a= -1, b=6 easily enough. I was told by my teacher that if a system of three equations has infinite solutions, one of the equations can be found from the other two. I multiplied equation 1 by 2 and subtracted the result from equation 3. This gave:

4x - y + 3z = 6

and so finding the values of a and b was pretty simple from there. Plugging the numbers into a calculator gave an infinite number of solutions.

My question is, how can I better explain how to get a and b from the provided data? My method just seems like an educated guess rather than solid evidence - I don't think it'd look very good to an examiner. Cheers.
 
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Your method is perfectly good. What you have shown is that a combination of the first and third equation give the middle equation if a= -1 and b= 6. It was clever of you to notice that "twice the first equation subtracted from the first" gives an equation that can be made exactly the same as the second if a and b have those values. I see no "guessing" there.

What I might have done (perhaps because I am not as smart as you!) would be to try to solve the system of equations and see what might prevent me from getting a single solution. If I subtract twice the first equation from the second, the "x"s cancel and I am left with (a- 8)y- 7z= b- 34. If I subtract 4 times the first equation from the third, again the "x"s cancel and I am left with -9y- 7z= 28. Now I can eliminate the z by subtracting that last equation from the previous one: [(a-8)+ 9]y= (a+1)y= b- 6. If a+1 is not 0, I can divide by it and get a single answer. In order that there not be a single answer, we must have a+ 1= 0 or a= -1. Of course, that would make the equation 0= b- 6. In order that that be true (for any y) we must have b= 6.

Frankly, I prefer your method.
 
Ewan, welcome to PhysicsForums. Halls: Nice humble response. I'm of Halls' mind here. I too tend to use a sledge hammer to force the solution to fall out where a little poking around would make the solution fall out with little effort. One of the PF 2007 guru award winners, Dick, is very good at poking around and making the solution fall out with little effort. (He also knows how to wield sledge hammers when needed.) Honing your skills with the heavy-handed techniques is important, but so is honing your skills at seeing a quick and easy way to solve the problem. There's nothing wrong with a bit of creativity.
 
Thanks for the responses. I guess I'll stick with my method in the OP then.
 

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