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Linear Algebra: Part 2. Some crazy jump in logic which I don't get.
Jimmy King
Registered User regular
Registered User regular
Alright, another problem similar to the one I asked about a few days ago. This is actually from the same set. There were 3 problems in the first homework set which were very difficult and everything else (I've done at least 100 different linear algebra problems now) has been simple and made total sense. I went back to those 3 trouble problems and see where they went with 2 of them (although how I should have known to do that is still questionable), but the 3rd makes no fucking sense.
Solve for K (a constant) so that this has no solution:
x + 2y + Kz = 6
3x + 6y + 8z = 4
My initial solution was that I could clearly see the first equation was 1/3 of the second equation except the answers... so set K to 1/3 of 8 and I've got equivalent equations with different answers and there's no solution. Problem solved.
The "worked" (for some questionable definition of worked) solution can be seen at http://www.calcchat.com/book/Elementary-Linear-Algebra-7e/ if you then select chapter 1, section 1, problem 81.
1. Reduce to row echelon form (Why would I have known to start there rather than doing what I did?)
2. The problem now becomes the following 2 equations (that's fine, if I know to do #1, getting it here is trivial)
x + 2y + Kz = 6
(8-3K)z = -14
3. The "worked" solution now jumps right to "if 8-3K = 0 then there is no solution". That trivially solves to K = 8/3, of course. But where did it make that jump? Where the did z and -14 go?
I could go 8-3K = (-14/z) and then -3K = (-14/z) - 8 and then K = (14/z) + (8/3)... but there's still that 14/z hanging out.
I've tried solving for z first and backfilling, but unless my brain has stopped doing math right for the morning (it's possible), that doesn't get me anywhere useful either, just to 1 = -1, eventually, which is not true and may show no solution, but doesn't get me where the book went with it.
Solve for K (a constant) so that this has no solution:
x + 2y + Kz = 6
3x + 6y + 8z = 4
My initial solution was that I could clearly see the first equation was 1/3 of the second equation except the answers... so set K to 1/3 of 8 and I've got equivalent equations with different answers and there's no solution. Problem solved.
The "worked" (for some questionable definition of worked) solution can be seen at http://www.calcchat.com/book/Elementary-Linear-Algebra-7e/ if you then select chapter 1, section 1, problem 81.
1. Reduce to row echelon form (Why would I have known to start there rather than doing what I did?)
2. The problem now becomes the following 2 equations (that's fine, if I know to do #1, getting it here is trivial)
x + 2y + Kz = 6
(8-3K)z = -14
3. The "worked" solution now jumps right to "if 8-3K = 0 then there is no solution". That trivially solves to K = 8/3, of course. But where did it make that jump? Where the did z and -14 go?
I could go 8-3K = (-14/z) and then -3K = (-14/z) - 8 and then K = (14/z) + (8/3)... but there's still that 14/z hanging out.
I've tried solving for z first and backfilling, but unless my brain has stopped doing math right for the morning (it's possible), that doesn't get me anywhere useful either, just to 1 = -1, eventually, which is not true and may show no solution, but doesn't get me where the book went with it.
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Edit: Like for K=8/3 the second equation becomes 0*z=-14 which is always false.
The reason for doing it their way instead of yours is that it works easier for equations that don't line up so neatly.
So, now my question is why would I have ever thought to reduce to row echelon form and go from there? Is that just something that is always going to be the first step if doing a problem like this, where you're trying to find a K which creates no solution? The book has no examples of finding the constant which causes a specific result, no discussion of how to go about it, etc.
I'm not familiar with linear algebra but it seems to be the standard way to do these problems. Normally you already know K and are just trying to find the x, y and z to solve the equation.
I guess when it comes time for the test I'll just have to review these to make sure I recognize them and hope that the pattern from the homework questions holds true on the test questions... or that questions like this just don't show up on the test.
The short explanation is that you would row reduce because you are looking to find out the number of solutions, except this case is harder because the number of solutions depends on K. By row reducing you get something like:
x + 2y + Kz = 6
0x + 0y + (8-3K)z = -14
and you know (hopefully) that this system has a solution if (8-3K) is not equal to zero, and has no solutions if (8-3K) is equal to zero.
But (simple version), there's infinitely many solutions if there's a row of all zeroes, there's no solutions if you get that 0 = not 0, and there's exactly one solution if it's a square matrix with a leading one in every row.