Mathmatical Logic Fails Me
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Basically, you can write your three equations as a matrixBaronAsh said:
(2 -4 -1) (x) = (10)
(4 -8 -2).
= (16)(3+1 1 ) (z) = (12)
or
- M.r = k
- r = M[sup]-1[/sup]k
Substitute your answers back inMonkeyman8 said:
2(23/4)-4(9/4)-(-33/4)= 10.75 <--(not 10)
4(23/4)-8(9/4)-2(-33/4)= 21.5 <--(not 16)
3(23/4)-(9/4)-(-33/4)=23.25 <--(not 12)
"I've got a baaaaad feeling about this..."
ActualyMonkeyman8 said:
4x - 8y - 2(2x - 4y -10) = 16
simplifies to
4x - 8y - 4x + 8y + 20 = 16
or
20 = 16
after you reduce
your error was multiplying -4y by essentialy a -2
4x - 8y - 2(2x - 4y -10) = 16
4x - 8y + (-2*2)x +(-4*-2)y +(-10*-2) = 16
4x - 8y - 4x + 8y + 20 = 16
putting those values into the first equation givesMonkeyman8 said:
(46 - 36 + 33)/4 = 10 Or 10.75 = 10
As for the solution, either two things is wrong OP. You've written down the equations wrong or the question is wrong, using all 3 equations together does not give an answer because they do not intersect at a single point.
Equation 1 and 2 are parallel and therefore will never have any value for x y and z that is true for both equations becuase they never touch.
all that work and you didn't bother checking your answer?Monkeyman8 said:
3x+y+z = 12
3(23/4) + 9/4 + -33/4 = 69/4 + 9/4 -33/4 = 45/4 = 11.25
I dropped it into a linear solver and there doesn't seem to be a solution. Either the OP copied the system wrong or the book screwed up.
EDIT: Epic triple ninja. -.-
What the hell is with all these long-ass posts involving matricies? It can be disproved by very basic algebra.
Look:
Equation 1:
2x-4y-z=10
2(2x-4y-z)=2(10)
4x-8y-2z=20
Equation 2:
4x-8y-2z=16
Solution:
4x-8y-2z=20
but
4x-8y-2z=16
16=/=20, therefore equations are not simultaneous and no answer can be found for any variables.
Look:
Equation 1:
2x-4y-z=10
2(2x-4y-z)=2(10)
4x-8y-2z=20
Equation 2:
4x-8y-2z=16
Solution:
4x-8y-2z=20
but
4x-8y-2z=16
16=/=20, therefore equations are not simultaneous and no answer can be found for any variables.
I'm even more diappointed, though that said I wish they'd teach more uni level maths at schools here in good old Blighty. I only covered this way of solving equations in Linear Algebra 1 in my freshers year. I'm in second year now doing Maths BSc and so I know this is impossible anyway, but we really should have done it at school.The_root_of_all_evil said:
That said, today we went through an extremely long and tedious proof of the existence of the 'cycloid' in Calculus of Variations, which is basically the curve that will lead to the fastest time for a particle to reach a point (x2,y2) when dropped from (x1,y1) and acting only under gravity (no other external forces). The proof took about 5 A4 pages. Back in A-Level our teacher showed us and although he didn't prove it, he gave us information on how to prove it ourselves, and it didn't take anywhere near as long as the university method...
x=10, y=2, z= for the first one.
This doesn't work for the second one, therefore is intentionally impossible. It's like factorising 12x^2-x-6=0, it can't be done as on a graph it doesn't cross the X-axis.
This doesn't work for the second one, therefore is intentionally impossible. It's like factorising 12x^2-x-6=0, it can't be done as on a graph it doesn't cross the X-axis.
What the hell are you on about? That makes no sense. We want all of them to equal 0.silver scribbler said:
Now, just because there is no solution, doesn't mean you can't explain what's going on there.
You have a system with two independent variables and one dependent variable, thus a system of planes, expressable as z=z(x,y). So, a single solution would be the point of intersection between these three planes.
Now, although it has been shown that there is no solution, you can still figure out what exactly is going on between the planes.
List of possibilities:
Single solution (nicest)
All planes equvilent (2-dimensions of infinite solutions)
Sheaf formed (three planes intersect at a line, giving infinite solutions. Note that two planes may be equivilent still)
Triangular Prism formed (no solutions, instead solutions to only equations 1 and 2, 2 and 3, 1 and 3 form parallel lines)
All planes parallel, not all three planes the same (self-explanatory)
Two planes parallel, other one not.
I _think_ these are all the possibilities, and it's possible to discern which solution is valid. Of course, I learned this two years ago, failed miserably at it, and thus can't explain how you detect which possibility is the valid one. Taking a cursory glance, I'd say you have two parallel planes and another plane which can form a line of solutions with either parallel plane.
Thank you.
You have a system with two independent variables and one dependent variable, thus a system of planes, expressable as z=z(x,y). So, a single solution would be the point of intersection between these three planes.
Now, although it has been shown that there is no solution, you can still figure out what exactly is going on between the planes.
List of possibilities:
Single solution (nicest)
All planes equvilent (2-dimensions of infinite solutions)
Sheaf formed (three planes intersect at a line, giving infinite solutions. Note that two planes may be equivilent still)
Triangular Prism formed (no solutions, instead solutions to only equations 1 and 2, 2 and 3, 1 and 3 form parallel lines)
All planes parallel, not all three planes the same (self-explanatory)
Two planes parallel, other one not.
I _think_ these are all the possibilities, and it's possible to discern which solution is valid. Of course, I learned this two years ago, failed miserably at it, and thus can't explain how you detect which possibility is the valid one. Taking a cursory glance, I'd say you have two parallel planes and another plane which can form a line of solutions with either parallel plane.
Thank you.