Problem 1:Prove that the following system of linear equations has unique solutions(sorry,I don`t know precise english term for this kind of system) and if it has,solve it:
x_1+x_2+x_3+x_4=4
x_1+2x_2+3x_3+x_4=7
x_2+x_3+2x_4=4
x_1-x_2+x_3-2x_4=-1
2x_1-2x_2+x_3-x_4=0
So, four variables,five equations.I don`t see how any method could be applied here:Gaussian elimination,Cramer`s rule,matrix method.Any hint just to get stared would be enough.
dirk_mec1
Aug22-08, 06:52 AM
Look for dependent equations and see if there are four independent equations then calculate the determinant.
R A V E N
Aug22-08, 06:58 AM
What is dependant equation?If I am right,it is one equation from the system multiplied by some constant to produce additional equation.But I do not see any.
R A V E N: stared
*started
HallsofIvy
Aug22-08, 08:53 AM
Problem 1:Prove that the following system of linear equations has unique solutions(sorry,I don`t know precise english term for this kind of system) and if it has,solve it:
x_1+x_2+x_3+x_4=4
x_1+2x_2+3x_3+x_4=7
x_2+x_3+2x_4=4
x_1-x_2+x_3-2x_4=-1
2x_1-2x_2+x_3-x_4=0
So, four variables,five equations.I don`t see how any method could be applied here:Gaussian elimination,Cramer`s rule,matrix method.Any hint just to get stared would be enough.
Well, try to solve it! Since you have 4 variables, you should be able to solve any 4 of the equations for those variables. Once you have done that, do they satisfy the fifth equation?
A slightly more formal way, though not any better, would be to set up the 5 by 4 matrix of coefficients and row-reduce it. If the last row turns out to be all 0's then there is a solution, otherwise, there is not.
R A V E N
Aug22-08, 09:30 AM
Yeah,choosing random four equations and solving system by using Cramer`s rule works.Thanks,HallsofIvy.
R A V E N
Aug22-08, 02:19 PM
Problem 2:Solve this equation(z - complex number):
(z^2-2)^6+z^6=0
Any hints?
Dick
Aug22-08, 04:07 PM
z is not a root. So it's safe to divide both sides by z^6. That gives you ((z^2-2)/z)^6=(z-2/z)^6=(-1). So z-2/z is equal to one of the six sixth roots of (-1). Solve z-2/z=k and set k to be each of the six sixth roots. You'll get a total of twelve roots (as you should, it's a twelfth degree equation). That's pretty tedious. But you can do it.
I think you slipped up on the trig identities, but you don't really need them anyway. You have a 1^(infinity) type limit. Take the log to change it to an infinity*0 limit and rearrange it to apply l'Hopital's rule.
R A V E N
Aug23-08, 02:14 PM
Yeah,instead of attempting anything and ramble in various faulty courses,it`s better to wait advice for correct approach.
On what basis did you concluded that it is 1^\infty type of a limit?
Dick
Aug23-08, 02:29 PM
No, it's better for you to ramble in faulty courses, otherwise I won't help at all. :) x-> 0, x+pi/4 -> pi/4. tan(pi/4)=1. 2x -> 0, cos(2x) -> 1, sin(2x) -> 0, cos(2x)/sin(2x)=cot(2x)->infinity. Hence 1^infinity.
R A V E N
Aug23-08, 02:54 PM
And it was that simple?(By the way,everything is simple if you have enough time and energy to invest into studying,thinking and looking for advice on various places).
I just wonder why those 800+ pages mathematical books which have zillion things in it don`t have any example which would for the moment escape rigorous mathematical world and demonstrate things like that and other in more human-friendly fashion?
For the sake of more successful and contented both teachers and students.
Dick
Aug23-08, 02:58 PM
And it was that simple?(By the way,everything is simple if you have enough time and energy to invest into studying,thinking and looking for advice on various places).
I just wonder why those 800+ pages mathematical books which have zillion things in it don`t have any example which would for the moment escape rigorous mathematical world and demonstrate things like that and other in more human-friendly fashion?
For more succesful and contented both teachers and students.
What I gave just shows you what KIND of limit it is. It doesn't solve the limit. Did you figure out the value?