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August 17th, 2017, 06:29 PM  #11 
Global Moderator Joined: Dec 2006 Posts: 19,739 Thanks: 1810 
From the first equation, 2x + 4ky = 2k. Subtracting that from the second equation gives (2k  2)x = 2k  2, so k = 1 allows an infinite number of solutions, where x has any value and y = (1  x)/2. If k isn't 1, the above equation is satisfied by x = 1. Putting x = 1 in the second equation gives 2k + 4ky = 4k  2, which implies 2ky = k  1. That has no solution if k = 0. If k is neither 1 nor 0, the system has the unique solution (x, y) = (1, (k  1)/(2k)). 
August 18th, 2017, 12:06 AM  #12 
Newbie Joined: Aug 2017 From: America Posts: 5 Thanks: 0  Maybe I'm wrong, but eventually I get this: (k1)(x1)=0 1. x+2ky=k multiply it by two then you get 4ky=2k2x 2. then the second one is 4ky=4k22kx 3. eventually kx=2k1kx is (k1)(x1)=0 Last edited by skipjack; August 22nd, 2017 at 10:34 AM. 
August 22nd, 2017, 10:41 AM  #13 
Global Moderator Joined: Dec 2006 Posts: 19,739 Thanks: 1810 
That's correct, but you need to continue.

August 23rd, 2017, 03:44 AM  #14 
Math Team Joined: Jan 2015 From: Alabama Posts: 3,261 Thanks: 894  
August 25th, 2017, 11:58 AM  #15 
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,559 Thanks: 113 
The system of n equations Ax = y has a solution if matrix [A] and augmented matrix [A:y] have the same rank r. If r = n, exactly one solution. If r < n, infinitely many solutions. Determine rank by reduction to reduced row echelon form using Gaussian elimination. Last edited by skipjack; August 25th, 2017 at 12:21 PM. 
August 26th, 2017, 05:22 AM  #16  
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,559 Thanks: 113 
This isn't done to give the answer, already given, but to demonstrate the principle: Quote:
2kx+4ky=4k2 [A:y] $\displaystyle \begin{bmatrix} 1& 2k & k\\ 2k & 4k &4k2 \end{bmatrix} $ $\displaystyle k=0, \begin{bmatrix} 1 &0 &0 \\ 0 &0 & 2 \end{bmatrix} $, no solution, rank [A:y] > rank [A] $\displaystyle \begin{bmatrix} 1 &2k &k \\ 2(k1) &0 &2k1 \end{bmatrix}$ $\displaystyle k=1, \begin{bmatrix} 1 &2 &1 \\ 0 &0 &0 \end{bmatrix}, y=\frac{1x}{2} $, infinitely many solutions, x arbitrary $\displaystyle k\neq 0,1, \begin{bmatrix} 1 &2k &k \\ 1 & 0 & 1 \end{bmatrix}, x=1, y=\frac{k1}{2k}$, one solution  

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