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 August 17th, 2017, 06:29 PM #11 Global Moderator   Joined: Dec 2006 Posts: 18,713 Thanks: 1532 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
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Quote:
 Originally Posted by Country Boy That does not answer the question asked.
Maybe I'm wrong, but eventually I get this: (k-1)(x-1)=0
1. x+2ky=k multiply it by two then you get 4ky=2k-2x
2. then the second one is 4ky=4k-2-2kx
3. eventually k-x=2k-1-kx is (k-1)(x-1)=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: 18,713 Thanks: 1532 That's correct, but you need to continue.
August 23rd, 2017, 03:44 AM   #14
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Quote:
 Originally Posted by Sequoia Maybe I'm wrong, but eventually I get this: (k-1)(x-1)=0 1. x+2ky=k multiply it by two then you get 4ky=2k-2x 2. then the second one is 4ky=4k-2-2kx 3. eventually k-x=2k-1-kx is (k-1)(x-1)=0
And what was the question asked?

 August 25th, 2017, 11:58 AM #15 Senior Member   Joined: Mar 2015 From: New Jersey Posts: 1,300 Thanks: 94 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
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This isn't done to give the answer, already given, but to demonstrate the principle:

Quote:
 Originally Posted by zylo 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.
x+2ky=k
2kx+4ky=4k-2

[A:y]
$\displaystyle \begin{bmatrix} 1& 2k & k\\ 2k & 4k &4k-2 \end{bmatrix}$

$\displaystyle k=0, \begin{bmatrix} 1 &0 &0 \\ 0 &0 & -2 \end{bmatrix}$, no solution, rank [A:y] > rank [A]

$\displaystyle \begin{bmatrix} 1 &2k &k \\ 2(k-1) &0 &2k-1 \end{bmatrix}$

$\displaystyle k=1, \begin{bmatrix} 1 &2 &1 \\ 0 &0 &0 \end{bmatrix}, y=\frac{1-x}{2}$, infinitely many solutions, x arbitrary

$\displaystyle k\neq 0,1, \begin{bmatrix} 1 &2k &k \\ 1 & 0 & 1 \end{bmatrix}, x=1, y=\frac{k-1}{2k}$, one solution

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