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August 17th, 2017, 06:29 PM   #11
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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)).
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August 18th, 2017, 12:06 AM   #12
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Quote:
Originally Posted by Country Boy View Post
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.
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August 22nd, 2017, 10:41 AM   #13
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That's correct, but you need to continue.
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August 23rd, 2017, 03:44 AM   #14
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Quote:
Originally Posted by Sequoia View Post
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?
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August 25th, 2017, 11:58 AM   #15
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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.
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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 View Post
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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