Posted by Anonymous on Friday, August 24, 2012 at 11:10am.
If
kx + y + z = 1
x + ky + z = k
x + y + kz = k^2
has unique solution
then k is not equal to 1 and W. What is W?

Math  Steve, Friday, August 24, 2012 at 11:23am
since the determinant of the system is
 k 1 1 
 1 k 1 
 1 1 k 
= k^3  3k + 2
= (k+2)(k1)^2
Then k cannot be 1 or 1

math  oops  Steve, Friday, August 24, 2012 at 11:24am
did I say 1?
I meant 2

Math  drwls, Friday, August 24, 2012 at 11:26am
The determinant of the three equation left sides cannot be zero.
(k^3 k) +(1k) +(1k) NOT=0
k^3k NOT= (k1)
k(k^2 1) NOT= k1
k(k+1)NOT =1
k cannot be 0 or 1
W = 1

Math  drwls, Friday, August 24, 2012 at 11:28am
Steve is right. I left out a 2
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