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)(k-1)^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) +(1-k) +(1-k) NOT=0

k^3-k NOT= (k-1)

k(k^2 -1) NOT= k-1

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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