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
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?
4 answers
did I say -1?
I meant -2
I meant -2
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
(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
Steve is right. I left out a 2