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?
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
did I say -1?
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
Steve is right. I left out a 2
To find the value of W, we need to solve the system of equations:
kx + y + z = 1 ----(1)
x + ky + z = k ----(2)
x + y + kz = k^2 ----(3)
We are given that this system of equations has a unique solution. For a system of equations to have a unique solution, it means that the determinant of the matrix formed by the coefficients of the variables must be non-zero.
Let's form the coefficient matrix A for this system as:
A = [[k, 1, 1], [1, k, 1], [1, 1, k]]
The determinant of A is given by:
det(A) = k(k^2 - 1) - (1)(k - 1) + (1)(1 - k^2)
= k^3 - k - k + 1 + 1 - k^2
= k^3 - 2k^2
For the system to have a unique solution, det(A) must be non-zero. Therefore, we have:
k^3 - 2k^2 ≠ 0
Simplifying this inequality, we can factor out k^2:
k^2(k - 2) ≠ 0
To find the values of k that satisfy this inequality, we set each factor equal to zero and solve for k:
k^2 = 0 ----(4)
k - 2 = 0 ----(5)
From equation (4), we get k = 0.
From equation (5), we get k = 2.
Therefore, the value of W is the set of numbers that make the determinant zero, which is k = 0 and k = 2.
Hence, W = {0, 2}