# What value of k does the following have an infinite number of solutins., unique number and or/no solutions. (1-k)x-y +z=0 2x +(1-k)y =0 2y-(1+k)z = 0

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### 1 Answer

You need to form the matrix of the system using the coefficients of variables x,y,z such that:

`A = ((1-k,-1,1),(2,1-k,0),(0,2,-1-k))`

You need to evaluate the determinant of the matrix A such that:

`Delta = [[1-k,-1,1],[2,1-k,0],[0,2,-1-k]]` = `(1-k^2)(k-1) + 4 + 0 - 0 - 0 - 2(1+k)`

`Delta = k - 1 - k^3 + k^2 + 4 - 2 - 2k`

`Delta =- k^3 + k^2 - k + 1`

You should remember that if `Delta != 0` , then the solution to the system is unique and if `Delta = 0` , then the system may have an infinite number of solutions or no solution.

You need to solve the equation `Delta = 0` such that:

`- k^3 + k^2 - k + 1 = 0`

You need to group the terms such that:

`- (k^3 -k^2) - (k- 1) = 0`

You need to factor out `k^2` in the first group such that:

`-k^2 (k - 1) - (k - 1) = 0`

You need to factor out `(k - 1)` such that:

`(k - 1)(-k^2 - 1) = 0`

`k- 1 = 0 => k = 1`

`k^2= -1` , there are no reals k

**Hence, if the system has an unique solution, then `k in R - {1}` and if the system has no solution or an infinite number of solutions, then `k= 1` .**