# For which values of k is this sytem have infinitely many solutions? Consider the following system of equations: {(kx+y+z=-8),(x+ky+z=-1),(x+y+kz=9):}

## Expert Answers

You need to consider two conditions for the system has an infinte number of solutions, such that:

`Delta = 0 =>` `[(k,1,1),(1,k,1),(1,1,k)] = 0`

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

`k(k - 2) - (k - 2) = 0...

## See This Answer Now

Start your 48-hour free trial to unlock this answer and thousands more. Enjoy eNotes ad-free and cancel anytime.

You need to consider two conditions for the system has an infinte number of solutions, such that:

`Delta = 0 =>` `[(k,1,1),(1,k,1),(1,1,k)] = 0`

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

`k(k - 2) - (k - 2) = 0 => (k - 2)(k - 1) = 0 => {(k-2=0),(k-1=0):}`

`{(k=2),(k=1):}`

You need to consider the next condition, such that:

`delta = [(k,1),(1,k)] != 0` => `k^2 - 1 != 0 => k != +- 1`

`delta_c = [(k,1,-8),(1,k,-1),(1,1,9)] = 0`

`delta_c = 9k^2 - 8 - 1 + 8k + k - 9 => 9k^2 + 9k - 18 = 0`

`k^2 + k - 2 = 0 => k_(1,2) = (-1 +- sqrt(1 + 8))/2`

`k_(1,2) = (-1 +- 3)/2 => k_1 = 1 ; k_2 = -2`

Hence, evaluating k for the system to have an infinite number of solutions yields `k = -1 ; k = +-2` and `k != 1.`

Approved by eNotes Editorial Team