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...

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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.`