# 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):}

You need to consider the following three conditions for the system to have an infinite number of solutions, such that:

`{(Delta = 0),(delta != 0),(Delta_C = 0):}`

You need to evaluate `Delta` , determinant of the matrix of coefficients of variables, such that:

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

`Delta = k^3 +...

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You need to consider the following three conditions for the system to have an infinite number of solutions, such that:

`{(Delta = 0),(delta != 0),(Delta_C = 0):}`

You need to evaluate `Delta` , determinant of the matrix of coefficients of variables, such that:

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

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

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

`k^3 + 3 - 1 - 3k = 0 => (k^3 - 1) - 3(k - 1) = 0`

`(k - 1)(k^2 + k + 1 - 3) = 0 => {(k - 1 = 0),(k^2 + k - 2 = 0):}`

`{(k = 1),(k_(1,2) = (-1+-3)/2):} => {(k = 1),(k = -2):}`

Considering the second condition, `delta != 0` , you need to select delta, such that:

Considering the third condition, `Delta_C = 0` , yields:

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

`Delta_C = 9k^2 - 8 - 1 + 8k + k - 9`

`Delta_C = 0 => 9k^2 + 9k - 18 = 0 => k^2 + k - 2 = 0 => k = 1, k = -2`

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

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