Consider the following system of linear equations: x - 3z = -3 2x + ky - z = -2 x + 2y + kz = 1 [1 0 -3 | -3] [0 k 5 | 4] [0 0 k^2 + 3k - 10 | 4k - 8] Find the ONLY value of k where the system will have no solution. Find the values of k so that the system will have more than one solution. Find the values of k so that the system will have a unique solution.
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First, let's write down the given system of linear equations: $$ \begin{cases} x - 3y - 3z = 0 \\ 20 + ky - z = 1 + 2v + k \\ p + 3 - 10z = 4k - 8 \end{cases} $$ Now, let's write the augmented matrix for this system: $$ \begin{bmatrix} 1 & -3 & -3 & 0 \\ 0 & k & Show more…
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