$$\begin{cases}
x − 2y + 5z = 2
\\
x + y + z = k
\\
2x − y + 6z = k^2
\end{cases}$$
For which value(s) of k does this system of linear equations have no solutions/unique solution/infinite sol?
Using Gauss elimination I end up with this matrix, not sure how to proceed.
\begin{bmatrix}1&-2&5&2\\0&3&-4&k-2\\0&0&0&k^2-k-2\end{bmatrix}
Best Answer
Using Gauss elimination, we get: $$ A u = b \iff \\ [ A \mid b ] = \\ \left[ \begin{array}{rrr|r} 1 & -2 & 5 & 2 \\ 1 & 1 & 1 & k \\ 2 & -1 & 6 & k^2 \end{array} \right] \to \left[ \begin{array}{rrr|r} 1 & -2 & 5 & 2 \\ 0 & 3 & -4 & k-2 \\ 0 & 3 & -4 & k^2-4 \end{array} \right] \to \\ \left[ \begin{array}{rrr|r} 1 & -2 & 5 & 2 \\ 0 & 3 & -4 & k-2 \\ 0 & 0 & 0 & k^2-k-2 \end{array} \right] \to \left[ \begin{array}{rrr|r} 1 & -2 & 5 & 2 \\ 0 & 1 & -4/3 & (k-2)/3 \\ 0 & 0 & 0 & k^2-k-2 \end{array} \right] $$ The last row determines, if there are solutions or not. To be consistent, we must have $k$ such that $$ k^2 - k - 2 = 0 $$ otherwise there will be no solutions to the system, as this last equation can not be fulfilled. We have $$ 0 = k^2 - k - 2 = (k - 1/2)^2 - 1/4 - 2 = (k - 1/2)^2 - 9/4 \iff \\ k = (1/2) \pm (3/2) \in \{ -1, 2 \} $$ Case $k = -1$: $$ \left[ \begin{array}{rrr|r} 1 & -2 & 5 & 2 \\ 0 & 1 & -4/3 & -1 \\ 0 & 0 & 0 & 0 \end{array} \right] \to \left[ \begin{array}{rrr|r} 1 & 0 & 7/3 & 0 \\ 0 & 1 & -4/3 & -1 \\ 0 & 0 & 0 & 0 \end{array} \right] $$ This gives the infinite many solutions $((-7/3)z, -1 + (4/3) z, z)$ where $z$ is an arbitrary real number.
Case $k = 2$: $$ \left[ \begin{array}{rrr|r} 1 & -2 & 5 & 2 \\ 0 & 1 & -4/3 & 0 \\ 0 & 0 & 0 & 0 \end{array} \right] \to \left[ \begin{array}{rrr|r} 1 & 0 & 7/3 & 2 \\ 0 & 1 & -4/3 & 0 \\ 0 & 0 & 0 & 0 \end{array} \right] $$ This gives the infinite many solutions $(2-(7/3)z, (4/3)z, z)$ where $z$ is an arbitrary real number.
Summary: