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Where $B$ is the matrix: $ \begin{pmatrix} 1 &1 &k\\ 2 &k &1 \\ k &2 &2 \end{pmatrix} $

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    Yes, that's correct for $k=1$. If you want to be sure I see a comment directed to me, you have to write @Gerry.2012-09-19

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The determinant of $B$ is $-6+7k-k^3.$ Hence $B$ is singular for $k \in \{-3,1,2\}$. By Cramer's rule, the unknown $x_1$ is indetermined if $k=1$, since $ \det \begin{pmatrix} 1 &1 &k \\ 2 &k &1 \\ 1 &2 &2 \end{pmatrix} =0 $ when $k=1$ (and $k=5$, but we do not care). Hence you have infinitely many solutions for $k=1$.

If you try to solve for $x_2$ and $x_3$, you will see that no solutions can exist for $k=-3$ or $k=2$.

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    Thanks for clearing it up, the explanation was helpful2012-09-19