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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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I would swap rows so that the last becomes the first one. Then I'd apply Gauss' reduction. –  Siminore Sep 18 '12 at 8:16
    
Roxana, are you saying that you know how to bring a matrix to reduced row-echelon form, but that you don't know how to read off the solutions once you've done that? –  Gerry Myerson Sep 18 '12 at 8:57
    
Gerry, would I be right in saying that where k=1, x1+x4=0 and x2+x3=0 so that there are infinitely many solutions but there are no solutions where k=2 and k=-3? –  Roxana Sep 18 '12 at 10:55
    
@Roxana Why do you have four unknowns? –  Siminore Sep 18 '12 at 11:42
    
Sorry, would x1=1 and x2+x3=0 be correct? –  Roxana Sep 18 '12 at 11:48

1 Answer 1

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 helpful –  Roxana Sep 19 '12 at 0:27

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