# Solve $Bx = [ 1 2 1 ]^T$ for each $k \in \{ 1, 2, -3 \}$?

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

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 $k=2$.