# Define parameter in linear system to get different number of solutions

For what values of the parameter $t$ does the following system of equations have

1. no solution
2. more than one solution
3. exactly one solution

$$(I):x+y+t\cdot z=-1$$ $$(II):3x+(t+1)y+(t-1)z=-1$$ $$(III):tx+2y+z=0$$

I'm not quite sure how to handle this. First of all I wrote that as an augmented coefficient matrix:

$$A= \begin{pmatrix} 1 & 1 & t &|&-1 \\ 3 & (t+1) & (t-1)&|&-1 \\ t & 2 & 1&|&0 \\ \end{pmatrix}$$ I suppose there'd be no solution if one of the lines showed inequality and exactly one if the rank of A is equal to the number of unknowns, but I don't know how to apply this here. Furthermore: what's the requirement for this system to have more than one solution?

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If the $\det A_0 \ne 0$ ($\operatorname{rank} A_0=3$), then the solution exists and is unique.
This gives you a cubic equation on $t$, whose solutions have to be investigated individually.
For each such $t$, the solution exists (and is not unique) if $\operatorname{rank} A = \operatorname{rank} A_0$ and there are no solutions otherwise.
Thanks so far sds! I'm not quite sure how to execute this tho, do I compute the determinante of the unaugmented coefficient matrix? $A_0$ puzzles me a bit – Rickyfox Jun 20 '13 at 15:08
yes, $A_0$ is the unaugmented coefficient matrix – sds Jun 20 '13 at 15:18