\begin{bmatrix} a & 0 & b & 2 \\ a & a & 4 & 4 \\ 0 & a & 2 & b \\ \end{bmatrix}
Find the values of a and b such that the system has:
a) A unique solution
So I think I got this although I'm not sure.
$|A| \not= 0$
reduced it down to:
\begin{bmatrix} 1 & 0 & b/a \\ 0 & 1 & 4-1/a \\ 0 & 0 & b-2 \\ \end{bmatrix}
Therefore $b\not=2$ and $a\not=0$ Do I mention anything else here?
b) A two-parameter solution
This is the one I'm having trouble with. I don't know how to answer it and can't find anything telling me how to.
I got this from simplifying everything down and assuming $y=v$ and $z=w$
$$x=\frac{4-av-4w}{a}$$
Therefore $a\not=0$ and $b$=all real