Consider the system of equations $x$,$y$, and $z$,
$$2x+3y-z=p$$ $$x-2z=-5$$ $$qx+9y+5z=8$$
where $p$ and $q$ are real.
Find the values of $p$ and $q$ for which this system has:
(i) a unique solution
(ii) an infinite number of solutions
(iii) no solutions
I understand part (i) Here is how I did it:
The system of equations has a unique solution when the determinant is not equal to zero.
$$\begin{bmatrix} 2 & 3 & -1 \\ 1 & 0 & -2 \\ q & 9 & 5 \end{bmatrix} \neq 0$$
$$2(18)-3(5+2q)-1(9) \neq 0 \implies q \neq 2$$
I believe $p$ can be any value.
So the answer is : $q\neq 2$
I know that when $q = 2$. Then the system of equations don't have a unique solution or no solution.
I don't know the condition that separate the part (ii) and (iii) and
Hence, I don't know how to find the values of q. Please help me.
if $q=2$ and $p=-4$ we have $y=\frac{1}{2}(-1-x),z=\frac{5+x}{2}$
if $q\ne 2$ we get $x=-\frac{3(4+p)}{q-2},y=\frac{1}{6}{(5+2p-3x)} ,z=\frac{5+x}{2}$