I was trying to solve the following exercise but I couldn't get anywhere:
Find the values of a and b for which the following system has infinite solutions:
$2x+3y-z=a$
$3x-by+z=1$
$ax-y-z=2$
I tried to solve the problem using the determinant, I mean we know that $Ax=b$ has infinite solutions if $|A|=0$ but at the end I get an expression that I'm not able to reduce.
I would really appreciate any help or advice you could give me.
Many thanks
Idea
Write $$A=\begin{pmatrix} 2 & 3 & -1 \\ 3 & -b & 1 \\ a & -1 & -1\end{pmatrix}, B=\begin{pmatrix} 2 & 3 & -1 & a \\ 3 & -b & 1 & 1 \\ a & -1 & -1 & 2\end{pmatrix}.$$
The system has infinitely many solutions if and only if $ran(A)=ran(B)<3.$ Since $A$ has a 2-order minor that non vanishes, $$A=\begin{pmatrix} \color{red}{2} & 3 & \color{red} {-1} \\ \color{red}{3} & -b & \color{red}{1} \\ a & -1 & -1\end{pmatrix},\begin{vmatrix}2 & -1 \\ 3 & 1\end{vmatrix}\ne 0,$$ the only posibility is $ran(A)=ran(B)=2.$ So, that happens if and only if
$$\begin{vmatrix} 2 & 3 & -1 \\ 3 & -b & 1 \\ a & -1 & -1\end{vmatrix}=\begin{vmatrix} 2 & -1 & a \\ 3 & 1 & 1 \\ a & -1 & 2\end{vmatrix}=0.$$ I hope you can finish.