Finding values of $a$ which this system of linear equations has one, no solution, or infinite solutions Given the following equation:
$x+ay -z = 1$
$x+y -az = a$
$2x-4y +2az = -2$
Determine the value of $a$ (if any) for which the system has no solutions.
Determine the value of $a$ (if any) for which the system has exactly one solution.
Determine the value of $a$ (if any) for which the system has infinitely many solutions
And in case the system has infinite solutions, find them
So I'd like to put the coefficients of $x,y,z$ into a matrix like the following:
$\begin{pmatrix} 1 & a & -1 \\ 1 & 1 & -a \\ 2  & 4 & 2a \end{pmatrix}$
And I want to do this by the method of checking when the determinant not equals to $0$.
So I calculate it and got to a place where $(a-1)(-4a-6)$ which means that $1$ or $-3/2$ will zero the determinant.
So I realize that if $a$ will be $0$ or $3/2$ then the system has no solutions,
but then I don't know how to continue,
How do I find out about the unique infinitely soulutions? I guess for unique solution $a$ needs to be not $0$ or $3/2$
but for some reason I can't put it all together..

Calculate the determinant of the system's matrix:
$$
\begin{vmatrix}
1 & a & -1 \\
1 & 1 & -a \\
2 & 4 & 2a \\
\end{vmatrix} = -4a^2+6a-2
$$
You must know that, if that determinant is different from $0$, then the system has unique solution. When it's $0$, it can either have infinite or none. Let's calculate the values which make the determinant be $0$:
$$-4a^2+6a-2=0 \ \implies \ a=\frac{1}{2},\ a=1.$$
So now you can conclude that the system has unique solution when $a\in\mathbb{R}\setminus\{\frac{1}{2},1\}$. Now the only work you have left is to study what happends when $a=\frac{1}{2}$ and what when $a=1$ separately.
