For a real number $\alpha$, if the system of linear equations $\begin{bmatrix} 1 & \alpha & \alpha ^{2} \\ \alpha & 1 & \alpha \\ \alpha ^{2} & \alpha & 1 \end{bmatrix}$ $\begin{bmatrix} x \\ y \\ z \end{bmatrix}$ $ = \begin{bmatrix} 1 \\ -1 \\ 1 \end{bmatrix}$ has infinitely many solutions, find $\alpha$.
Let
$A=\begin{bmatrix} 1 & \alpha & \alpha ^{2} \\ \alpha & 1 & \alpha \\ \alpha ^{2} & \alpha & 1 \end{bmatrix}$
$X=\begin{bmatrix} x \\ y \\ z \end{bmatrix}$
$D=\begin{bmatrix} 1 \\ -1 \\ 1 \end{bmatrix}$
The equation can be written as $AX=D$
I learnt that this system(for $D \neq 0$. i.e. non homogenous system of equations):
has a unique solution if $A$ is a non-singular matrix($|A| \neq 0$)
has infinite number of solutions if $A$ is a singular matrix ($|A|=0$) and $adj(A)*D=0$.
has no solution if $A$ is a singular matrix ($|A|=0$) and $adj(A)*D \neq 0$.
Solving for $|A|=0$, we get $\alpha = \pm 1$.
But here comes my doubt. In this case, $adj(A)=0$ (null matrix) for both $\alpha = \pm 1$. So $adj(A)*D=0$ for both those values. So according to what I learnt, for both $\alpha = \pm 1$ there should be infinite solutions to this system of equations.
But on expanding the matrix into a system of linear equations, it is easily observed that for $\alpha=1$ there are no solutions and for $\alpha=-1$ there are infinite solutions.
I feel that the condition $adj(A)*D$ equal to or not equal to 0 for infinite or no solution respectively is valid only if $adj(A) \neq 0$. Is this correct?
How can we deal with this case without having to expand it into a system of linear equations and verifying with values? (as that can get very tedious at times)
P.S: I have verified all my calculations with a calculator as well so there are no calculation mistakes.