Let $A \in \Bbb{R}^{n \times (n-1) }$ be of rank $n-1$, let $b\in \Bbb{R}^n$.
Let the system $Ax=b$ be incompatible.
Prove that $C^kx=0, C=[A,b]$ is determined for all $k\in \Bbb{N}$.
I can't use properties of the determinant here, and I kind of can see why this is true, but I have no idea on how to start a proof, hints would be appreciated.
The assumption $rank(A) = n-1$ implies that the columns $a_1, \dots, a_{n-1}$ of $A$ are linearly independent. Then since $Ax = b$ is incompatible, the vectors $a_1, \dots, a_{n-1},b$ are also linearly independent.
Otherwise one could find numbers $\alpha_1, \dots, \alpha_n$, not all zero, such that $$ \alpha_1 a_1 + \dots \alpha_{n-1}a_{n-1} + \alpha_n b = 0 $$
or in matrix notation $A \tilde \alpha = - \alpha_n b$, where $\tilde \alpha = (\alpha_1, \dots, \alpha_{n-1})^T$.
Then $\alpha_n$ cannot be zero since the columns of $A$ are linearly dependent. But if $\alpha_n \ne 0$, the system $Ax = b$ has the solution $x = -\alpha_n^{-1} \tilde \alpha$, so that is also impossible.
Consequently the matrix $C = [A,b]$ has rank $n$ and is invertible. It follows that all powers $C^k$ are also invertible and the systems $C^kx = 0$ all have only the trivial solution $x = 0$.