I'm solving a quiz and there is a question where I can't decide between two answers: Let $Ax = b$ be a system of equations with a square matrix $A$. Which of the following is true? (the other 2 are nonsense)
a) If $Ax = b$ has exactly one solution, so does $(A \cdot A)x = b$.
b) If $Ax = b$ doesn't have any solution, same applies for $(A \cdot A)x = b$.
It's easy to prove that once a matrix is regular, no power of it can make it singular. Yes it makes sense but what about b). I can't magically make the matrix contain $b$ in it's image or can I?
About (a): Since $Ax=b$ has exactly one solution, the kernel of $A$ is equal to $\{0\}$. But then $A$ is invertible. But then also $A^2$ is invertible, and $A^2x =b$ has exactly one solution.
The difference between (a) and (b) is the quantification 'exactly one' in (a) and 'no solution' (which is the negation of 'there is at least one') in (b).