linear algebra - Is there always a mapping from invertible $A$ to any $B in M_n(Bbb R)$?

Let $A, B$ be $n\times n$ matrices, then
$1)$ If $A$ is invertible then for every $B$ exists a matrix $X \in M_n(\Bbb R)$ such that $AX = B$.
$2)$ If for every $B$ there exists a matrix $X \in M_n(\Bbb R)$ such that $AX = B$ then $A$ is invertible.

For $1)$ I started with: $AX=B \implies X=A^{-1}B$

But I'm not sure I can do this:
$$A(A^{-1}B)=B$$
$$(AA^{-1})B=B$$ --> not sure about this
$$IB = B$$

Am I right about the first part?

How should I prove the second part? thanks

Answer

The steps $A(A^{-1}B) = (A A^{-1})B = IB = B$ are okay because matrix multiplication is associative.

For the second part choose $B = I$. Now there exist $X$ s.t. $AX = I$, so $A$ is invertible.