Choose the correct(s): $A_{m*n}$ , $B_{m*n}$ and $X_{n*1}$ are matrices with real entries.
A) If $AX=O$, $\forall X$ , then $A=O$ (here $O$ represents zero matrix.)
B) $AX=BX$, $\forall X$ , if $A=B$.
C) $AX=BX, \forall X$ , only if $A=B$.
D) If $A_{n*n}$ and $X_{n*1}$ , then $AX=O$ and $A'X=O$ are of same nature in terms of trivial and non trivial solutions.
I found that D is true as $\rho(A)=\rho(A')$ so their homogeneous systems are of same nature. But I'm confuse for A,B,C.
EDIT ($1$):$A'$ is transpose of matrix $A$ and $\rho(A)$ is rank of $A$.
A)$\forall X,\ AX=0 \rightarrow A=0$
Proof:
The hypotesis tells us that $\ker(A)=n$. Thus, using rank's theorem: $\rho(A)=0$, which is true iff $A=0$
B)$A=B\rightarrow \forall X,\ AX=BX$
Proof:
$\begin{align} (A-B)X=0X=0\\ AX-BX=0\\ AX=BX \end{align}$
C)$\forall X,\ AX=BX \iff A=B$
Proof:
We have already proved half of the statement in B). It remains to show only the if part:
$\begin{align} AX=BX\ \forall X\\ (A-B)X=0\ \forall X\ \end{align}$
Thus the matrix $A-B$ satisfies the hypotesis of A), and thus $A-B=0$, and the statement is proved