Let A = [a_ij]_n×n and B = [b_ij]_n×n Then which of the following is/…

The correct answer is Option (2) → (B) and (C) only

Given square matrices $A=[a_{ij}]_{n\times n}$ and $B=[b_{ij}]_{n\times n}$, evaluate each statement.

(A) $AB=BA$

$\text{Not necessarily true. Example: } A=\begin{pmatrix}0 & 1\\[4pt]0 & 0\end{pmatrix},\ B=\begin{pmatrix}0 & 0\\[4pt]1 & 0\end{pmatrix}.$

$AB=\begin{pmatrix}1 & 0\\[4pt]0 & 0\end{pmatrix}\neq BA=\begin{pmatrix}0 & 0\\[4pt]0 & 1\end{pmatrix}.$

(B) $(AB)^{-1}=B^{-1}A^{-1}$

If $A$ and $B$ are invertible (so $(AB)^{-1}$ exists), then

$ (AB)(B^{-1}A^{-1}) = A(BB^{-1})A^{-1} = A I A^{-1} = I $

and similarly $(B^{-1}A^{-1})(AB)=I$. Hence the identity holds whenever $A,B$ are invertible.

(C) $(AB)^T = B^T A^T$

Transpose identity is valid for all conformable matrices. Direct verification:

Entry at $(i,j)$ of $(AB)^T$ equals entry $(j,i)$ of $AB$, which is $\sum_k a_{jk} b_{ki}$, while entry $(i,j)$ of $B^T A^T$ equals $\sum_k b_{ki} a_{jk}$, the same sum. Thus equality holds.

(D) $AB=0 \Rightarrow A=0$ or $B=0$

False in general. Counterexample:

$A=\begin{pmatrix}1 & 0\\[4pt]0 & 0\end{pmatrix},\ B=\begin{pmatrix}0 & 0\\[4pt]0 & 1\end{pmatrix}.$

$AB=\begin{pmatrix}0 & 0\\[4pt]0 & 0\end{pmatrix}$ while $A\neq 0$ and $B\neq 0$.