If $A=\begin{vmatrix}a & d & l\\b & e & m\\c & f & n\end{vmatrix}$ and $B=\begin{vmatrix} l & m & n \\a & b & c\\d & e & f\end{vmatrix}$, then

A= B

The correct answer is Option (1) → $A=B$

$A=\begin{vmatrix}a & d & l\\b & e & m\\c & f & n\end{vmatrix}$

let $A=A^T$ so $A=A^T=\begin{vmatrix}a & b & c\\d & e & f\\l & m & n\end{vmatrix}$

so $C_1↔C_2=-\begin{vmatrix}l & m & n\\d & e & f\\a & b & c\end{vmatrix}$

$C_2↔C_3=-1×(-1)\begin{vmatrix}l & m & n\\a & b & c\\d & e & f\end{vmatrix}=B$

$⇒A=B$