But note that $I_4$ is nothing but $n$x$1$ vectors stacked side by side
$I_4 = \begin{bmatrix} 1 & 0 & 0 & 0\\ 0& 1& 0& 0\\ 0& 0 &1 &0 \\ 0& 0 &0 &1 \end{bmatrix}$
Also note the behaviour on passing these $n$x$1$ vectors to an arbitrary matrix transformation $A=\begin{bmatrix} a & b& c & d\\ e & f & g & h\\ i & j & k & l \\ m&n &o &p \end{bmatrix}$
$Then \; A\begin{bmatrix} 1\\ 0\\ 0\\ 0 \end{bmatrix} = \begin{bmatrix} a & b& c & d\\ e & f & g & h\\ i & j & k & l \\ m&n &o &p \end{bmatrix} \begin{bmatrix} 1\\ 0\\ 0\\ 0 \end{bmatrix} = \begin{bmatrix} a\\ e\\ i\\ m \end{bmatrix} $
$Similarly, A\begin{bmatrix} 0\\ 1\\ 0\\ 0 \end{bmatrix} = \begin{bmatrix} b\\ f\\ j\\ n \end{bmatrix} .. and \; so \; on$
So if we interchange first two rows of the $I$ matrix, we simply interchange the first two column vectors of $I$ and hence of A..
So the answer is $B$
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