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So each matrix has $N^{2}$ elements, and so just by comparing each element we would be doing $O(N^{2})$ operations. Is there any other way to compare these two matrices such that the number of operations is less than $O(N^{2})$ or is the matrix comparison lower bound also $\Omega(N^2)$?

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    $\begingroup$ What kind of comparison are you looking at? $\endgroup$
    – Raphael
    Commented Feb 22, 2016 at 13:43

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A very simple adversary argument shows that when comparing two vectors of length $M$ (in your case, $M = N^2$), you must query (in the worst case) all positions of both vectors to know whether they are equal. That takes time $\Omega(M)$.

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