Consider the following circuits:

Which one of the circuits A, B, C, D (if any) are equivalent to the one at the top?

With thanks to Elmar for simplifying matrix printing here's how we proceed:

So that's the matrix we need to match. We repeat with each circuit:

So we see B is equivalent, while A is not. We look at the other two:

This shows that neither C, nor D, are equivalent with the initial circuit.

A question to ponder: can we do the same by hand (e.g., in the last case).

The answer is: yes, via matrix multiplication (but we have to be careful, a little).

Now let me state here two exercises for practice:

This should be easy, as it's exactly like the one above.

This, on the other hand, brings up new topics for discussion.