I had to provide an example of an idempotent matrix. That's the kind of matrix that yields itself when multiplied to itself. Much like 0 and 1 in scalar multiplication (1 x 1 = 1).
It is not so easy to predict the result of a matrix multiplication, especially for large matrices. So, instead of settling with the naïve method of guessing with trial and error, I explored the properties of a square matrix of the order 2.
Since 0 cannot be divided by 0, I could not divide 0 by either term unless it was a non-zero term. Thus, I had two possibilities, to which I called case A and B.
As you can see, I could not use the elimination method in an advantageous manner for this case.
I couldn't get a unique solution in either case. That is because there are many possible square matrices that are idempotent. However, I don't feel comfortable to intuit that every 2 X 2 idempotent matrix has one of only two possible numbers as its first and last elements.
Others’ take on it
So given any 2 X 2 idempotent matrix and its first three elements, you can find the last element unequivocally with this formula.
Conclusion
I wonder if multiples of matrices that satisfy either case are also idempotent. Perhaps I will see if I can prove that in another post.
In the next lecture, professor Venkata Ratnam suggested using the sure-shot approach of a zero matrix. And I was like “Why didn’t I think of that”?