I’ve watched this video a few times, and I still can’t understand the why the solution for system 1 doesn’t also apply to systems 2, 3 and 4.

The instructor says for systems 2 and 3 that *c* is zero because changing it from 2 to 3 doesn’t affect the result of zero. He then goes on to say that *a = -b* because *a + b* must also equal zero. Why couldn’t *a =0* and *b = 0* too, just as for system 1? That would also result in zero. A similar argument could be made for system 4.

Also, when he moves onto the matrices for systems 2 and 3, he says “The matrices are singular or non-singular based on if the systems are singular or not singular,” which seemed a bit circular.

Apologies if the answer is obvious, but i found the explanation in the video very confusing.

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Hi @B_Dapaah!

You are correct that, in system 2 and 3, a = 0 and b = 0 is a solution. However, different from system 1, there are other solutions as well (any pair (a,b) such that a = -b). Note that a = b = 0 is a particular case of a = -b.

System 4 is a bit different in argumentation, because now in every system, there are different constants in each variable, but they are all multiples of the first system. Note that a = b = c = 0 is also a solution in this case, but not the only one.

About the definition of singular matrices, it is not circular because it is defined singular system as a system with no solutions, or infinite solutions.

There is no need to apologize! I hope this answer helps you. Please let me know if you still don’t understand.

Cheers,

Lucas

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Hi, @lucas.coutinho

Thank you for taking the time to reply to my question. Your explanation has made it somewhat clearer for me. I will sit down with a pad and work through it. If I have any other questions, I’ll come back to bother you if that’s ok

i am not still getting the same confusion