PLEASE HELP ?!?!? MATRIX QUESTION!!!

One thing about the multiplicative inverses is if you multiply them you get the identity.

could you show me? I'm still a lil confused with this

Well, do you know how to do matrix multiplication?

no I don't think so

I made one small mistake in there I need to fix, but that is the principal of matrix multiplication.

There are two ways to find if two matrices are inverses. 1) Multiply them and see if you get the identity. 2) Take one of them, find its inverse, and see if that is the other one. Both work just fine.

Most inverses do not look much like the original. Option A and D are too similar, so I would test them last. B and C look more different, like inverse pairs normally do, so I would test them first.

So c?

Did you multiply it and see? Multiplication test: \(\begin{bmatrix} a&b\\c&d \end{bmatrix} \begin{bmatrix} e&f\\g&h \end{bmatrix} = \begin{bmatrix} 1&0\\0&1 \end{bmatrix}\) If that is true, then they are inverses.

For a 2X2 you can just plug it in by positions: \(\begin{bmatrix} a&b\\c&d \end{bmatrix} \begin{bmatrix} e&f\\g&h \end{bmatrix} = \begin{bmatrix} ae+bg& af+bh\\ ce+dg& cf+dh \end{bmatrix}\) If you just make that right hand side and it becomes \(\left[\begin{smallmatrix} 1&0\\0&1 \end{smallmatrix}\right]\) then you have found the inverse.

Wow my brain is still a little bungled could you walk me through this question? been trying to figure it out since yesterday

Well, you said C. So, if I take the abcdefgh thing, that would be: \(\begin{bmatrix} (5)(\frac{1}{2})+(8)(-\frac{4}{5})& (5)(-\frac{1}{2})+(8)1\\ (5)(\frac{1}{2})+(10)(-\frac{4}{5})& (5)(-\frac{1}{2})+(10)1 \end{bmatrix}\)

Now, the great thing is, you DON'T have to do all that math. Only need to find one failure and you stop. Well, right there I see a problem.

\(5(-\frac{1}{2})+8\implies -\frac{5}{2}+8\) That, in the upper right, does not look like 0. Therefore it can't be C.

ok I got it so it's A right? C was my second choice but I wasn't sure

Not A. I have to head out now. The joys of work... But if you put the others into that formula, you will find one that comes out as the identity matrix.

hmm so beetween B and D

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