[Solved]

dexter

Somebody, please help me find the inverse of this matrix. I've tried so many times and keep getting nonsense solutions. The matrix is {{5},{4},{3}}. Those are the rows ie {{A11,A12,A13},...}.

Please, if anyone would be so kind as to give me the first few row operations then I would be so very grateful. So far I've tried:
R1-R2, R3-8*R1, (1/3)*R2, R3-12*R2, R1+R2, R1-(2/3)R3, R2+(4/3)R3.

I also tried R1-R2, R3-R2, R2-R3, R2-4*R1 and so on...

I have tried both of these orders of operations multiple times and it's driving me crazy. I've checked the actual matrix multiple times to make sure I have written it correctly. Any help would be greatly appreciated.

Xego

Unfortunately, this inverse is going to be a little messy. The first thing I would do is replace R1 with R1 - R2 so you have a 1 in the first column. Then do R2 - 4R1 and R3 - 8R1 and that will reduce the first column. After that you're going to have to divide a 7 out of R2 which'll make it (0,1,-12/7) which isn't fun to deal with but it's what you have to do. Then just do what you have to do to finish reducing the matrix and make sure you're doing identical operations to the identity matrix to get your inverse.

tarnok

It's been a very long time but iirc what you're trying to do is change A in the augmented matrix into the identity matrix by line operations and then I should have changed into the inverse matrix, right?

If so these operations should do it, assuming I didn't make an arithmetic mistake.

R2-(4/5)R1
R3-(8/5)R1
R1-(10/7)R2
R3-(4/7)R2
R1 -(10/67)R3
R2-(84/335)R3
(1/5)R1
(5/7)R2
(7/67)R3

You could make the arithmetic a little easier by starting out with (1/5)R1, (1/3)R2 and (1/5) R3, but where's the fun in that?

dexter

Thank you, thank you, thank you! That was such a pain in the neck, I always fuck up these simple operations. Thanks again!