Hey guys what the best type of filter for my pet goldfish, he’s in a 15 gallon tank and I’m thinking about getting him a pet snail

Oh I realized my mistake, I falsely tried to distribute an exponentiation right there in step one lmao.

Thanks vulc!

Follow-up question: Do I take it then that there is actually no way to compute this without knowing matrix A? Is the only way to solve this problem to guess the answer intuitively?

I'd show it like this :
X*(A+I) = I+A^-1
X*(A+I) = (A^-1)*(A+I)
Which means X = A^-1 since you can multiply both sides by (A+I)^-1 on the right.
EDIT : A+I has to be invertible for this to be true, if not then I don't see how you can solve from that without knowing A.

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Been a few years since I did algebra, but closest I can get it by multiplying by A on the right. then you have

X (A^2 + A) = A + I . Since A commutes with itself you can write this as
XA(A + I) = A + I

[edited bc I suck. trippa got it right below me, unless A = -I (in which case any X solves the equation) this solves to XA = I]

X(A+I)=I+A^-1
factor out A from the first part
XA(I+A^-1)=I+A^-1
hence XA = I

quick edit: this assumes I + A^-1 is not 0
infact if A = -I the original equation is:
X(-I+I)=I + -I
or X0 = 0 so every X is a solution, so X = A^-1 will technically always work

edit im retarded

ok after thinking some more. if A is invertible, then it is row equivalent to I. you would think adding two row equivalent matrices would be row equivalent so A would be row equivalent to (A+I), but theres actually an exception here...when A has a row which is equal to the same row in -I, then adding that to I will mean that to get to (A+I) from A, you must perform the row operation of adding the negative of a row to itself, which is the same as multiplying by 0. this isnt a valid row operation so they aren't row equivalent. however this is the only exception.

so this means that either (A+I) is invertible, or it has at least one row of all 0's.

plugging this back into the equation XA(A + I) = A + I doesn't really help though. if all of the rows are 0 (A=-I), then it obviously solves for all X, but if only some rows are 0, I can't think of anything that would mean, especially since A+I is on the right side of the multiplication. if it was on the left side then that would mean an entire row of the product would also be 0 which would be equal to A+I, but on the right side that doesnt seem to necessarily be the case. you can use distributive property on other side to prove (A+I)A has the same zero rows, but you’re still left multiplying that by X so it doesnt help.

https://i.imgflip.com/262r6i.jpg

wait a second once you get XA(A + I) = A + I isnt that just the solution

doesnt matter if A+I is invertible or not, you know XA=I because I is the only matrix that would solve IC=C for C=(A+I) unless C=0

so since XA=I you can multiply both sides by A^-1 and get X=A^-1

i made this way too complicated in my last post rofl

of course theres still a case where A+I=0 where X can equal any matrix it wants

thats what i said, i just explained the C = 0 case.

Thanks everyone! Yeah aside from the distribution error I made to arrive at my wrong results, I was mostly confused why I couldn't solve it algebraically all the way to X=A^-1. But I guess I was just in the commutative mindset that every linear system should be computable... but I understand now that with matrices you have to make assumptions such as (A+I) is invertible to be able to do that. The problem was probably designed to trick you into doing a whole lot of unnecessary steps and then realizing you should just try out a few simple values first.