Given $A,B$ square matrices and $AB = I + B$ : Prove $AB=BA$

Question

Given $A,B$ square matrices of order $n$ greater or equal to $2$ , $AB = I + B$ : Prove $AB=BA$.

what I am trying to do is :
$AB = I+B \implies AB-B=I \implies (A-I)B = I$

but then I got stuck XD any ideas?

Does this answer your question? How can i prove that $AB=BA$? the hypothesis is the same if you exchange $A,B$ and replace one of them with its opposite. — Anne Bauval, Apr 12 '23 at 08:22

score 2 · Accepted Answer · answered Apr 12 '23 at 08:15

2

$AB=I+B$

$AB-B=I$

$(A-I)B=I$

So $(A-I)$ and $B$ are inverse of each other

$\implies B(A-I)=I$

$\implies(A-I)B=B(A-I)$

$\implies AB-B=BA-B$

So, $AB=BA$

answered Apr 12 '23 at 08:15

It was a duplicate (up to notations). – Anne Bauval Apr 12 '23 at 08:23
for (A−I) and B to be inverse of each other we must show that B(A-I) also equal to I and not assume it, don't we? – nicolachter Apr 12 '23 at 08:26
$XY=I$ already shows that $X^{-1}=Y$, i.e both are inverses, implying that $XY=YX$. There is nothing we are assuming here. – Apr 12 '23 at 08:28
on my textbook we are told that To prove that two matrices are inverses of each other, you need to show that when you multiply them together in either order, you get the identity matrix. i.e AB = BA = I, so just one direction is not enough it says – nicolachter Apr 12 '23 at 08:36
1

https://math.stackexchange.com/questions/3852/if-ab-i-then-ba-i should help you there – student91 Apr 12 '23 at 08:38

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