# Thread: AB=A+B;Prove A and B commute

1. ## AB=A+B;Prove A and B commute

If $A$ and $B$ are two square matrices of the same order such that $AB=A+B$ then prove that A and B commute i.e. AB=BA

2. Umm working on it.

$AB = A+B$

$A(BB^{-1}) = (A + B)B^{-1}$

$A = (AB^{-1} + I)$

$BA = BAB^{-1} + B$

...

3. Originally Posted by pankaj
If $A$ and $B$ are two square matrices of the same order such that $AB=A+B$ then prove that A and B commute i.e. AB=BA
If $AB=A+B$ then $(A-I)(B-I)=I$ (I is the identity matrix). So $B-I$ is the inverse of $A-I$ and therefore commutes with it. Thus $(B-I)(A-I) = I$, from which $BA = A+B = AB$.

4. So if $CD=I$ does it always mean that $C$ and $D$ are inverse of each other.What if any one of $C$ or $D$ is singular matrix

5. Originally Posted by pankaj
So if $CD=I$ does it always mean that $C$ and $D$ are inverse of each other.What if any one of $C$ or $D$ is singular matrix
If $CD = I$ then C and D cannot be singular. One way to see this is to notice that the determinant is multiplicative: $\det(C)\det(D) = \det(I) = 1$, so that $\det(C)$ and $\det(D)$ cannot be 0.