If $\displaystyle A$ and $\displaystyle B$ are two square matrices of the same order such that $\displaystyle AB=A+B$ then prove that A and B commute i.e. AB=BA
If $\displaystyle A$ and $\displaystyle B$ are two square matrices of the same order such that $\displaystyle AB=A+B$ then prove that A and B commute i.e. AB=BA
If $\displaystyle AB=A+B$ then $\displaystyle (A-I)(B-I)=I$ (I is the identity matrix). So $\displaystyle B-I$ is the inverse of $\displaystyle A-I$ and therefore commutes with it. Thus $\displaystyle (B-I)(A-I) = I$, from which $\displaystyle BA = A+B = AB$.
So if $\displaystyle CD=I$ does it always mean that $\displaystyle C$ and $\displaystyle D$ are inverse of each other.What if any one of $\displaystyle C$ or $\displaystyle D$ is singular matrix
So if $\displaystyle CD=I$ does it always mean that $\displaystyle C$ and $\displaystyle D$ are inverse of each other.What if any one of $\displaystyle C$ or $\displaystyle D$ is singular matrix
If $\displaystyle CD = I$ then C and D cannot be singular. One way to see this is to notice that the determinant is multiplicative: $\displaystyle \det(C)\det(D) = \det(I) = 1$, so that $\displaystyle \det(C)$ and $\displaystyle \det(D)$ cannot be 0.