Practicing Success
If A and B are two non-singular matrices which commute, then $\left(A (A+B)^{-1} B\right)^{-1}(AB) =$ |
$A+B$ $A^{-1}+B$ $A^{-1}+B^{-1}$ None of these |
$A+B$ |
$\left(A (A+B)^{-1} B\right)^{-1}$ $=B^{-1} (A+B)A^{-1}$ $[∵(ABC)^{-1}=C^{-1}B^{-1}A^{-1}]$ $=B^{-1}AA^{-1}+ B^{-1}BA^{-1}$ $=B^{-1}I+IA^{-1}=B^{-1}+ A^{-1}$ $∴\left(A (A+B)^{-1} B\right)^{-1}(AB) =(B^{-1}+ A^{-1})AB$ $=B^{-1}(AB) +A^{-1}(AB)$ $=B^{-1}(BA) +A^{-1}(AB)$ $[∵AB=BA]$ $=(B^{-1}B)A+(A^{-1}A)B$ $=IA+IB= A + B$. |