ISI MStat PSB 2006 Problem 1 | Inverse of a matrix

This is a very beautiful sample problem from ISI MStat PSB 2006 Problem 1 based on Inverse of a matrix. Let's give it a try !!

Problem- ISI MStat PSB 2006 Problem 1

Prerequisites

Solution :

We are given that A,B,A+B are all invertible real matrices . And in this type of problems every information given is a hint to solve the problem let's give a try to use them to show that \( A^{-1}+B^{-1}\) is also invertible.

Observe that ,\( A(A^{-1}+B^{-1})B= (B+A) \Rightarrow |A^{-1}+B^{-1}|=\frac{|A+B|}{|A| |B| } \) taking determinant is both sides as A+B , A and B are invertible so |A+B| , |A| and |B| are non-zero . Hence \(A^{-1}+B^{-1} \) is also non-singular .

Again we have , \( A(A^{-1}+B^{-1})B= (B+A) \Rightarrow B^{-1} {(A^{-1}+B^{-1})}^{-1} A^{-1} = {(A+B)}^{-1} \) , taking inverse on both sides .

Now as A+B , A and B are invertible so , we have \( {(A^{-1}+B^{-1})}^{-1}=B {(A+B)}^{-1} A \) . Hence we are done .

Food For Thought

If \( A \& B\) are non-singular matrices of the same order such that \( (A+B) \) and \( \left(A+A B^{-1} A\right) \) are also non-singular, then find the value of \( (A+B)^{-1}+\left(A+A B^{-1} A\right)^{-1} \).

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