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# 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

Let A and B be two invertible $$n \times n$$ real matrices. Assume that $$A+B$$ is invertible. Show that $$A^{-1}+B^{-1}$$ is also invertible.

### Prerequisites

Matrix Multiplication

Inverse of a matrix

## 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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