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 $(A+B)^{-1}+\left(A+A B^{-1} A\right)^{-1}$ .

Similar Problems and Solutions

ISI MStat PSB 2008 Problem 10 — Outstanding Statistics Program with Applications

Outstanding Statistics Program with Applications

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 $(A+B)^{-1}+\left(A+A B^{-1} A\right)^{-1}$ .

Similar Problems and Solutions

ISI MStat PSB 2008 Problem 10 — Outstanding Statistics Program with Applications

Outstanding Statistics Program with Applications

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