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Suppose $A$ is an $m \times n$ matrix and there exist $n \times m$ matrices $C$ and $D$ such that $C A=I_{n}$ and $A D=I_{m} .$ Prove that $m=n$ and $C=D .[\text { Hint: Think about the product } C A D .]$

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Algebra

Chapter 2

Matrix Algebra

Section 1

Matrix Operations

Introduction to Matrices

Campbell University

Baylor University

Lectures

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okay for this problem. We have a is an m by n matrix, see as an n by M matrix and D as an n by M matrix, Where, um, see a is equal to the and identity matrix and a D is equal to the M identity matrix. We want to prove that C equals D and n equals M. So what we can do here is consider the product C a d. We'll see a d we know that that could be done as c times A than whatever see times a is times D. Well, you know that C times a is the an identity matrix. So we have i n times d Now we know that i n times d is always going to be equal to D because it's multiplying. Um a n by n matrix by a compatible e dimension, uh, identity matrix. So we have that c a. D is just going to be the same thing as deep then. Also, you can think about C A. D as being see left multiplying. Whatever a times deeds, we know that eight times D is the m dimensional identity matrix. So that means that C A d is also equal to see. But if C equals ch and D equals C a d, then that means that C equals D Then additionally, what we can think about is that yeah, we had that c a d equals the n dimensional identity matrix which equals see times the m dimensional identity matrix. So we know that since C is a n by n matrix and the I am is a compatible that mentioned identity matrix, then that means that I m d is equal to i m times c But we also know that d equal c So we have that i m d is equal to I am times see or sorry. Yeah, that I m d rather is equal Thio I am times d then means that we have to have that I m is equal to I am means that and has to eat

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