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In Exercises 9 and $10,$ mark each statement True or False. Justify each answer. a. In order for a matrix $B$ to be the inverse of $A,$ both equations $A B=I$ and $B A=I$ must be true.b. If $A$ and $B$ are $n \times n$ and invertible, then $A^{-1} B^{-1}$ is the inverse of $A B .$ c. If $A=\left[\begin{array}{ll}{a} & {b} \\ {c} & {d}\end{array}\right]$ and $a b-c d \neq 0,$ then $A$ is invertible.d. If $A$ is an invertible $n \times n$ matrix, then the equation $A \mathbf{x}=\mathbf{b}$ is consistent for $\operatorname{each} \mathbf{b}$ in $\mathbb{R}^{n}$ .e. Each elementary matrix is invertible.

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Algebra

Chapter 2

Matrix Algebra

Section 2

The Inverse of a Matrix

Introduction to Matrices

Dk B.

February 5, 2022

she is just telling to refer to theorems. no explanation of the solutions.

Harvey Mudd College

Baylor University

Idaho State University

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for this problem. We have a Siris of five statements and we need to show whether they're true or false, and we will justify each answer. So let's start with the first one. In order from Matrix B two b. The inverse of a both a times B has to equal I and B times A has to equal I Is that true? Well, yes, yes, it iss if both a times B and B times a result in the identity matrix, then that means that B is the inverse of a That is how we're defining the inverse. Um, you can see that specifically if you look at the text on page page 105 you'll see that it's highlighted there for you. And as long as both ways give you the identity, it is the inverse. Now you do need to show both. So it has to work both ways. But this one is true. Okay, Our second one says, if a and B are both square matrices and by ends and in vertebral, then the inverse of a times B equals the inverse of a times the inverse of B. Now this one is false and to see that you could look a the're, um six in your book, dear, um, six tells you what the order should be for this, which is, if a and B are convertible matrices when you take the inverse of the product, it should be the inverse of a times B is the inverse of b times the inverse of a This is what it should be. And as you can see, these are backwards and the one that were given. So the way it is in the in this problem is false. We need to flip thean versus being a on the right hand side. That would make it true. Okay, Next, I have a two by two matrix on the entries. They're going to be a, B, C and D. Now, if a B minus c D does not equal zero, then a is in vertebral in vertebral. Is that true? No, it's false to see that you can look at the're, um, number four, which tells you the correct order. It should be a D minus B. C. We're looking at these diagonals so that the product of aid times D minus B times C. That's what we're comparing 20 If that isn't zero, then a is convertible. In this case, though, we have the variables in the wrong order. So this one is false. Hey, next part D says if a is an in vertebral and by N matrix, then the equation a X equals B is consistent for each be in my, uh, in my domain and that one is true. In order to see that you can look at the're, um, number five from the book and the're, um number five tells us that this equation has a unique solution for each be within, uh, within my real numbers there. So this one is indeed true. Now, our last one, each elementary, each elementary matrix is in vertebral. And yes, this one is also true. In order to see that, you can turn to page 109 in your text and you will see fairly close to the top. It says that each elementary matrix e is in vertebral. And not only that, but the inverse of e is the elementary matrix of the same type that transforms e back into I. So yes, every elementary matrix is indeed convertible. So these are our five statements and which ones are true and which ones are false.

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