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Exercises 15 and 16 concern arbitrary matrices $A, B,$ and $C$ for which the indicated sums and products are defined. Mark each statement True or False. Justify each answer.a. If $A$ and $B$ are $2 \times 2$ with columns $\mathbf{a}_{1}, \mathbf{a}_{2},$ and $\mathbf{b}_{1}, \mathbf{b}_{2}$ respectively, then $A B=\left[\mathbf{a}_{1} \mathbf{b}_{1} \quad \mathbf{a}_{2} \mathbf{b}_{2}\right] .$b. Each column of $A B$ is a linear combination of the columns of $B$ using weights from the corresponding column of $A .$c. $A B+A C=A(B+C)$d. $A^{T}+B^{T}=(A+B)^{T}$e. The transpose of a product of matrices equals the product of their transposes in the same order.

a) false b) false c) true d) true e) false

Algebra

Chapter 2

Matrix Algebra

Section 1

Matrix Operations

Introduction to Matrices

Campbell University

McMaster University

Baylor University

University of Michigan - Ann Arbor

Lectures

01:32

In mathematics, the absolu…

01:11

07:15

Exercises 15 and 16 concer…

06:37

In Exercises 17 and 18 , $…

04:42

03:09

Exercises $17-20$ refer to…

02:55

02:38

In Exercises 27 and $28, A…

02:09

04:21

00:23

In Exercises 21 and $22, A…

03:53

Mark each statement True o…

okay for this problem. We have a few statements that we need. Thio either prove as being true or false. So our first statement is that if a and B are two by two matrices where a is equal to a one vector A to vector and B is equal to be one vector B to vector in a times B is equal to a one vector B one vector and a two vector B to back. So that is false. Uh, we think about matrix multiplication as a superposition of transformations. If you have matrix a times matrix B times X So we have B act on X first on Vector X first and then act on the result with a then we'd have a times well, the result of multiplying a vector by a matrix or matrix by a vector. Um, we end up getting that would have be one vector times x one. Let's be to vector and X two plus dot, dot, dot Yeah, So when we are actually, that should result in matrix itself. My mistake. Actually, that should that should be. Plus, that should be those air, the columns of matrix. So we have Collins of The Matrix going up to E n vector x n. Then when we multiply in that matrix A You end up with a times B one x one up to you a times b n x n, which is equal to a Times B one vector up to a Times B n vector times x one dot dot dot x n as a vector of its own, which is equal to a Times B one vector dot dot dot a times B n vector. She's equaling or times X as a vector, which we said from the very start was equal to a B and X factor, which means that a B has to equal a Times B one vector data dot up to eight times bien vector. So that is false. Be that says each column of a B is a linear combination of the columns of Be using weights from the Correspondent column of A This is false as well. We have that a B that that statement would imply that a B is equal to be times a one up to be tens a N, which we just showed as not being the case. It's the exact opposite way around for see. We want to prove that a Times B plus a Times C is equal to a Times B plus C. We consider the element the A B plus a C that is equal to the element of a B plus the I J element of a C, which is equal to the sum from K equals one up to N a i k b k j plus the sum from K equals one to n of a i k c k j. Now we can put those two sums together into a single some. So we get the some from K equals one up to N a i k b k j plus ai que c k j. Next, we can factor out those ai K's if the some from K equals one up to end ai que times B k j yeah, plus C k J, which you can then recognize is B k J for C K J. That is the same thing as right down here. The KJ element of Matrix B plus C, which then means that this all together is equal to using the definition of the row column major multiplication that we know that that is equal to the Hi Jake element of a Times matrix B plus C. Yeah, so that is true for D. We then have that a transpose plus be transposed equals a plus b transposed. You think about it a transpose or if we look at the individual elements, So the i j the element of a transpose a transpose plus g be transposed. Excuse me equal to the I J element of a transpose, plus the element of the transpose. The I J element of a transpose is the same thing as the J either element of a similarly for B. So we get a J I plus B J I. Then if we consider the J. I element of a plus B all transpose I'm sorry, the I j element other i J that is equal to the element J I of a plus B element J I of a plus B is a J I plus BJ. So that shows our statement is true. Lastly for E, it's given as sort of sentence in the textbook, but I'll make it as an equation. We need to show whether or not. It's true that a be transposed is equal to a transposed Be transposed So this is false. We can prove this by the i J. The element of a transposed be transposed is equal to the sum from K equals one up to n of a transpose element i k and be transposed element KJ, which then is going to be equal to the sum from K equals one up to n o a transpose element I i ke is the same thing as a tricks a element k I be transposed element. KJ is the same thing as matrix B element J K. Now, since these are scale er's weaken swap the order that was a bad son K equals one up to n You have bjk I'm a k i. And now from the Row column definition of matrix multiplication that is the same thing as the J I element of be tens A with J I sten means that it is equal Thio the hi Jake element of be a let me make this nicely. Yeah, that means that it is the same thing as the I to me Hi, Jake. Element of be times a transpose. So we have that a transposed be transposed is equal to be a transpose, not a be transposed So the order is reversed.

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