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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 $3 \times 3$ and $B=\left[\mathbf{b}_{1} \mathbf{b}_{2} \mathbf{b}_{3}\right],$ then $A B=$ $\left[A \mathbf{b}_{1}+A \mathbf{b}_{2}+A \mathbf{b}_{3}\right]$b. The second row of $A B$ is the second row of $A$ multiplied on the right by $B .$c. $(A B) C=(A C) B$d. $(A B)^{T}=A^{T} B^{T}$e. The transpose of a sum of matrices equals the sum of their transposes.

a) Falseb) Truec) Falsed) Falsee) True

Algebra

Chapter 2

Matrix Algebra

Section 1

Matrix Operations

Introduction to Matrices

Missouri State University

Oregon State University

Harvey Mudd College

Lectures

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Okay. This question gives us a number of true false statements, and they want us to determine whether they're true or false and justify. So the 1st 1 says, if a and B or three by three matrices and B is the column vectors B one, B two and B three and A B is, uh, the column vector a B one plus a column vector a B two plus the column. Vector a B three and this is false. Um, and despite the definition of matrix will publication A. B is equal to a B one, as you call him. Maybe too, as a column in a B three, as you call him. Ah, this matrix here is actually a one by one ah matrix, which is just equal to like a scaler. It's just a scaler. It's not actually a matrix. Okay, The 2nd 1 is says that the second row of a B is the second row of a multiplied by the right on the right by B. And the trick here is that we're talking about the second row of a B, not the second column of baby. And so this is actually true and we can see that hereby, if I just do, I just do a two by two matrix, for example. It's a 11 A 12 a 21 and then a 22 multiplied by the matrix B 11 be want to be to one and b 22 Um, the second row here will be equal to 8 to 1 times. Um, be 11 plus 8 to 2 times be to one, and then the other entry will be a 21 times be one too. And our plus a 22 times beat you too. And you can see here that, um, second row entries of a have been multiplied by every element of B on the right. So the statement's true. Okay. And the 3rd 1 a b times C is equal to a C times be. This is false again. Um, and it's because matrix multiplication is not communicative. Which means that a B is not equal to B C or ah b A. Um and this is necessary thio for this equality to see that weaken do an actual formal proof and we'll do it proved by contradiction. And so we suppose it's true. All right, so we suppose that a B time see is in fact equal to hey c a Times B, and we're also going to assume that the inverse of a exists, um, which is not necessarily always true. So this isn't really a really good proof of this, but, um, in the case that a inverse does exists weaken, continue the proof. And so, by competitivity, this statement can become a times or by associative ity. So we're not competitivity associative me. This statement could become a times b c is equal to, uh, a time C b, um And then if we multiply on the left by a inverse both sides, um, we these cancel out because a inverse times a is just the identity and we get left with BC is equal to C b. So here this is showing that it's necessary for ah, majors want complication to be communicative in order for this equality toehold. But since matrix multiplication is not communicative, this equality doesn't hold, Which means that Arthuis assumption that we made supposing it's true must be false because everything that we did here was equal. And so the full statement then is false. And that's Ah, a quick entrance of the proof by contradiction method. Okay, so the 4th 1 is this statement here eight times be the transport of a times B is equal to the transport of eight times the transfers would be, and this is false again. And I can prove this by just doing basically a size analysis of each The Matrix is the matrix or the matrices on each side of the equality. So let's assume that a is, um and a que by our matrix. And then, um, for a b to be defined, we need that, um, be is, um, on our by s Matrix because we need this number to match this number. So then that means that a B is a, um, Cube. I ask majors. I shouldn't even need to do this because, um, if we just look at what a transpose is if a is Q B r than a transpose is, um our bike you and similarly be transpose is going to be ah s by our and so a transpose times be transposed would be a our bike. You times s by our And so since these numbers don't necessarily match up, this multiplication is not to find. And so the statement is false as well

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