exercises-15-and-16-concern-arbitrary-matrices-a-b-and-c-for-which-the-indicated-sums-and-products-a

Question

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 $2 	imes 2$ with columns $\mathbf{a}_{1}, \mathbf{a}_{2},$ and $\mathbf{b}_{1}, \mathbf{b}_{2}$ respectively, then $A B=\left[\mathbf{a}_{1} \mathbf{b}_{1} \quad \mathbf{a}_{2} \mathbf{b}_{2}ight] .$
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.

Lucas Finney · Accepted Answer

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&#x27;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&#x27;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&#x27;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&#x27;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&#x27;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&#x27;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&#x27;s given as sort of sentence in the textbook, but I&#x27;ll make it as an equation. We need to show whether or not. It&#x27;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&#x27;s weaken swap the order that was a bad son K equals one up to n You have bjk I&#x27;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.

Mike Verwer · Answer

Okay. This question gives us a number of true false statements, and they want us to determine whether they&#x27;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&#x27;s just a scaler. It&#x27;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&#x27;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&#x27;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&#x27;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&#x27;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&#x27;ll do it proved by contradiction. And so we suppose it&#x27;s true. All right, so we suppose that a B time see is in fact equal to hey c a Times B, and we&#x27;re also going to assume that the inverse of a exists, um, which is not necessarily always true. So this isn&#x27;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&#x27;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&#x27;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&#x27;t hold, Which means that Arthuis assumption that we made supposing it&#x27;s true must be false because everything that we did here was equal. And so the full statement then is false. And that&#x27;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&#x27;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&#x27;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&#x27;t necessarily match up, this multiplication is not to find. And so the statement is false as well

Subham Jyoti Mishra · Answer

in this question. We have a Marcus. This came close and metrics. And we have a set off statements which I read up to Workforce. You will check them one by one. What statement is the rule? Space off is the same as the common space. Off transpose. No, those space off is a total carbon. Yes, off a transport. This is true. Hmm. Programs was off road space off. Is the total number off non zero this panel? They&#x27;re not non zero, though. Viktor&#x27;s. Uh huh. Metrics. But when is transports toe transport that always become columns? So therefore, this is equal and well by work. Government. So the So there is this true second statement? Is he? He&#x27;s in a form of here and his 39 year groups. Because do those people on the beach who a now here, living under be used, uh, take alone normal here. Not really different up here. And and yeah, and that number off non zero of those visible the tree, so it may not be necessary that the Post she rose? Uh huh. Norm zero. Those two roles, maybe, uh, independent. So therefore, we cannot create necessarily say back. Uh, well, studios mama B. C&#x27;s off the circus sports? No. From the whole shipment we have when the dollar statement. We have the diamonds channel. The rose space up here is equal to payments. Um, off column space off here, given he iss not a square metrics. And it is true from the rank here, Um, and also the diamonds and off role space here is equal number. All none. Zero rules in the assist. Ekotto&#x27;s number four columns. So therefore, number all favorite problems. So that days ago, diamonds and off problem space, then in the pork statement we have is some of dimensions of the rules. No space off going toe, remember, Oppose. Mm. Which is diamonds and, uh, no. Is this dimension of Cruz, please? Here has your work diamonds and most is here less diamonds and all column space care. Since dimension of those places, dimensional space here has thank off by rank and which is number off columns is a nautical EMS response with this statement. Mhm miss. On a computer role, prisons can change the train called metrics. You are the true on computer He did can be his grounding. Yeah, So the word is true There in the the rank of metrics can change enroll prisons and computer

Runpeng Li · Answer

problem problem. 18. We start with part. Hey, it says, um if bees any edge inform off A than the people call himself, it be former basis for the column. Space of A. So this Damon actually is false. The reason is that the column space the basis optimum off the com space off A is actually not people Columns A B is the corresponding columns in a or his funding columns in a. So, actually, we&#x27;re taking the columns of A instead of the national home. Okay. Okay. So the next R B, we have Roe operations preserved the win your dependence relations about the rose off, eh? When it isn&#x27;t stroke, all&#x27;s well didn&#x27;t steal false. Why? Well, because we we&#x27;ve performed the, uh We performed the road operations we interchange. You didn&#x27;t inter change in the rose. Then everything will be messed up. Interchanging, rose, mess up things right? Because we are interchangeable. Rosen. We cannot tell which Rosie. It&#x27;s exactly the the linear dependence relations we want we&#x27;re looking for Okay. Hard to see so exas. The dimension of the non space of a is the number of columns of a that are not people. Columns Well, this this statement is true. You guess by our wrecked the room we have mentioned up in all space off a is and minus drank a So if we know the dimension of the no space of a number of the columns off, eh? There are not people columns, but rank of A actually keeps our the number of people columns. So those are not people. Columns is actually the columns that the total comes minus the people call him. So this is a you correctly that not people columns. All right. So hearty, it says means the rose space up. Hey, transposed the same as a common space of a Well, this is also true because that&#x27;s how we perform the transpose operation. Just a bite transpose operation. Okay, Because by the transpose operation, the rose becomes the columns, columns becomes the rose. So the the rose face the game will be the same as a con space. When you do the transpose. Okay, our e So it says if a and B are really equivalent, then they&#x27;re roast bases are the same. But it&#x27;s also true. The reason is that you know the what we have to really prevalent people in the matrices that it&#x27;s actually how we do how we find tree, how we find Rose face so we don&#x27;t

Michael Jacobsen · Answer

in this video. We&#x27;re considering the Matrix be further. So be was given to be this four by four matrix And what we&#x27;ve done already is we have read role reduced it into its echelon form. We found that row one, row two and row three in its role reduced form our pivot columns So we can say for the following that be in particular does not have a pivot in every row since the fourth row is a zeroed out row. Now this not having a pivot in every row causes certain problems. For example, we do also know as soon as there is not a pivot Every row that there exists vectors be in our four Where are four comes from? The four rows of this matrix p So there exists Exists Victor&#x27;s being are for that are not linear combinations of the columns of be This is an important thing to consider because it means that in a vector equation such as B X equals, say why would not have a solution for every vector. Why? Let&#x27;s consider a different situation. We know the set evolving your combinations is also the span of the columns of B and we also know that we have three pivots altogether. So let&#x27;s consider the claim. The columns of be spans are three. This is an interesting claim because it really feels quite right. Sure, we don&#x27;t have a pivot and every row, so the comes of B cannot spend our four. But if there are three pivots, could the calm spend our three? Well, Sadly, this claim is false. The columns of be do not span or three and this claim is false irrespective of the number of pivots. So if we had no idea what the number of pivots were, if we were just back to the original matrix, it would still be faults. And the reason is since B is in are four by four. In other words, this four here is the most important thing to consider and it causes a mismatch with our three. So the columns of beak on lee have a chance of spanning are for not are three, not are two and so on

Michael Jacobsen · Answer

in this video, we have a four by four matrix a provided. And our goal here is to analyze the pivot positions. This can be done by first placing this matrix into its echelon form. Let&#x27;s make that our first steps. So I&#x27;ll copy of the first row 13 03 And our goal is to go to this position here and make all entries below zero. So we can add Rwanda Row two and record the result. And that will give us a zero then too negative. One and four. Next. We have a zero here, so I&#x27;ll just copy this road, then it zero negative for two and negative eight. Next. We need to eliminate this too. So multiply Row one by negative too. At it to row for And we have zero negative. Six three and negative seven. Now, since we&#x27;re going for echelon form, I&#x27;m going to pivot off this, too. We see here and go for an elimination of the entries. Negative four and negative six. But let&#x27;s ignore the three. So our next step will say that this is row equivalent to first. Let me copy the 1st 2 rows. 13 03 and zero to negative 14 Now let&#x27;s multiply row two by two, adding it to row three so we can eliminate that entry will have 000 and zero. So this is working out nicely next to eliminate the negative six. Multiply road to buy three and add it to row for we&#x27;ll obtain zero zero. Let&#x27;s see we have zero again and three by four is 12. My seven gives us a positive five. So at this stage are matrix is not quite initial on form. Let&#x27;s take that one extra step to place it there. I&#x27;ll have 13 03 in row, one row twos just fine, so it&#x27;s copies your to negative 14 The problem is Row three. We would prefer to have the zeroed out rose towards the bottom. So it&#x27;s interchange Row three with Roe for then our matrix. Let me try better with the zero that are matrix is now in echelon form. So we&#x27;re going to be able to analyze the pivots by saying here, here and here are our pivot position. So by this work it follows that three rows of a contain a pivot position and our work is complete

Problem

Exercises 15 and 16 concern arbitrary matrices $A…

Problem 15 Easy Difficulty

Answer

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

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David C. Lay, Steven R. Lay, Judi J. McDonald

Linear Algebra and Its Applications

Anna Marie V.

Alayna H.

Caleb E.

Kristen K.

Absolute Value - Example 1

Absolute Value - Example 2

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

Linear Algebra and Its Applications

Anna Marie V.

Alayna H.

Caleb E.

Kristen K.

Absolute Value - Example 1

Absolute Value - Example 2

David C. Lay, Steven R. Lay, Judi J. McDonaldLinear Algebra and Its Applications

Anna Marie V.

Alayna H.

Caleb E.

Kristen K.

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

David C. Lay, Steven R. Lay, Judi J. McDonald

Linear Algebra and Its Applications