in-exercises-9-and-10-mark-each-statement-true-or-false-justify-each-answer-a-in-order-for-a-matrix-

Question

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 $n 	imes n$ and invertible, then $A^{-1} B^{-1}$ is the inverse of $A B .$ 
c. If $A=\left[\begin{array}{ll}{a} & {b} \ {c} & {d}\end{array}ight]$ and $a b-c d 
eq 0,$ then $A$ is invertible.
d. If $A$ is an invertible $n 	imes n$ 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.

Jacquelyn Trost · Accepted Answer

for this problem. We have a Siris of five statements and we need to show whether they&#x27;re true or false, and we will justify each answer. So let&#x27;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&#x27;re defining the inverse. Um, you can see that specifically if you look at the text on page page 105 you&#x27;ll see that it&#x27;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&#x27;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&#x27;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&#x27;s false to see that you can look at the&#x27;re, um, number four, which tells you the correct order. It should be a D minus B. C. We&#x27;re looking at these diagonals so that the product of aid times D minus B times C. That&#x27;s what we&#x27;re comparing 20 If that isn&#x27;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&#x27;re, um, number five from the book and the&#x27;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.

Angelo Rendina · Answer

so a. He&#x27;s false because the inverse of a product, let&#x27;s say a B members is the product off. The inverse is in the reverse order, so it&#x27;s being worse times a embers. Now, for me, that is true, because if we have a times a m birth, he called it entity. Then from here, we need that A is the inverse off a mers and at the same time, a inverse is the inverse. Okay, now for C the stroke In fact, the inverse off the two by two matrix A B C. D is given by the Formula One over its determinant, which turns out to be a d minus. BC. And then we stopped the elements on the men diagonal and we negated the elements on the second. There there are gone, so we have d n a swapped on their minus B minus C. And of course, this happens only if a d minus BC is non zero. So if Eddie is no NBC now D is true as well. In fact, we have to go to them. We put the metric say, and then we put their identity. Now, if we can&#x27;t use these metrics to identity in some metrics, be on the right. Thes means that be is the evers of a and so is in vegetable finally e That is false. Because if we reduce, um I think say toe there. Entity. Yes, I&#x27;m reduction. We can ride. This is a product off mattresses as something like E A is equal to They&#x27;re entity where is called Elementary metrics Now, these falls because the same operations do not reduce chambers to the identity. In fact, we see that E a mers can only be that entity if now we multiply by a on the right If he&#x27;s equal to a while instead, this condition here says that E is equal to a inverse. And in general, the Matrix is not equal to it in verse.

Bobby Barnes · Answer

right. So they want us to determine if these statements are going to be true or falls. Um, so this first one, it says essentially, that we have the zero vector and H. So this is a subspace of our in and this is going to be false because it doesn&#x27;t say anything about the other two properties that we need to have, which is it Closed under sums and closed under scaler multiplication. So false. Because we do not No. If actually, let me beat the Patriots. Bigger. Do not know if closed under some and scaler multiplication. Okay, so we have a there now for B. They say we have this set of factors, and the set of all linear combinations is a subspace. And so this one is going to be true, because if we just think about, um, what this is so it would be something like see one V one plus c two, b, two plus dot, dot dot all the way up to C p v p. So we can get the, um zero vector by saying all see, I equal to zero for every eye. So that&#x27;s fine. Um, we can get any scaler out of this by saying, see, hi equals zero except for some value. Um, but I is equal to and where it is, like the vector that we want. So that gives us the scale of property and three, some wolf. I mean, some just kind of comes from the definition. So some is, um, true by definition of linear combination linear combo. So that&#x27;s how we end up getting true for B. See, I believe this one is also true, but let&#x27;s go ahead and kind of draw it out. So if we have over here some Mbai in Matrix, remember what the knoll space really is? It&#x27;s saying, Well, what collections of vectors here output the zero vector. And, um, in order for us to multiply these we needed in by one matrix, which means we would have in entries here and then this is an element of our in. So then the null space has to be a subspace of our because it will have the zero vector on. We&#x27;ll also have where if we add two things by the linearity of matrix, that should be fine. And then if we just multiply this, that should also be fine. Um, well, actually, let&#x27;s go ahead and write all those out. So eso let&#x27;s say just so we can kind of be a little bit mawr exact right? So Ah, we have some matrix A with the vectors. Let&#x27;s just call it X Y elements of the knoll of a All right. So if we want to do the some property So let&#x27;s look at a Times X plus y Well, then this is going to be a X plus a y and by definition of both of these being in the null space, this is the zero vector plus zero vector which is zero vector. And so then that implies X Plus y is also an element of nor a Okay, so we have that one and then first scale er&#x27;s. So this is some. So that&#x27;s good. I guess we should have also did the zero vector. But that one&#x27;s kind of true just by definition. So this is zero element of Norway. Alright, so that&#x27;s good so far and then lastly we need these scaler. So we have a times c of X, remember axes so in the north space. But we can go ahead. And do you see? Actually me, Right This over here. So this is going to be C a X like that so we can apply community since he is just some constant. And then we&#x27;ll that c times zero vector, which is equal to zero vector. So then that implies C X is also in the knoll. So then the constant multiple or the scaler also holds. So, um, I don&#x27;t know if you really need to be that exact, but, I mean, it doesn&#x27;t really hurt now for D the column space of Matrix A is the set of solutions. A X is equal to be all right. And so this is actually going to be, uh, false. And the reason why this is false is because it&#x27;s not necessarily just for the solution of one. It&#x27;s supposed to be the span of all of the columns I&#x27;m sorry of. Ah, yeah, the span of all of the columns. Um, so, yeah, that&#x27;s why this one isn&#x27;t necessarily true. Because it&#x27;s not talking about a solution set and then for E. So this is saying that B is the echelon form of a matrix A in the pivot columns of B form a basis for a And so this one, um, is also false Because it&#x27;s not the pivots of B, but the pivots of call a Arab the pivots of a instead. So this should actually say a here on beacon, actually kind of show that off on the side really fast. Eso Let&#x27;s come down here and do that, right? So let&#x27;s say we have some matrix here. And so this is kind of like you can use, like, a counter example as to why this wouldn&#x27;t be the case. Um, so we have, like, 123 uh, then 567 13, 14, 15. And then, I don&#x27;t know, negative 112 So hopefully if I reproduce this year Ah, this gives me 111 000 And if it doesn&#x27;t just kind of tweak this so it does. But I believe this will give us that. Okay, well, if we look at it, this is a and then this is the echelon form. Be over here. Well, if we were to look at this, though, over here, this is saying if we take each of these are fourth position of our vector is always going to be zero. But over here, notice There are numbers here. So we wouldn&#x27;t even be able to get out the, like, original columns here. So because of that, um, we would need to use the columns for a as opposed to the columns of being So Yeah. So that&#x27;s kind of like the rationale behind. Why that? Why that ISS? Yeah. So s it looks like the first statements. False second statement. True, They&#x27;re true. And then the last two are going to be false.

Alexandra Embry · Answer

in this problem, we wonder if to find X we would multiply a by the inverse of B. However, if we want to find X, what we can do is we can multiply both sides of this equation by a inverse so that a inverse times a is just the identity. As we get the identity Times X equals a and bursts times be so that means that X is equal to a inverse times Be so no, that statement is false. We don&#x27;t want to multiply a by the inverse it be We want a most play a inverse times be

Amy Jiang · Answer

Okay, so a is false because this is just the absolute value to determine it. He is also false cause determinants of a is equal to the determinant of 80. We notice from my term. Uh, but see, it&#x27;s joke. Murthy is false.

Chris Trentman · Answer

were given statements and were asked to market these statements, tour falls and to justify each answer statement. A Is there are symmetric matrices that are not orthe orginally diagonal Izabal. The statement is false. This is by fear, um, to from this chapter call it the arm to says that a is a symmetric matrix if and only is is or thought Nelly Bagnall Aysel statement B is If the Matrix B is equal to p d. P transposed he transpose equals P in verse and D is a diagnose matrix, then be is a symmetric matrix. This statement is true. See where this statement is true. Recognize that the requirement that B equals P d. P transposed repeat transpose equals P inverse India&#x27;s a diagonal matrix. This means that be is or Thorgan Aly Diagonal Izabal and again by fear, um, to from this chapter we have that be must be symmetric statement see is an orthogonal matrix is orthogonal e diagonal Izabal. This statement is false. This indeed may not be the case Recall that the definition of Martha Graham matrix it&#x27;s matrix who is rowing column vectors or since of worth the normal vectors. So for example. Let&#x27;s take the matrix A to B. Matrix one. Okay, it&#x27;s 2111 one 01 No, 100 01 zero. Yeah, maybe a simpler example. One of her, too. One of a route to and negative one of a route to one of a route to now, just looking at this matrix we have that if we take you one to be the vector of one of a route to one of her too. And you to be the column vector negative 1/2, 1/2. We see that both of these are unit vectors. And that you won dotted with you too, is equal to zero so that they&#x27;re also orthe ogle to each other. So that&#x27;s you one. You too is an Ortho normal set. Likewise, if the one is the column Vector one of a route to one of a route to in v two. Sorry, the row vector, whatever to negative One of her two and V two is the row vector one of the route to one of a route to then we have that V one dotted with V two is also equal to zero and that view and envy tour. Both unit vectors do that V one and V two Foreman Ortho Normal said as well. And so we have that all is that a is an Ortho normal or distant orthogonal matrix. But clearly they transpose is equal to whatever route to one of two on the diagonal. And then we swap negative one of a route to with one of them too. This is not equal to a so that a is not symmetric. Therefore, again, by theory, too, we have that a is not or Thorgan Aly Diagonal Isa ble statement D says the dimension of an Eigen space of a symmetric matrix is sometimes less than multiplicity of the corresponding night in value. This statement is false to see why for to the spectral fear, um, for symmetric matrices, which is three earned three in this chapter and we

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Mark each statement True or False. Justify each a…

Problem 9 Easy Difficulty

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Grace H.

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Absolute Value - Example 1

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

Grace H.

Kayleah T.

Caleb E.

Michael J.

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

Linear Algebra and Its Applications