in-exercises-11-and-12-mark-each-statement-true-or-false-justify-each-answer-a-the-set-of-all-affine

Question

In Exercises 11 and $12,$ mark each statement True or False. Justify each answer.
a. The set of all affine combinations of points in a set $S$ is called the affine hull of $S$ .
b. If $\left\{\mathbf{b}_{1}, \ldots, \mathbf{b}_{k}\right\}$ is a linearly independent subset of $\mathbb{R}^{n}$ and if $\mathbf{p}$ is a linear combination of $\mathbf{b}_{1}, \ldots, \mathbf{b}_{k},$ then $\mathbf{p}$ is an affine combination of $\mathbf{b}_{1}, \ldots, \mathbf{b}_{k}$
c. The affine hull of two distinct points is called a line.
d. A flat is a subspace.
e. A plane in $\mathbb{R}^{3}$ is a hyperplane.

Lucía Guerrero · Accepted Answer

and it&#x27;s exercise a statement. A is true. In fact, what we have here is the definition off their fine hall on this can be verified by going to page for 139. A statement V is false on. We can consider the following counter example in our to consider the vector who isn&#x27;t resort to to This is a linear combination, but not on a fine combination. Off the victor 11 statement see is true. When we have two vectors there, I find combinations Have the following form on this is just the equation of a life. So statement see is true, then statement d is false. I remember that Ah Flat is a translation of a safe space in this, uh, growing w is a flat and we can see that the victory zero is not contained in W Therefore, w cannot be a super space sort of statement. Thes falls on finally statement e is true. The dimension of a plane is two on two is only one less than the dimension of our three. So in our three planes are also hyper planes

Gideon Idumah · Answer

cinemas. All we have that the A part is force because I find off s equal toe s and they can see uh, the park of below the definition on page for 59 number B parts. This is true. Um, scenes, The fine all off s always contents s. We know these. So if, as contains, it&#x27;s fine all it&#x27;s fine. All So we have a friend all always convinced ISS. So if s contents are fine, all this would imply that s with after Beatie quote So they find all off a fine of s on DDE bites urine Sue serum sue because s a course to a fine of s. This implies that s is our friend Number C is also true. You can see the Parker off below the definition in page for 42. Question is force This statements the statement is only true is all the true when floods he&#x27;s in our three on then is true because floods is a trust leads off subspace. So this means that so if we translate it back if we translates, it&#x27;s Bach to the origin. It becomes a subspace ce good

James Chok · Answer

Okay, so this question will be answering with a serious off a tree and forces. So it was a if b is a basis for subspace hated in each. Better in hate can be ridden in only one way as linear combination of factors in B. This fact is true, which is, quite literally, the definition of a basis. And if you need more help on this question, if you go to the text book page 1 55 the beginning off section to 2.9 on the second paragraph, it says, is basically verbatim. Question be if B is a basis for suspicion of hate in the correspondence, makes hatred look and act the same as our p. So this is also true, which is literally told bourbon in page 1 57 See the dimensions off mo off a number of variables in the equation X because is there this is false is equal to the number off free variables in a executed the dimension off The column Space off A is rank A. This is true because this is that that this is the definition off, Frank. Hey, if head is a p dimensional subspace, there linearly independent set off the vectors in hedge is a basis for Paige. This is true by the base system.

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.

Doruk Isik · Answer

in this trouble were given a few statements and were asked to determine whether the statement is true or not. So I started important. A standard basis for our end will be open for 10 zeros. Um, and we will have in terms 010 zero. And finally we&#x27;ll hire and minus one zeros and entering will be Won&#x27;t so in a factory can be written a a linear combination off our standard basis. Specter&#x27;s right. So let&#x27;s say we have be and that is equal for everyone. Be too up to wi n. So this is we want times 10 plus B two times +010 plus dr dot the end times zeros and the anthem is again worm. So then the coordinates back to is the vector formed by this collection. He&#x27;s right. So don&#x27;t the corner factory in the spaces B one B two up to eat them. But this is actually the vector B. It&#x27;s so so they&#x27;ve given a statement is true in Party of the statement, this school&#x27;s because Tina X, into the Carter director, not the other way around. This is not right. Seeing party given salmon is true as long as the plane passes through the orbs. So it is true

Runpeng Li · Answer

Okay, So this problem, we&#x27;re going to check whether the following statements are true or false. So the first statement says, um, if axis a nontrivial solution up off X equals zero than every entry in access none. Zero. So this is clearly false because we can consider a solution. Consider such a solution that it&#x27;s x three one and zero x three. So in this case, ex three&#x27;s a pre bearable. But, um, the specter asked, contains zero, which is the second entry here. So what? That means our our first aim in this wrong. So that&#x27;s we will mark falls here. All right, the second statement, Now we have ext because x two times you us x three times fee. So we need to check that, um, whether this is the plane that that is passing through the origin. So now remember how we describe a plane as how we describe a lying in ah are to space. So that is acts equals you times you plus tee times feet where you and V are vectors and tease is ah, real number. So similarly, we can express playing in our three space. That is you times as a plus B plus the ice tea class W your W ys accounts tend the casting term, and we have two variables here that determines the plane playing in the, uh that determines the plane are three space. But right now, the key thing here is the casting term. W now recall that the constant, constant terms the cats in terming are two space that represents the The point is that the line is passing through, and so in the arteries are three spacing the plane. So if we have the constant that is zero, that means that means that this plane is passing through the already origin. So that means in this case, we have x two times you and x x three times V. But we don&#x27;t have any. We don&#x27;t have any constant terms. So that means the council terms is the reflector. So that that implies this plane is passing through the origin. So we&#x27;re done. All right, this third a statement. So excess the equation. There he acts equal speeds homogeneous. See if the zero vector is a solution. No. Now recall the detonation of homogeneous equation. That means, uh, we have a homogeneous equation. So That means the baby has to be zero because we don&#x27;t allow any consented. None zero constant. On the right hand side, we have a homogeneous equation, so as chatty If this is true, so zero is a solution. So that means we plugging zero. So a times zero vector will be zero. And that is exactly be, because no matter better what the matter was solution. We put in our equation in a basket this council be, but, uh, here we put in zero, so it must be zero or so. So that implies he is a zero constant. And so and so we have a tax equals zero. So this is a homogeneous system, and we&#x27;re done. All right, The fourth statement. So it says, um effect off Adam P two a vector is to move the with the factory in a direction parallel to Pete. No. Oh, before we start 1/4 1 out just right. Right down. True or force here. So 2nd 1 is true? Yeah, the first ones Paul&#x27;s 2nd 131 are all true. So the 4th 1 So right now here we&#x27;re given I say we&#x27;re given a vector. Say, b how Drove it in pictures. Um, I&#x27;m trying to be this way. So we&#x27;re given the vector the and also we are adding p to this vector. Me. So let&#x27;s say P is, uh is this this is Pete. So to perform the vector edition withdrawal such a thing here. And finally we will get this and this is the plus peak. And we can we can see from a picture that we&#x27;re moving be from this point directly to this point. And it is along the direction off p. We&#x27;re moving straightly, right? Poor. My word in this direction the direction of P. And so that means it is It is exactly just like a translation here. So it&#x27;s true.

Problem

In Exercises 11 and $12,$ mark each statement Tru…

Problem 11 Easy Difficulty

Answer

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

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a) trueb) falsec) trued) truee) true

Linear Algebra and Its Applications

Anna Marie V.

Heather Z.

Michael J.

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Vectors Intro

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

Anna Marie V.

Heather Z.

Michael J.

Joseph L.

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

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

Linear Algebra and Its Applications