in-exercises-23-and-24-mark-each-statement-true-or-false-justify-each-answer-a-another-notation-for-

Question

In Exercises 23 and $24,$ mark each statement True or False. Justify each answer.
a. Another notation for the vector $\left[\begin{array}{r}{-4} \ {3}\end{array}ight]$ is $\left[\begin{array}{cc}{-4} & {3}\end{array}ight]$ and
b. The points in the plane corresponding to $\left[\begin{array}{r}{-2} \ {5}\end{array}ight]$ and
$\left[\begin{array}{r}{-5} \ {2}\end{array}ight]$ lie on a line through the origin.
c. An example of a linear combination of vectors $\mathbf{v}_{1}$ and $\mathbf{v}_{2}$
is the vector $\frac{1}{2} \mathbf{v}_{1}$ .
d. The solution set of the linear system whose augmented
matrix is $\left[\mathbf{a}_{1} \quad \mathbf{a}_{2} \quad \mathbf{a}_{3} \quad \mathbf{b}ight]$ is the same as the solution
set of the equation $x_{1} \mathbf{a}_{1}+x_{2} \mathbf{a}_{2}+x_{3} \mathbf{a}_{3}=\mathbf{b} .$
e. The set $\operatorname{Span}\{\mathbf{u}, \mathbf{v}\}$ is always visualized as a plane
through the origin.

Priyanka Sadarangani · Accepted Answer

okay. In this question, there&#x27;s a set off five questions from A to E. And we have to find that these statements are true or false so fast, The statement is the question is another notation off the vector minus 43 is is minus 43 Okay, so it is false because vectors are represented as call the metrics is okay. And this is two dimensional, two dimensional vector. And this is set off toe one dimensional mattress mattresses. Okay, two dimensional vectors. So this is false. How come Toby be the caution be is the points in the plane corresponding toe minus 25 and minus five to lie on a line through the origin. Okay, so you will check that this past for the reason or not So minus two and five is hair. So this will be the line and minus five. And who will be here? So this will be line, and it does not pass to the or is not. So it is false. Okay, now we should see an example off a linear combination of vector even. And we do. Is Victor halfway one? Okay, so we can say that a linear combination is defined as the sum of vector. We won we 233 and so on We end multiplied by these killers C one c two c three, as seen So so. Therefore, as for the ocean half even Okay, so half even is a linear combination is a linear combination. How? Because it is the same value off half, even plus zero. Well, okay, here&#x27;s even is even toe half and sit close to zero. So this statement is true. Okay, now the caution be on the day The solution set off Linear system has documented metrics is a one a two a three b has the same solution. Seems as the solution set off equation Excellent even plus x two, a two plus x 383 equals to be so. Yes, it is true. Why? Because the solution says are saying because the augmented metrics off this equation is even a two, a three and b So this solution off this documented metrics and because they are same, then the solution will be also same. So this will be true and the last one e the sad span you ve is always visualized as a plane through the reason is always we realize as plain through the or reason and this is false. Why? Because if these factors are linearly dependent, then they respond. Will be will produce a line. Okay, note a plane so line will be correct. Plane will be not correct hands. The statement is false. So we have the final answers here. A false B false see True D Do you on e false? So this will be our final answer. Thank you.

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.

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.

Ashley Boni · Answer

eso part, eh? You don&#x27;t be minus b dot You equal zero. Well, we know that the dot product is commuted s. So that means that you dot b is equal to read about you. And these two things are part B for any scaler. See, uh, the norm of CVI is equal to see times the normal b Well, so we want to see if, uh, we can pull a scaler out of the norm. And if this holds true, So let&#x27;s take c t equal Negative one on our vector Be to just be, say, 11 Let&#x27;s compute uh, these two values So the norm of CV is equal to the norm of negative one negative one which is equal to the square root. Um, negative one squared postnegative one squared, which is a squared. But now, if we take, uh, c times the normal V, we get negative one times the norm of 11 which is negative one times the square root of one square post one squared, which is negative, squared or two. So these things aren&#x27;t the same once he&#x27;s negative. So this one iss who else? This person is true. Uh, next for part C. If X is or thought inal toe every vector in a subspace W then X is in, um, the Ortho Wagnalls base to W. So that&#x27;s just the definition. Um, the space here, This just means that all the vectors that or thought minal to the space w closes just true. By definition, party says, if the normal view squared plus the norm of these squares is equal to the norm of you plus b squared then you movie or are orthogonal. So if we know that this is true, let&#x27;s write out what this means. So the normal view squared is you, don&#x27;t you? Normal v squared is we Dock B now on the right We have you close to me dot You cost me So we have you, don&#x27;t you? Plus B, that me is equal to, um this very hand side we can essentially like foil it out So we have you got you Plus to you don&#x27;t be plus me dot B um, so you don&#x27;t use could cancel be Darby&#x27;s can cancel. So we get zero equals two times you don&#x27;t me. So this means that you don&#x27;t be has to equal zero, which means that you and v our orthogonal. So this is true. And lastly, part e ah, for an m by n matrix A vectors in the null space of a our earth Ogle two factors in the roast base of it. Well, if we look in the section on here, um, three. It states that everything orthogonal to row space of a is equal to mall space of a. So this is just another way of saying exactly what the statement says. Um, so this is true because this is just stating they&#x27;re on three, which is on page 3 55

Chris Trentman · Answer

Well, it&#x27;s like a different album. Were given a statement and were asked to mark each statement True or false and justify our answer. V is a vector space. Mm. In part a, the statement is Yeah, our square. It is a two dimensional subspace of are acute dominance of Yes, So they&#x27;re in there. They&#x27;re like, you know, podcast. They got a very dialectic. This is kind of a tricky question, but we know there are two is not a subset of our three because the vectors in R two of two entries and vectors in R three of three entries, however, are too. Is I so more thick to a subset of R three? Therefore, since our two is not even a subset, the statement is false. Then in part B Mm. The statement is the number of variables in the equation. X equals zero equals the dimension of the null space of a Mm. Oh, well, we know that the dimension of the null space of a is actually equal to the number of three variables. Therefore, if the number of free burials variables is not the same as the number of variables which is possible, the statement is false. So the statement is false. So get dick kebab. Pray in part C. Something like statement is a vector. Space is infinite dimensional. If it is spanned by an infinite set, read it. Suck dick Just legal out Grace. Well, in fact, I actually want it&#x27;s true that any and finite character doesn&#x27;t make any sense. Vector space is spanned, and it&#x27;s not like I don&#x27;t I don&#x27;t bad job by an infinite set as well. Green Any non empty finance sector space, I should say. Therefore, to solve your requirement is that an infinite vector space cannot be spanned said by a finite set. And in this way, an infinite vector. Space is different from a finite vector space because finite victor spaces must be spanned by a finite set. Therefore, the statement is once again false. What happens to it finally in part? Mm. Actually, that was part C in part d. The statement is if the dimension of V equals n and s spans V then s is a basis of the Yeah, but you say types that we&#x27;re talking about fucking face. Yeah, this is actually mhm. Not necessarily true. Well, if the dimension of V equals n and s is equal in s spans, V then it follows. That s is a basis for V if and only if the number of vectors and s is also in. Therefore the statement could be false. We could choose a set s which spans V, which has more than or less than n vectors. In fact, it has to have more than n vectors. Finally, in part E were given the statement. The only three dimensional subspace of r three is ours. Three itself. Mhm. All the well, we know that the dimension of our three is three. Yeah, therefore span of our three must contain three. Mm, but nearly independent factors. And therefore he didn&#x27;t finish mhm r three. It&#x27;s happening right now, right? Okay, that&#x27;s true. All right is the only three dimensional subspace of our three. Therefore, the statement is true. Would have

Doruk Isik · Answer

in this problem. We&#x27;re given a few statements and were asked for work. It&#x27;s that where it is true or false. So the first statement, uh, schools, Because the set of backers not only should be linearly dependent, it should also coincide which age? Hey, you could check the definition off base base. Second statement is true, and I&#x27;m gonna report you refer to the, uh, spending said here. Now, the statement is statement, you see, is also true. And this time I would refer to the section named two years off a basis. Uh, Indy, begin the statement. This falls because this standard method of producing a spend settle now a is actually always Cordish iss a re nearly Bennett sent. And it actually always produces a basis for no hey and less Damon. ISS also pulls because we need Thio look at D or use the different elements or give its Collins off matrix A not the magics B, which is the Rebecca Formal, eh? Because some columns of B mike nights be in the column space. Oh, okay. So we should also we should always use the pivot columns off our orginal matrix

Problem

In Exercises 23 and 24, mark each statement True …

Problem 23 Easy Difficulty

Answer

FALSE, FALSE, TRUE, TRUE, FALSE

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

Linear Algebra and Its Applications

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Absolute Value - Example 1

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FALSE, FALSE, TRUE, TRUE, FALSE

Linear Algebra and Its Applications

Heather Z.

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Caleb E.

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Absolute Value - Example 1

Absolute Value - Example 2

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

Heather Z.

Kayleah T.

Caleb E.

Kristen K.

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

Linear Algebra and Its Applications