determine-which-sets-in-exercises-1-8-are-bases-for-mathbbr3-of-the-sets-that-are-not-bases-determ-6

Question

Determine which sets in Exercises $1-8$ are bases for $\mathbb{R}^{3}$ . Of the sets that are not bases, determine which ones are linearly independent and which ones span $\mathbb{R}^{3}$ . Justify your answers.
$$
\left[\begin{array}{r}{1} \ {2} \ {-3}\end{array}ight],\left[\begin{array}{r}{-4} \ {-5} \ {6}\end{array}ight]
$$

Doruk Isik · Accepted Answer

in this problem. We&#x27;re given a set of that whistle of a form. Our system constant. Becker&#x27;s that is one to make it three and negative. Four and five and six. So ready. So I went for this system can be written as well. 00 and 010 So, as you can see, we have the rolls, but two holes. So that means that, um, weekend ever give it element means roll right is the number of homes across. It means that this set cannot form a basis, and it can not spend this space. But it is a bit element in each column. So it means that this given set, this records are the nearly independent.

Michael Jacobsen · Answer

for this example, we have a set of three different vectors provided encircle here is to determine if they form a basis for our three. Let&#x27;s start off by letting A be a matrix formed by setting V one, V two and V three to be its respective columns. Then A is going to be 10 negative, too. Three to negative four and then negative five or excuse me, negative three, then negative 51 So let&#x27;s reduce this to echelon form so that we can determine the pivots from this pivot position. One. We&#x27;re going to start out by eliminating the negative to which is in the third row. Saul Copy Row one and row to to start out, then multiply Row one by positive to add the results to Row three will have 02 and negative five from that role operation. But as soon as we have a matching bro, Row three is going to be eliminated. You could visualize multiplying row to buy negative one and then added it, ending it to Row three. That would eliminate Row three. So now it&#x27;s a row of zeros. Row two is zero to negative five and 13 negative three for row one, then with their echelon matrix Initial inform. We have a pivot here in here, but there is not 1/3 pivot. So it&#x27;s now state our conclusions. We now know that the columns of a are linearly dependent and they do not span or three. So how do we know this? This is by the in vertebral matrix. The&#x27;re, um all right, I m t for murder convertible matrix dirham. And this is since A, which is of size three by three, has on Lee to pivots that we found from its echelon form. So with 2 52 pivots, we are linearly dependent and we do not spend our three when we&#x27;re using this matrix A. But to form a basis are matrix which could consist of called vectors. V one V two V three would have to have linearly independent columns and they would have to spend our three. So our conclusion here is that the set the one V to V three is not a basis. Had we found three pivots, we would have concluded that it is a basis

Michael Jacobsen · Answer

in this example, a set of vectors have has been provided. A miracle here is to determine if they form a basis for the space are three. First, let&#x27;s look at the set V one, V two, and notice that we have a set of exactly two vectors. In this case, we can determine linear independence as follows. Let&#x27;s determine if they could potentially be multiples of each other. In other words, could be multiply V one by some constant and obtained V to that constant being non zero. Well, it turns out that there is a major challenge to this. Our entry zero here and our entry five here in the corresponding vector V two tells us that no matter what value we put in from the V one toe, multiply the zero. We will never get a value five so automatically for the set V one V two. The set is Lee nearly independent, since V one and V two are not multiples of each other. If they were multiples of each other, we would say that the layer linearly dependent. So now we know we have linear independence. If these two vectors span are three, then they would also form the basis to resolve spanning. Let&#x27;s start by setting a matrix, Let the Matrix a be given by V one and V two. Then this matrix, if we write it out, is negative. 230 six, negative one and five and without riddled row reduction, we know that a cannot have a pivot in every row. The reason for this being is if we have a pivot here, then in the best case, we have to pivots another pivot position here in an echelon form. But we could never obtain 1/3 pivot unless we had 1/3 column. So since they cannot have a pivot in every row, week now can conclude based on this statement, that the columns of A, which are V one and V two do not span, are three. So because we cannot have a pivot in every row, the columns do not span are three itself. So let&#x27;s conclude this example. We know that the set is linearly independent but does not span. So our conclusion here is that the set of vectors V one and V two is not a basis. Four are to excuse me are three. If it spend that, it would have been a basis

Doruk Isik · Answer

in this problem. We&#x27;re given a set of that truth and this system using that word. What? There&#x27;s looks like this. So this system and ready to roll echo for your 010 001 And, Nick, you&#x27;re too negative. One hell do you want? Uh, so we have three rows and or colmes. And as you can see, how long given element here won&#x27;t give it element here. Element here. So we have a deal element for each role. So it&#x27;s me. In fact, this sad actually forms a basis, and it spends the the space. And as you can see, we have three defense furs, second and third column in the rock Registrar. Excellent form. So it means that 1st 2 factors birth three or the nearly

Doruk Isik · Answer

this problem we&#x27;re giving us set up. So let&#x27;s 1st 4 worst system from the given lectures to negative to one one negative 32 and 8754 So dis metrics in reduced role. That quote for can be written as 100 01000 Did you guess you from here? 33 different positions for patrol. So that means that this different set space art.

Michael Jacobsen · Answer

in this exercise, a set of four different vectors are provided, and our goal here is to determine if these vectors form a basis for our three. Let&#x27;s start off by making a knob survey shin that if we form a set the one v to V three N V four. That this set is automatically Lee nearly dependent and it&#x27;s linearly dependent for really two reasons. But let&#x27;s go with the big glaring reason This is linearly dependent. Since V three is equal, toothy zero vector we know any set that contains a The zero vector is automatically a linearly dependent set. It&#x27;s also linearly dependent because we have four vectors with three entries and we also know a set with more vectors than entries per vector must also be dependent. So now that we know it&#x27;s literally dependent, it also follows that if we call it s, that&#x27;s the set s does not form ah, basis four r three And the reason for this is our basis must be linearly independent, and so obtaining linear dependence tells us we do not form a basis in this case. Now, another question we might have for this particular set is sure that it vectors are linearly dependent. They don&#x27;t form a basis. But would they spend our three? Well, one waited to determine that would be to let a be a matrix formed from V one V two V three end before the vectors in question. Now, if we ro reduce this matrix, we find that it&#x27;s row equivalent to 100 negative to 30 The zero vector remains unchanged. B three and we have zero negative. Three and five. So this is an echelon form. When we wrote reduce analyzing this echelon form, we see a pivot here, here and here so we can conclude that since A has a pivot in every row, we now know that s which consists of these vectors. Spans are three. Or in other words, if we spend s that contained the four vectors, we will have our three. So the set is linearly dependent does not form a basis. But that does span

Problem

Determine which sets in Exercises $1-8$ are bases…

Problem 6 Easy Difficulty

Answer

the set of vectors in $(1)$ does not form a basis for $R^{3}$ .

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

Linear Algebra and Its Applications

Lily A.

Caleb E.

Kristen K.

Michael J.

Vectors Intro

Vector Basics - Example 1

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the set of vectors in $(1)$ does not form a basis for $R^{3}$ .

Linear Algebra and Its Applications

Lily A.

Caleb E.

Kristen K.

Michael J.

Vectors Intro

Vector Basics - Example 1

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

Lily A.

Caleb E.

Kristen K.

Michael J.

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

Linear Algebra and Its Applications