determine-which-sets-in-exercises-15-20-are-bases-for-mathbbr2-or-mathbbr3-justify-each-answer-lef-6

Question

Determine which sets in Exercises $15-20$ are bases for $\mathbb{R}^{2}$ or $\mathbb{R}^{3}$ . Justify each answer.
$$
\left[\begin{array}{r}{1} \ {-6} \ {-7}\end{array}ight],\left[\begin{array}{r}{3} \ {-4} \ {7}\end{array}ight],\left[\begin{array}{r}{-2} \ {7} \ {5}\end{array}ight],\left[\begin{array}{l}{0} \ {8} \ {9}\end{array}ight]
$$

Gideon Idumah · Answer

to check if they&#x27;re Lee. If the former busies so thought the check. If the vetters from a busies we need to check when it&#x27;s a check. If the linen independence if they our linearly independence. Ah, Now do we go about checking these? So we do these by checking if the matrix formed by these vectors, that is our 11 minus two minus five minus one. Sue 705 We check if this is in vegetable because if it is inviting boot and shows that the columns are legally independent So how do we go about checking days? So we tried to make sure that we have three pie votes was after the pie votes. Then it is in vegetable. So let&#x27;s rock. Let&#x27;s reduce these. Who do? I found it on everything on. Then it&#x27;s 10 So we do are too equal toe arsu minus are worn. And then our three or Dr Road 3/4 or all three lost suit. Times rule warned. When you do that, you&#x27;re gonna off want 00 minus 54 minors. EADS seven minus seven nine. Yeah, that&#x27;s where you&#x27;re gonna get so next aren&#x27;t on these 20 So what I do is I I do, um Or three. He called so rude. Three blows So rude, Sue. So you have won&#x27;t minus 57 04 minus seven 00 minus five. So we don&#x27;t even need to reduce this tree. Rachael on form. We just ask that this trip I vote. There&#x27;s a pile of this pie votes So we see that this matrix as Chery Pie votes on dance is impossible and therefore says it&#x27;s in vegetable. They are literally independence. I didn&#x27;t have the independence that implies that&#x27;s the former business.

Gideon Idumah · Answer

this question. We want to check whether minus four, comma six. And so with minus three wanna check? Is this busies for our TSU? That is the question. So now recall. That&#x27;s for this to be a busies For this to be a business one kite era is that they have to be Linda Independence. They have to be linearly independence. I have from these from this too with see Dad&#x27;s I can write minus four and si ves as sue more deployed by two minor story. So I&#x27;ve been able to express these as a linear combination of days. So this implies that they are not really gnarly. Independence got dependence. So memorize the dependence on this are not a business, so they can be in business because they depend on the charter.

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

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

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

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In Exercises 21 and $22,$ mark each statement Tru…

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Linear Algebra and Its Applications

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

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

Linear Algebra and Its Applications