in-exercises-27-30-use-coordinate-vectors-to-test-the-linear-independence-of-the-sets-of-polynomia-3

Question

In Exercises $27-30$ , use coordinate vectors to test the linear independence of the sets of polynomials. Explain your work.
$$
(1-t)^{2}, t-2 t^{2}+t^{3},(1-t)^{3}
$$

Priyanka Sadarangani · Accepted Answer

okay. We have to test the linear independence of this sort of polynomial. And the questions that are given is one minus T Squire on D minus to D squared plus D u and one minus T to leave our whole cube. And as we simplify that, we will. We can write this one minus two p plus t square. And this is simplified already. So t minus two t square plus teach you by simplifying that we can write it one minus three D plus three d Squire minus D cube. Okay, now we can. Alright, directors off these equations. These are Tzeitel. It&#x27;s coefficient off. Did you? This is coefficient off the square. This is coefficient off T. And this is constant. So same here. It will be one minus two, one and zero. And in this question, it will be minus 13 my industry and one. And if we have to check the linear dependence independency, we have to put these metrics in these vectors. In this equation, he x equals to zero. And if zero is the only solution, the vectors are linearly dependent. So now we will put the values in the metrics. It will be zero one minus 21 one minus 210 minus 13 minus three and one is one Here and the solution will be zero Zito, Zito! Zito Okay, now we all this all this First we will do, um in rotary we will go to our two plus r three and in drove for we will do minus r one plus r four So we will get Zito one minus one Luisito one My next to Titi Zito Zito Monastery Titty Zito A Tzeitel Tau minus 20 Okay, now again, we have to solve it again And we will do in r three r three equals two Terry R one plus r three and in our four they will do minus two r one plus are four and we will get zero Who won my next one Brazil one minus 230 Zito Zito Zito Zito 00 Zito Zito Okay, so in this equation from the road three and RAV four, we can say that. Okay, First, it can say that X two minus X three equals to zero. So next week was toe x three Because there will be so many way lose off extra hair and from the raw, too from the road to means this means this role. We can say x one minus two x two plus three x three equals to zero. That means Exxon. It was 22 x two minus three X three. Okay, that will be again. I will put the value two x two s two x three minus three x three equals two minus x three So finally, finally X t is very variable. Ex tree is free Variable. So so the sad is not linearly independent. This is all fine alongside. Thank you.

Doruk Isik · Answer

in this problem, we&#x27;re given a set of point no meals. They were asked to determine whether these are the nearly independent or not. So let&#x27;s rename our polling Ulises people on P two and he three. And let&#x27;s look at the corner of directors for each aluminum so that people would be We&#x27;re being 10 negative to one. So this is the collection talk T zero power first Poverty, second poverty and power Bi. So Anandhi Corning, back to before p two would look like zero 102 and according vector for P. D. With the play 11 name, too. And zero Um, this should be the second. Okay, so let&#x27;s write the augmented majors for my desk. Orton back there. So listen by electric, eh? Trade will be People want you to pee pee so that this 10 negative negative. 10102 and 11 day to zero. And let&#x27;s look at the register or echo or all our men takes a and that is 100001000010 Now, as you can see, we have one favorite element here Want people element here. So basically, we have a favorite element for each hole. It means that, uh, this bull no meals. Are we nearly independent? Why? Three columns, three event fellows.

Doruk Isik · Answer

in this problem we&#x27;re given a Senate. Were test those for linear get fancy. Let&#x27;s first right. Let&#x27;s first explain these pollutants or this is R P one p to a This is B B C P. One is eight minus 12 T plus 60 spread minus you. And if you were to write the basis so I write the important factor for people in that would be age negative 12 6 and 91. So the first time is basically the corruption, but he&#x27;s your firm t one t 2 23 So dumpy to would be an expanded on written is nine minus 60 plus t squared. So they recorded. Lecture her pee, too. It would be nine negative 61 and zero and P three is already an expanded form. So one plus 60 minus five piece fried. Let&#x27;s take you so it only Cordant back through four p three with little A, um 16 and 8 to 5 for one. So let&#x27;s write. Be augmented Major&#x27;s phone like these porn in vectors. We have a JJ negative 12 6 negative one mine. 96 10 in 16 negative five and one a CZ you can see from here. Okay, First, let&#x27;s rename those lectures us, B, B one, B two and B. As you can see, we can write be three as a combination of the term you want. So what does B B three&#x27;s even privy to minus We want, since we can write one of the factors as a combination of others were going to say, That piece hold on the over this system is linearly depended.

Doruk Isik · Answer

in this problem were given instead of, well, no meals that were asked to determine order to set off a basis. So let&#x27;s start with writing. Recording it vectors for Colonial. So were the 1st 1 is the look by this five negative three for two. So basically the surgical options off. Zero t wan t to three. So important. 2nd 1 it will look like this. 918 a six and 1/3 1 would be six negative to buy here and actually have one board that is t cubed. And this would look by 000 Celeste first, right. The fomented system. Momentous made suits for my inspectors that it&#x27;s five negative three for two 918 Negative. 6692250 and 0001 Let&#x27;s name this. Make it says a Let&#x27;s find the regional Chekalin form of amenities, and that is 10000100 Keyboard one or 00 and 0010 No, we have one PM of element here, and we have another favorite element here. You don&#x27;t have anything of it elements of this wrong, And we, uh at Hubert element here. So we had three cola. Sorry, he had 12344 columns, but he had t give it&#x27;s Ellen&#x27;s. So this means that these vectors so in Viktor&#x27;s do not spend four, it means that given colonials do not spend, he threw.

Christian Otero · Answer

Okay, so we&#x27;re gonna go ahead and show that, um, these three vector functions X one x two and x three are ah Leonard dependence on each other. So we do normally is evaluate the Ron skin matrix, the model of Ron&#x27;s key indeterminant. So in this case, it&#x27;s the determinant of the following matrix. And since we have three, three, um, different vectors, we&#x27;re gonna have a three by three matrix and so that I can split up into three individual parts. So we&#x27;re headed right, then known and, um, evaluate each one separately. So this is what we end up with. So when we have, um, go ahead and have this part first and then this and finally this. So we have these three determinants to evaluate, and so we&#x27;ll go ahead and do that and notice how they&#x27;re zeros in the and these matrices. So, um, will have an easier of time evaluating it. So here the diagonal was nine to you to the fifth and the other diagnose zero and second, our diagonals three teeth of the fifth and are the diagonal zero. And for the last one, our first Bagnall zero and our souther diagnose is, uh, negative. Ah, three t to the fits. And then since it&#x27;s the opposite diagonal will have another negative. So it sends a positive three to fit. And then we end up with, um, teeth, the sixth terms for each one. So when we have factoring out the teeth of six, we have nine minus six, minus three or zero. So since we evaluated the round skin of, ah, three vectors and turns out to be zero, um, for all time t on the interval, Negativity to infinity. Ah, we can say that all three of these vectors are linearly deepened.

George Dekermenjian · Answer

in this video, we&#x27;re going to try to see if three functions air linearly independent on an interval. The three functions are the 1st 1 is one over one minus t squared. The 2nd 1 is one over to minus T squared, and the 3rd 1 is won over three minus t squared. Here. The squares are for the quantity. Okay? And we want to see if these air linearly independent on the interval, minus 11 Okay. Where linear independence means that a linear combination of these functions is equal to zero. So here&#x27;s a linear combination. C one times the first function, plus C two times a second function plus C three times the third function. This is equal to zero. Um, if and only if all these seas are zero means that the functions are linearly independent. Okay, so we start by saying, Let&#x27;s take a linear combination. Said it equal to zero. And let&#x27;s see what see onesie to see three have to be. Do they have to be zero, or can they be not zero. What we&#x27;re gonna end up with, we&#x27;re gonna end up showing that they actually have to be zero. And so we&#x27;re gonna conclude that the functions are linearly independent. So first, what we&#x27;d like to do is we&#x27;d like to multiply both sides of that equation. By the factors one minus t squared to minus t squared and three minus t square. That way we get rid of all the fractions. And so we end up with Here is C one times to minus T squared times three minus t squared plus C two times one minus T square times three minus t squared plus C three times one minus t squared, times two minus t squared. And we make this equal to zero. Okay, so what we have on the left side is a polynomial of degree four. So, as you know, a polynomial of Decree four has some coefficient for T to the fourth, plus some other coefficient for tea to the three. Plus some other coefficient of t squared. Another coefficient of tea, plus the constant. Okay. And we know that this is equal to zero. And so the only way that could happen is if each of these are actually equal to zero for every value of tea between minus one and one. Okay. Now, since we only have three unknowns. We only care about three of these will pick the three easiest ones to work out. So we&#x27;ll start with the coefficient of tea to the fore. Well, the coefficient of tea to the four has to take the square term of this times The square term of this The square term of the 1st 1 has a coefficient of one. The square term of the 2nd 1 has a coefficient of one. Therefore, the constant here is C one times one Okay, plus. Or maybe I should write it down here with me. Erase this. Okay, So the coefficient of teacher the four C&#x27;s c one plus for the second term we have the coefficient is again one for T squared coefficient is 11 times one is one. So the coefficient of T to the fourth from the C two times expression is just see two times one. And similarly, the coefficient of teacher the four in the from the third factor is equal to see through. Okay, now let&#x27;s take a look at the coefficients of T. Cute. Okay, Now, how do we get a T cubed term to get a T cubed term, you could either multiply the quadratic term of office times the linear term of this or vice versa. The quadratic term of the first expression we said, was one. The linear term of this expression will be negative six. And so the product will be negative. Six. So let&#x27;s start by writing that will have negative six c one. The other way is to take the linear term off the first factor, which is negative four times the quadratic term of the second factor, which is minus for. So we also have minus four C one. Okay. Plus, if we repeat the same procedure here we take the quadratic term times the linear term again we get minus 62 and then if we reverse that, take the linear term from here, which is minus two times the quadratic term from here, which is one we just get minus to see to and repeating the same thing forward from the front third factor, you end up with minus C three Linus to see to I&#x27;m sorry. Minus four C three minus two C two. And this whole thing is the coefficient of tea cube. Okay, now, let&#x27;s just take this constant term because that&#x27;s super easy. How did we get a constant Well, we just take the constant term of this, which is two squared or four times the costumes. Term of this, which is 99 times Force. Of course, 36. So we have something times t squared, which we don&#x27;t care about. Plus something Times T plus Now the constant term again from this factor, See, one is just four times three or four times nine, which is 36 c one plus the constant term that multiply. See, too will come from to Causton. Term of one times a constant term of nine. So here we have nine C one plus the constant term here is one to constant term. Here&#x27;s four. So we get four c three. This is sissy too. And that&#x27;s the constant term, and this is equal to zero. Okay, so we know that each of these coefficients must be equal to zero. So starting with this one, we know this must be equal to zero. So we have an equation that we can change a rate from that next. If I combine this, I get negative. 10 c one minus eight. C two minus 62 or C three Rather Okay, and that&#x27;s also equal to zero. And finally the last parentheses here, which is the constant term, must also be equal to zero. Okay, so if I combine these three, I have a system of equations with three unknowns, and it&#x27;s not hard to verify that The only way this has a solution is this has a unique solution when see one equal see two equals C three, which is equal to zero. And as we said before, since these coefficients are all forced to be 0 to 3 functions, we started out with must be linearly independent.

Problem

In Exercises $27-30$ , use coordinate vectors to …

Problem 29 Easy Difficulty

In Exercises $27-30$ , use coordinate vectors to test the linear independence of the sets of polynomials. Explain your work.
$$
(1-t)^{2}, t-2 t^{2}+t^{3},(1-t)^{3}
$$

Answer

$(1-t)^{2}, t-2 t^{2}+t^{3},(1-t)^{3}$ are not linearly independent.

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

Linear Algebra and Its Applications

Anna Marie V.

Kayleah T.

Michael J.

Joseph L.

Vectors Intro

Vector Basics - Example 1

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$(1-t)^{2}, t-2 t^{2}+t^{3},(1-t)^{3}$ are not linearly independent.

Linear Algebra and Its Applications

Anna Marie V.

Kayleah T.

Michael J.

Joseph L.

Vectors Intro

Vector Basics - Example 1

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

Anna Marie V.

Kayleah T.

Michael J.

Joseph L.

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

Linear Algebra and Its Applications