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Point) Determine which of the following pairs of functions are linearly independent. 1. f(t) = ext cos(ut), g(t) = ext sin(ut) , u # 02. f(t) = 3t, g(t) = It|3. f(x...

Question

Point) Determine which of the following pairs of functions are linearly independent. 1. f(t) = ext cos(ut), g(t) = ext sin(ut) , u # 02. f(t) = 3t, g(t) = It|3. f(x) = x2, g(x) = 4lxl24. f(t) = 4t2 + 28tg(t) = 4t2 28t

point) Determine which of the following pairs of functions are linearly independent. 1. f(t) = ext cos(ut), g(t) = ext sin(ut) , u # 0 2. f(t) = 3t, g(t) = It| 3. f(x) = x2, g(x) = 4lxl2 4. f(t) = 4t2 + 28t g(t) = 4t2 28t



Answers

In each of Problems 1 through 8 determine whether the given pair of functions is linearly independent or linearly dependent. $f(x)=x^{3}, \quad g(x)=|x|^{3}$

Hello and welcome to problem one of chapter 3.3 linear independence and their own skin. And this problem we have F F T equals t squared plus five T and G M t equals t squared minus five. To our question is to find out if these are linear independent functions to do this. We use the Ron skin, Which is um a two x 2 determinant of um between two functions involving the derivatives in this case will be F 15 whatever the first function is. Um F property the bottom on the top right. We'll have GDP and uh to prime of T. All right. Um If this is equal to zero, then uh we will have a dependent too dependent equations. But if it's anything but zero, you'll have two independent solutions. So let's plug in these functions and see what our answer is. Um F F let's just plug it right in T squared past 15 is the top left. Its bottom part here will be to T plus five. Alright this top right will be T squared minus five to And this bottom right will be two T minus five. Great. Um the trick for two x 2 determinants is we multiply these two together. Then we subtract it from these two. So let's do that now. We'll have t squared plus five two Times The Quantity to T -5. Yeah minus this. Um Two squared minus five to times To T Post five. All right. Now if we look at it these are very similar. Um So we are going to have to multiply it all out and see what the answer is. I'll do this part in green. So he squared in two team to t cubed five T. And two T. Will have 10 T squared minus five T. So it's a plus five C squared And subtract 25 T. Great. Let's do this next part here. So we'll do this big truck sign and over this thing. So we'll have to t cubed. So that will definitely cancel minus 10 T. Plus five T. So this would be minus five T squared. Um And to do a -25 T. All right. Now looking at them Uh these two parts can see that this to teach future will cancel this. 25 T. Will cancel. But this this part here will not cancel. I will have the solution to Tennessee square and 20 square is only equal to zero and T equals zero. So let's look at at what that means. That means when um F F. T. Equals zero and G M. T equals zero. They are linearly um Mhm literally dependent. Which is of course that is true because zero does equals equals zero. So the solution of this is linearly independent.

Hello and welcome to the next problem where we have F F. T equals E. To the lambda T. Times co sign of beauty in GFT. Who was either the lambda tetum sign of beauty, very similar equations. Our problem is to find out if these are um literally independent or literally dependent. And as all of the problems that we do, we have to in this section we're going to have to uh using the wrong skin. All right. And that's the tuba to determinant of the two functions and their derivatives, which means we have to find the derivatives of these two yuck. Uh huh. That will be a challenge. Um Let's do it. Uh We're going to start off with the F 50 so F prime of T. We'll use the product rule here and um sort of with E. To the lambda T. Well, he elevated the drift of of the first part times co sending me T. So it'll be lambda Eat the land of two times. Co sign of beauty. Um What's the driver of the second one? That's minus limb minus mu sign of beauty. New sign of Meaty times. This uh he to the land to great and this secondary which is going to look pretty similar. G. Prime too. And that's equal to well take the derivative of this them to eat to the land. Yeah. I love to eat bland at teatime sign of beauty. And we're going to add that to all this derivative is a co sign positive coastline to be um You co sign from UT times E. To the land. It too perfect. Now what's uh let's plug this into the runs he and and see if things cancel out. Yeah. I'm guessing that there's going to be some sort of cancellation because there's a co sign and signs and there's usually ah can cancel out to be sine squared cosine squared but ah you only know when he tried out so so the runs can hear uh F. F. T. Times uh do you priority? That's going to be very long. But let's just uh yeah let's just write it out. Not um not distribute but just write it out either landed. T. Co sign of um. Ut times well this term. Mhm. Be a lambda E. To the land of T. Sign lambda T. Or Senate beauty plus this new co sign of Me Too. Times each of the land of T. What's up long parts and this is going to be subtracted from um equally long part. It's rather the next line but yes I minus E. To the blender too. Sign with Me Too. Times this part groups shouldn't be like that. Um To multiply this by the next part. All right. Yeah. Uh It's a nice linda E. To the land of T. Sign. Oh Lindy 2 λ T. Cosign. Because we're doing this part. Co signing with me. T. Let this new sign of mut E. To the land itty. Let's see. Do any of these terms cancel out. And I'm most concerned with the sine squared and cosine squared because those can cancel. So I'm looking at it and I see a uh this coastline coastline will fly by that coastline to get a cosine squared. And um we will have a I'll just highlight that in red. Who's on there? And that goes on there. Then we'll have a um minus sign an a minus sign. So become a positive sign. Um That would go like that. So that means this part will cancel. Um Well the rest of it though I'm not sure. And so I think we're going to have to simplify a little bit. So we'll have the wrong skin. Well writing and red now I guess. Um Let's multiply this out. We're going to get um linda E. To the tool entity Khorasan son. I'm not going to write out the whole thing taking up way too much time. Um Then the next part is this lambda sign and co sign. So yes this is going to uh cancel linda heated to lambda T. Sign co sign. Which means this part will cancel. And this pearl cancel out right. Um So we can cross it out. Um That means we have this part left. So uh run skin is equal to um Will be uh oops. No more equals. This is Reddit. I'm here. Um. E. To the lambda T. Mu times. I will be 200 meeting because another blend into their New Times. Co sign squared mut. Um And that's going to be added to E to the to land t. New Times. Sine squared. Uh. Beauty. Great. Um So as we can see, we factor at this uh, ah E to the 2 λ T. Mu. Uh that is the simplified run skin. And um I was wrong. This uh, is not equal to zero Unless musical to zero or um yeah, just unless musical zero. So that means this is literally independent.

Here, we need to determine whether the vectors are linearly independent are linearly dependent. In Puerto. Pito denotes that of all people in Wales of degree less than equal to do. And it is a factor space over some field. F. So the first past first part of this question, as the victors, everyone is equal to two minus X plus four X square. Then we do is equal to three plus six X minus three plus two X square. Then we three equal to two plus 10 X minus four X square. Now we need to determine whether these vectors are linearly independent or linearly dependent. So the solution for this part is solution. So the standard way to check whether the vectors are linearly independent or dependent is as follow. So let given K two and K three are any skill. Urz, our scholars from field from some field, F no given times of even plus K, two times a way to plus K. Three times of the three is equal to zero. Now we need to solve this which is equal to so which imply Kevin into V one is minus X plus four X square plus Cato Into video is three plus six X plus two X square plus K three into V three is two plus 10 X minus four. X square Is equal to zero. So taking this constant term of this polynomial common. So we have two K one plus three K two Plus two K 3. Then the coefficient of excess minus given plus six, kato. Last 10. K three into X. Yes, taking the coefficient of x squared common. We have four K. One plus two K to minus 43 Is equal into excess square is equal to zero. No. Okay, Here we have opponent polynomial of degree two which is equal to zero. So on comparing the coefficient we have on comparing you get so comparing the coefficient we get two times of Caven Last three times of K two plus Okay, three is equal to zero. Yeah then minus given plus six, kato. Last 10. K three Is equal to zero. Then four K 1 plus two, Kato. Yes minus times of four K three is equal to zero. We can write this system of linear equations in matters form as. So the metrics will be the first race to three two Then -16. Then than four two minus four. Into the skill er column. Victory is given you too and Katri which is equal to 000 mm. Now we need to solve this system of homogeneous linear equation. Yeah. No it's all. Mhm. Yeah. This system or homogenous. Yeah. Mhm. Do you need equation as follow? So the the humane tid metrics for the system will be The 1st race. 2, 3, 2 and zito The Second race -16 and zero. Then the 3rd row is for two minus 40 Now we're applying the elementary row operations. We have the first operation is Yeah or do is replaced by our two minus minus times. Are won by two of our win. So the output is yeah 232 then 0 15 by 2015 by two than 11. It is 00. The 3rd row is full. Do -4, zoo playing. Another through operation is R three is replaced by okay Our 3 -2 times of our win. So the output is 232 0 15 x two 11 30. Is you? The third row is zero minus food -8 and zero. Again the cooperation is R three is replaced by our three minus Minour times of eight x 15 into root. That is harder. So the output is 232. Okay zero. Then the secondary 0, 15 by two 11 zero. Then the 3rd row is 00 -32 x 15. Zero. Yeah, no from this vineyard. The vision is so the linear equation are So the first original two K 1 plus three K two plus two. K three is equal to zero. Then the second division is 15 x two. K 2 Plus 11. K three is equal to zero. And the last equation is -32 x 15. K three is equal to zero which imply So value of K three is equal to zero substituting this value. In just our equation we get 15 by two. K two less zero is equal brazil which imply care two is equal to zero, substituting the value of K two and K three in the first division Weird two times of Cuban less 00 is equal to zero, which implies Given is equal to zero. Thus here the solution is so the solution is kevin is called kato is equal to K three is equal to zero, which is a trivial solution, which is a trivial solution. Does the system have a trivial solution? Which implies there the given set of vectors. We one, V two and V three R linear linearly independent. Hence the vectors we will be too and we three. All right, linearly independent. All right. So this is the solution for our first part. Now moving towards the second part of this question is so part B be part B contains the vectors even. So let us call the vectors as usual. Which is even is equal to 1-plus 3 x Plus three x square. Then we two is equal to x plus four X square. Then we three is equal to 5-plus 6 x Last three X Square. Then V four is equal to seven plus two X minus x square. Now we need to determine whether this vector is linearly independent or linearly dependent. The solution for this part is ehlert given K two, K 3 and K four are any skill er skills from field. If no Given into the one plus K two in two, V two plus K three and 23 plus K four in tv for is equal to zero, which implies that given into one plus three X plus three X square mosquito into explodes for X square. Yes, K three into five plus six X plus three X square plus K four into seven plus two, X minus x square is equal to zero. Now taking all this killers or all the constant term of this polynomial common period. Gay one plus five K three. Last seven K 4 plus taking the coefficient of X. Common bigger three Cuban. Las kato. Last six K 3. Let's do K four into X. Now again picking the common terms of Yeah, coefficient of X square common. We have three K one plus for kyoto. Last three. K 3 plus -1 into K4 in two. X squared is equal vision. No comparing Yeah the coefficient Viet on competing Bigot. Yeah kevin plus five K three plus seven. K four is equal to zero. Three. K one plus care two plus six K three plus two. K four is equal to zero. Then the last equation is 3K 1 plus four kato. Let's try to get a minus care for- Key for is equal to zero. This system. Can we return in matters form as the first stories? 10 Why seven? The secondary is three 162 The 3rd row is three for three minus one. Into this killer victories. K. one. You too K three careful which is equal to 000 Now we solve this system as follow. No it's all the bible system of homogeneous equation using elementary operation. Mm. Elementary operation. As for you. So, so the augmented matrix is 10 57 zero. The second rule is 3 1 6 to 0. The 3rd row is three for 3 minus one zero. Which employ city. Now we use the elementary row operation which is our two year is replaced by road to R. Two is replaced by Our 2 -3 are one. So the output is 10 57 zero when -9 -19 It is 00. The thorough is three for 3 -10. Now again we use the demented operation, which is R three is replaced by Our 3 -3 times of our win. So the output is 1057, then zero one minus nine -19, food minus 12 And -22. And here it is 000. Then the elementary operation used is artery is replaced by Our 3 -4 times of art. The output is 10. The first race 10 57 zero. The second rule is 0 1 -9 minus 19 0. And the 3rd row is zero 0 24 54 zero. No, we get the linear equations are the linear equation odd. So gay one Plus five K 3. Last seven K 4 is equal to zero. Can do -9 K three -19. is equal to zero, 24. K three Plus 54. K four is equal to zero, No name it as equation first equations second an agreement to now we use this equation to solve and to find The values of Kevin Kentucky three as well. So from a vision 3rd we have from third, 24. K three. Last 54. K four is a 4 to 0 vision plie Each employer Chetry is equal to is equal to -9 x four. Okay for Subsequuting this value of K three in equation too, substituting value of key three in to get. So if you get X two is equal to nine, sorry, the variable is kito kato is equal to 93. Last 19 K four which is equal to nine into now, starting the value of key three here, figure nine Into -9 x four give or less 1984 which is equal to. After solving we will get minus fiber for okay for so kato is equal to minus five by four, careful substituting the value of K two and K three in first, substituting the value of K two and K three invest. Do you get So you get K three. K 1 is equal to -5 K one Sorry -5. G three minus seven K4 which is equal to -5 times of subsiding developed country. We have -5 times of -9 x four, careful minus seven, careful is equal to Now solving this week. It my 17. We got 17 x four, careful this vigor. Thus we get the solution Given is equal to 17 x four K 4. Kato is equal to minus fiber four K four, K three is equal to minus nine by four K four, and K four is equal to care for here, therefore is free variable. Now we see that this system has nontrivial solution. Yeah. The system as nontrivial solution. Does the set of this vectors does the vectors, We won TV two, we three before our odd linearly dependent. All right. The nearly dependent. Yeah. So this is the required solution for our caution.

Hello and welcome to a new different equation problem. And this one we're trying to find the E 23 X and E. To the three times quantity x men is one are literally independent uh equation uh equations. So we're going to use the run skin here to buy two determinant to find out if that is the case. Uh The first step was using the wrong skin is to find the derivatives of F. Of X and G of X. So I'm going to do that now. F prime of X Is simply three E to the three X. And um G prime of X here is equal to E. All three E to the three times of 20 x -1 began appear. Remember this is just the chain rule? Um So yes, let's um apply the brown skin. The skin is F. Of X. G. Prime of X minus G. Of X. F. Prime of X. Where it doesn't really matter which uh to uh which order we choose F. Of X. And the O. Excellent because either way we don't want them to equal zero. If they do, then the equation is linearly dependent. So let's do that. Now if the vaccine is deep prime of X will be um three E. to the three x. Time to eat to the three Times X -1 and G of X. In terms of products will be three e. to the through reacts. I'm doing f prime of x 1st times G of X, which is E. To the three times x -1. In fact, these cancel and or equal to zero. So this means ffx and Vfx are linearly dependent. That's the problem.


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