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

This system has no free variables, hence no nontrivial solutions, so the given set is linearlyindependent.

Calculus 3

Chapter 4

Vector Spaces

Section 4

Coordinate Systems

Vectors

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I'll give you a little friend. Does the linear independence on the sets on the given equations are one plus two cube mhm two plus D minus three D square and minus T plus two T Squire minus take you so we can write down in these s coordinate vectors as 2001 zero minus 312 and minus one to minus 10 Okay, now we will put them in the matrix. X equals to zero. If zero is the only solution, then the vectors are linearly independent. Okay, Now we are going to solve that. We will put them value in the metrics. So this will be 2001 zero minus 312 My next one to minus 10 The solution will be 0000 Okay, so now we will first. Yeah. Live are four will be minus work R one plus four. And for this we will get to Zito. My next one. Zito zero minus 3 to 0 on 01 my next 10 And in the other one, we will get one. Do sorry. He don't to the 0.50 point five and Tzeitel Okay. Now we will do two patients simultaneously, then in our two level. No, that theory are three plus R two and R four. We will minus 3.5 artery plus are for And by this they will get Tau minus three minus one zero Zito, Zito, my next. 10 01 minus 10 0040 Okay, so for this we can see that four x three equals to zero. It means X three equals to zero. And from the second row, even we can. From this rule, we can say that X two minus x three will be zero and extra is only re zero then X two is also zero Okay. And from the first role, we can say two x one minus three x two minus X three equals to zero and t's and this is zero then excellent is also zero. So zero that there is only solution. So the vectors are linearly independent. Okay. And I can say that in these three and this system, you can say that this system has nor three variables okay? No, very, very well. So on Lee travel solution, it means no, no Any non travel solution. Okay, so given, sir, it is so the given sad is linearly independent. Thank you

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