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Find the interpolating polynomial $p(t)=a_{0}+a_{1} t+a_{2} t^{2}$ for the data $(1,12),(2,15),(3,16) .$ That is, find $a_{0}, a_{1},$ and $a_{2}$ such that$a_{0}+a_{1}(1)+a_{2}(1)^{2}=12$$a_{0}+a_{1}(2)+a_{2}(2)^{2}=15$$a_{0}+a_{1}(3)+a_{2}(3)^{2}=16$

$7,6,-1$

Algebra

Chapter 1

Linear Equations in Linear Algebra

Section 2

Row Reduction and Echelon Forms

Introduction to Matrices

Campbell University

Harvey Mudd College

Idaho State University

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for this exercise, we're going to be solving for the degree to polynomial p a t such that this polynomial is going to pass through these three given points. Let's start by using the function notation as follows. We know that p of one must be equal to 12 p of two. Must be equal to 15 and p evaluate at three must give us back 16. Then, if we look at the left hand side, P of one is just placing a one here and here, for example, so it would become the equation. A not plus a one times one for T plus a two times one for T squared is equal to the right hand side. 12. Let's do this again by placing to here and here and setting it equal to 15 all together, we would obtain the second equation that is a not plus two times. A one two is then going to be squared. Soul go plus four times a two, and that's going to be equal to 15 likewise, replacing three into pft and we get the equation that a not plus 3 +81 plus 9 +82 must be equal to 16. Now, at this stage we have a system of three linear equations in three unknowns. So if we air consistent, then will be able to determine a solution for Pft. Let's next take this to an augmented matrix for this matrix we see here. This portion has been augment has been placed here, and the right hand side has been augmented with the 12 15 and 16 Let's begin role reducing to see if we have a solution. So for the first reduction step, I'm going to take negative one times. Row one added to row to replacing row two. So first we have 111 12 in Row two will become zero one, three and three now for Row three would do the same operation multiply row one by a negative one added to row three and record the result here. The result is going to be zero to eight and four. Now for the next step in solving this system, let's pivot off this entry here. We're going to take to row operations like before, But first, let me copy that middle row, which is 0133 Now. Our goal is to eliminate this entry above next. So multiply wrote to that. We see here by negative one at it to row one and place the result here we'll obtain one zero negative two and a nine. Likewise, to eliminate this entry multiply row to buy negative to And at the result to row three, we'll obtain zero 02 and negative too. Now, we're just about there for the next operation. Let me take an extra step where we divide this entry by two, thereby obtaining an entry of one or record. The result down here will have 001 negative one. And you can do row operations at this step filling out these entries here. I feel like that's a lot of work to do at once. So let me just copy in the remainder of the Matrix. So now we have two objectives like before pivoting off of this entry here. We're going to eliminate first this entry so we can say that this is robe equivalents to the following first. Copy that last row 001 negative one. Then multiply Row three by negative three and add the results of Row two will have 01 zero by design and then six here for the next operation. We want to get rid of this negative, too. So if we multiply row to buy positive too, add it to row one will achieved. That result will have 10 zero here, and a seven at this stage were ready to record what we were able to solve for recall that the unknowns are a not a one and a two. So by our form of the system, a not is equal to seven. A one is equal to six kind of missed their six and A to this is a zero here is equal to negative one. So that tells us that the required polynomial is p f. T equals a knot, which is seven plus a one, which is six times t plus a two, which is negative one times t squared. So this is a required polynomial that goes through all three points

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