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In Exercises 7 and $8,$ find the barycentric coordinates of $\mathbf{p}$ with respect to the affinely independent set of points that precedes it. $$\left[\begin{array}{r}{1} \\ {-1} \\ {2} \\ {1}\end{array}\right],\left[\begin{array}{l}{2} \\ {1} \\ {0} \\ {1}\end{array}\right],\left[\begin{array}{r}{1} \\ {2} \\ {-2} \\ {0}\end{array}\right], \mathbf{p}=\left[\begin{array}{r}{5} \\ {4} \\ {-2} \\ {2}\end{array}\right]$$

$(-2,4,-1)$

Calculus 3

Chapter 8

The Geometry of Vector Spaces

Section 2

Affine Independence

Vectors

Johns Hopkins University

Oregon State University

Harvey Mudd College

University of Nottingham

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In this example, we have an F finally independent set that's from been provided, consisting of vectors, V one, V two and V three. Let's define a new vector here to start out where P is going to be in our for a swell, and it's determined to be 54 negative, too. And to circle now with this vector p just defined is to find the very centric coordinates of this vector relative to this finally independent set s consisting of the vectors V one through V three. Well, to find those very centric coordinates, the first step is to consider a matrix a the Matrix A is going toe, have homogeneous forms v one till day, the two till day V three till day and then we augment with the vector p in question With such a matrix say, augmented in this way, what we're really doing is considering the following a vector equation will be unknown. X one times V one till day plus x two times V two till day plus x three times V three. Till day equals P till day where there should also be a till day here. So let's solve this system this vector equation using the augmented Matrix A or Matrix A will have the following form. First, let me put in the vectors V one, V two and V three. We have one negative 1 to 1 2101 that one to native to zero. And then we're augmenting with the vector p, and this is has entries. 54 negative two and two Now because we're looking at the home Longinus forms, we have to have ones at the less entries for each vectors. That's what the till days here are conveying. So let's roll. Reduce this matrix A In a calculator, this road reduces to the following We'll have 100 negative two in row one row two is 0104 Row three is 000 negative one and the remaining rose are all zeros. So let's copy those in carefully and recall that we've also augmented at the same time where the unknowns air given to be x one x two and x three x one through x three Our fourth this equation where we have homogeneous forms. But it turns out that now that we know X one is negative to weaken right. Negative. Two times the original vector V one plus four, which comes from here X two times V two minus V three where the negative one comes from here The solution to x three and this is now all equal to the vector p. Then finally, what this tells us is that the very centric ordinance off P relative to S, which has finally independent vectors, is negative to four and negative one. So the major key to this entire process is knowing already that we have a and finally independent set. If that hypothesis is not verified than this method would not work.

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