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In Exercises $1-4,$ find the vector $\mathbf{x}$ determined by the given coordinate vector $[\mathbf{x}]_{\mathcal{B}}$ and the given basis $\mathcal{B} .$$$\mathcal{B}=\left\{\left[\begin{array}{r}{3} \\ {-5}\end{array}\right],\left[\begin{array}{r}{-4} \\ {6}\end{array}\right]\right\},[\mathbf{x}]_{\mathcal{B}}=\left[\begin{array}{l}{5} \\ {3}\end{array}\right]$$

$x=\left[\begin{array}{c}{3} \\ {-7}\end{array}\right]$

Calculus 3

Chapter 4

Vector Spaces

Section 4

Coordinate Systems

Vectors

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here. In this example, we have a set be that consists of two vectors which forms a basis for our to and what we like to do here is given a particular coordinate vector for X relative to this basis be is coordinates five and three. We would like to determine what the vector X itself ought to be. To do this, let's call the first vector here V one and the second vector V two. Since we already have it The coordinate vector relative to the basis be it then follows that the vector X itself is going to be equal to first take five which comes from the first cornet times, the first basis vector V one, then add to it the second coordinates three coming from the coordinate vector times the second basis vector B two. So this is our vector X and all we're doing now is determining what this is equal to. So we have five times V one, which is three negative five plus three times V two, which is negative +46 So working this out bit by bit. The first vector scaled by five is 15 negative 25 and the scaler multiplication of three times. This vector is negative. 12 and 18. So altogether. After performing the vector addition, we see that X itself is equal to three negative seven, provided that has this as it's coordinate vector.

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