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In Exercises 9 and $10,$ find the change-of-coordinates matrix from $\mathcal{B}$ to the standard basis in $\mathbb{R}^{n} .$$$\mathcal{B}=\left\{\left[\begin{array}{r}{3} \\ {-1} \\ {4}\end{array}\right],\left[\begin{array}{r}{2} \\ {0} \\ {-5}\end{array}\right],\left[\begin{array}{r}{8} \\ {-2} \\ {7}\end{array}\right]\right\}$$

$\stackrel{\mathrm{r}}{P}_{B}=\left[\begin{array}{ccc}{\mathrm{r}} & {\mathrm{r}} & {\mathrm{r}} \\ {b_{1}} & {b_{2}} & {b_{3}}\end{array}\right]=\left[\begin{array}{ccc}{3} & {2} & {8} \\ {-1} & {0} & {-2} \\ {4} & {-5} & {7}\end{array}\right]$

Calculus 3

Chapter 4

Vector Spaces

Section 4

Coordinate Systems

Vectors

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in his stroller were given a basis B and B No d lectures that that face is so we have the one we have be too and we have B feet and we're us to find the change of ordinance matrix. So we know that in a vector, X is equal to PB times the coordinate vector and this PV is the change of ordinance. Man picks on PB is actually the augmented mentions on by the factors that forms basis beat So pv will be b one b two b three and it is be negative. 1420 negative 58 negative to seven

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