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In Exercises $15-20,$ write a formula for a linear functional $f$ and specify a number $d,$ so that $[f : d]$ is the hyperplane $H$ described in the exercise.Let $A$ be the $1 \times 5$ matrix $\left[\begin{array}{ccccc}{2} & {5} & {-3} & {0} & {6}\end{array}\right] .$ Note that Nul $A$ is in $\mathbb{R}^{5} .$ Let $H=$ Nul $A .$

$\begin{aligned} f &=2 x_{1}+5 x_{2}-3 x_{3}+6 x_{4} \\ d &=0 \end{aligned}$

Calculus 3

Chapter 8

The Geometry of Vector Spaces

Section 4

Hyperplanes

Vectors

Harvey Mudd College

University of Michigan - Ann Arbor

University of Nottingham

Boston College

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Okay, so we want to find the hybrid thing described by when is given by this measure is right here. So then the Vigna functional if functional ex Victor, it's just a s. So this is just to x one us five x two Mars by three x three plus by six Next five Now this is effectively the same thing as Ryan x one x two x three x four, which is either to to x one plus by five x two Minds by three x three Suspects six x full. So the biggest difference between these two is that here using X fire and here, using X four now in describing hybrid friends, this wouldn't really matter so much because he just switcher to access is so It doesn't really matter too much now, since F since hedges and null space off, eh? It follows that this f should be equal to zero. So your linear functional is given by this thing with D is equal to zero

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