University:
Amherst College
Course:
MATH 271 | Linear Algebra
Academic year:

2020
Views:

241

Pages:

8
Author:

Jarrett Brock

Problem 1 Given: Point: (2, 4, 1) Line: x = 5 - 3t, y = 2 + t, z = 1 - 6t Steps: Find the direction vector for the line: (\vec{v} = <-3, 1, -6>) Find another point on the plane: Let t = 0, resulting in the point (5, 2, 9). Find another vector parallel to the plane: (\vec{w} = <5 - 2, 2 - 4, 9 - 1> = <3, -2, 3>) Find the normal vector to the plane: (\vec{n} = \vec{v} \times \vec{w} = <-9, -9, 3>) Write the equation of the plane: -9(x - 2) - 9(y - 4) + 3(z - 1) = 0, or -3x - 3y + z = -17 Problem 2 (Continued) Interior: (T_{x} = 2x = 0 \Rightarrow x = 0) (T_{y} = 4y - 2 = 0 \Rightarrow y = 1/2) Critical point: (0, 1/2) (T(0, 1/2) = 2(1/4) - 1 = -1/2) Boundary: (g(x, y) = x^2 + y^2 - 4) (∇f = λ∇g) (2x = λ(2x)) (4y - 2 = λ(2y)) Either λ = 1 or x = 0 If λ = 1: (4y - 2 = 2y) (y = 1) (x^2 + 1 = 4) (x = ±√3) (T(±√3, 1) = 3 + 2 - 2 = 3) If x = 0: (4y - 2 = 2y) (y = 1) (T(0, 1) = 8 - 4 = 4) (T(0, -2) = 8 + 4 = 12) Multivariable Calculus and Linear Algebra Practice Exam 3 200 8. Let A be the matrix A111 101 (a) Define what it means for a real number AER to be an eigenvalue of A. A real number is an eigenvalue of A it there exists nonzero vector such that Av = λν. a (b) Find all eigenvectors of A. det (A-1) = 2-1 0 = 2-11-1 ۱-۸۱ 0 = (2-1) [(1-x)²-0] 2 = (2-1)(1-1)2 =0 0 +0 1=2,1 12 ()()()⇒⇒ => Let x=2=1 y=2 y = 2x ()()()> X=Z=0 Let 8=1. Then the eigenvectors of A where xiy FIRE03. are the vectors X 2X X and (c) Find a diagonal matrix D and an invertible matrix P such that A PDP, or show that no such matrices exist. Since the dimension of the eigenspace corresponding to 1=1 is one while the multiplicity of the eigenvalue 1=1 is two, we know that A b not diagonalizable. Hence, there are no such matrices D and P.

Linear Algebra Practice Exam 3 Solutions

Related Documents

Description

Related Documents

Free up your schedule!

Take 5 seconds to unlock