University:
The University Of Western Ontario
Course:
CALC 1000A | Calculus I
Academic year:

2020
Views:

449

Pages:

4
Author:

Kyuwon J.

Advanced Matrix Practice Test Instructions: Solve the following problems. Show all steps clearly. Partial credit may be awarded for correct methodology. — Problem 1: Eigenvalues and Eigenvectors (20 points) 4 2 Let A = . 1 3 1. (10 points) Find the eigenvalues of A. 2. (10 points) For each eigenvalue, find a corresponding eigenvector. — Problem 2: Singular Value Decomposition (30 points) 3 1 Let B = . 1 3 1. (15 points) Compute the singular values of B. 2. (15 points) Find the matrices U , Σ, and V T in the singular value decomposition of B. — Problem 3: Transformation and Matrix Powers (25 points) 0 1 Let C = , which represents a linear transformation in R2 . −1 0 1. (10 points) Describe the geometric effect of C on a vector in R2 . 2. (15 points) Compute C 10 using properties of matrix powers. — 1 Problem 4: Real-World Application: Markov Chains (25 points) A company models customer behavior with the following transition matrix: 0.6 0.4 P = 0.3 0.7 1. (15 points) Find the steady-state vector of the Markov chain. 2. (10 points) Interpret the steady-state vector in terms of customer behavior. — 2 Solutions Problem 1: Eigenvalues and Eigenvectors 1. Eigenvalues: Solve det(A − λI) = 0: 4−λ 2 A − λI = 1 3−λ det(A − λI) = (4 − λ)(3 − λ) − (2)(1) = λ2 − 7λ + 10 = 0 Factorize: (λ − 5)(λ − 2) = 0 Eigenvalues: λ1 = 5, λ2 = 2. 2. Eigenvectors: For λ1 = 5: −1 2 A − 5I = 1 −2 2 Solve (A − 5I)x = 0, giving v1 = . 1 For λ2 = 2: 2 2 A − 2I = 1 1 −1 Solve (A − 2I)x = 0, giving v2 = . 1 — Problem 2: Singular Value Decomposition 10 6 1. Singular Values: Compute B T B = . Find eigenvalues of B T B: 6 10 det(B T B − λI) = (λ − 16)(λ − 4) = 0 Singular values: σ1 = 4, σ2 = 2. 2. Matrices U , Σ, and V T : Use eigenvectors of B T B and BB T to find: # " # " √1 √1 √1 √1 − 4 0 2 2 . Σ= , U = √12 √1 2 , V T = √1 √1 − 0 2 2 2 2 2 — 3 Problem 3: Transformation and Matrix Powers 1. Geometric Effect: The matrix C represents a rotation by 90◦ counterclockwise in R2 . 2. Matrix Powers: Note C 2 = −I (negative identity matrix). Thus: C 10 = (C 2 )5 = (−I)5 = −I Result: C 10 = −I. — Problem 4: Markov Chains 1. Steady-State Vector: Solve P x = x with x1 + x2 = 1: 0.6 0.4 x1 x = 1 0.3 0.7 x2 x2 Solve to find: 0.43 x= . 0.57 2. Interpretation: In steady state, 43% of customers will stay in the first state, and 57% will transition to the second state. — 4