University of British Columbia Mathematics 152 Exam April 2006: Linear Systems, Exams of Linear Control Systems

The april 2006 mathematics 152 exam from the university of british columbia covering topics on linear systems. The exam includes questions on matrix multiplication, eigenvalues and eigenvectors, determinants, and systems of linear differential equations.

Typology: Exams

2012/2013

Uploaded on 02/21/2013

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April 2006 Mathematics 152 Page 2 of 9 pages
Marks
[12] 1. [Short answer]
(a) [3] Let A=xy
y x and B=st
t s . Is it true that AB =BA for all choices of x,
y,sand t?
(b) [3] Write down a 2 ×2 matrix with real entries but with complex eigenvalues.
(c) [3] For which values of adoes 1a
0 1 have only one eigenvector (up to scalar multiples)?
(d) [3] If A=1 1
0 1 what is A100 ?
Continued on page 3
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Marks

[12] 1. [Short answer]

(a) [3] Let A =

[

x −y y x

]

and B =

[

s −t t s

]

. Is it true that AB = BA for all choices of x, y, s and t?

(b) [3] Write down a 2 × 2 matrix with real entries but with complex eigenvalues.

(c) [3] For which values of a does

[

1 a 0 1

]

have only one eigenvector (up to scalar multiples)?

(d) [3] If A =

[

]

what is A^100?

[12] 2. For which values of a and b are the vectors

 and

a b

 linearly independent?

[12] 4. Consider the quadratic function

f (x, y) = 2x^2 + y^2 + 2xy − 8 x − 6 y + 16

Find the minimum value of f and where it occurs.

[10] 5. Let T be the linear transformation from three dimensionsal space R^3 to R^3 with

T

 , T

 , T

What is the matrix for T?

[16] 7. The matrix P given by

P =

1 2

1 1 3 0 4

1 3

1 1 2 4

1 3

1 2

contains the transition probabilities for a random walk on three sites. The eigenvalues of P are 1, 1/3 and 0.

(a) [8] Find the eigenvector of P corresponding to the eigenvalue 1.

(b) [8] If the initial probabilities are given by a vector x with positive entries that sum to 1, find the limiting probabilities limn→∞ P nx.

[15] 8. Solve the system of differential equations

x′ 1 (t) = −x 1 (t) +2x 2 (t) x′ 2 (t) = − 2 x 1 (t) −x 2 (t)

with initial conditions x 1 (0) = 1, x 2 (0) = 1. Write your final answer in a form that does not involve complex numbers.

The End