Solutions to the Attendance Quiz for the Linear Algebra Class
1. Find all the eigenvalues of the matrix
and determine a basis for each eigenspace.
Sol.: The Characteristic matrix is
Taking its determinant, we have
Solutions to the Attendance Quiz for Lecture 8
1 Find the Fourier series of f (x) = 2x on the interval (2, 2).
First Sol. (By transfroming to the interval (, ) that has nicer-looking formulass)
We transform the problem to the standard interval, (, ), by c
Solutions to the Attendance Quiz 1
1. Using the definition find the Laplace transform Lcfw_f (t) (alias F (s) of
3, if 0 t 1;
f (t) =
et , if t 1.
Sol.: By definition
est f (t) dt .
Lcfw_f (t) =
Since this is a discontinuous function defined different
Solutions to the Attendance Quiz #2
1. Find L1 cfw_ s3s+1
Sol.: We first break-up the numerator:
3s + 1
s + 25
s + 25 s + 25
s + 25 s + 25
We next consult the tables:
= cos kt
Solutions to the Attendance Quiz for Lecture 4
1. Find Lcfw_t sin 2t
Sol.: We use the formula
Lcfw_tn f (t) = (1)n
F (s) .
Here n = 1 and f (t) = sin 2t. From the tables (or from our heads), F (s) =
Taking the first derivative, we have, by the chai
Solutions to Attendance Quiz for Lecture 3
1. Use the Laplace Transform to solve the following initial-value problem.
if 0 t < 2;
y + y = f (t), y(0) = 0, wheref (t) =
1, if 2 t < .
Solution to 1.: We first write f (t) in terms of U(t).
f (t) = 3(U
Solution to the Attendance Quiz for Lecture 5
1. Solve the initial-value problem
y 00 + y = (t 2) ,
y(0) = 0
y 0 (0) = 1 .
Sol.: First, we apply L:
Lcfw_y 00 + y = Lcfw_(t 2)
Using Lcfw_y 00 = s2 Y sy(0)y 0 (0) we have Lcfw_y 00 = s2 Y s01 = s2 Y 1. O
Solutions to the Attendance Quiz for Lecture 6
= x + y ,
= 2x ,
x(0) = 0 , y(0) = 3
Sol.: We first apply the Laplace transform, and a always, write X = Lcfw_x, Y = Lcfw_y.
Lcfw_x0 (t) = Lcfw_x + y
Lcfw_y 0 (t) = Lcfw_2x
Solutions to the Attendance Quiz for Lecture 7
1. Show that the given functions are orthogonal on the given interval.
f1 (x) = sin x
f2 (x) = cos x
[0, ] .
Sol.: We have to show that
(f1 (x), f2 (x) =
sin x cos x dx = 0 .
But by trig. identity sin