Math 2660-125
Exam 1
09/08/2011
PRINT Name:
Each problem counts 5 points.
1. Solve the following system of linear equations, applying elementary row operations on
the augmented matrix to get strict upper triangular form rst and then applying back
substitu
function LAB04ex1
t0 = 0; tf = 40; y0 = [-1;0];
[t,Y] = ode45(@f,[t0,tf],y0,[]);
y = Y(:,1); v = Y(:,2); % y in output has 2 columns corresponding to u1 and
u2
figure(1);
plot(t,y,'b-+'); ylabel('y');
plot(t,v,'ro-'); ylabel('v');
legend('y(t)','v(t)=y'(t
Exercise 1
(a)
function LAB04ex1
t0 = 0; tf = 40; y0 = [-1;0];
[t,Y] = ode45(@f,[t0,tf],y0,[]);
[t,Y]
y = Y(:,1); v = Y(:,2); % y in output has 2 columns corresponding to u1
and u2
figure(1);
plot(t,y,'b-+',t,v,'ro-'); % plots t,y
legend('y(t)','v(t)=y'(t
Exercise 1
The blue curve represents y=y(t) and I know this since it starts at 0.1 due to its initial
displacement.
(a)
(b) The period is pi.
= pi.
Omega0 is sqrt(4) and period = 2pi/omega0
(c) No. This is because this graph is showing harmonic motion
and
Exercise 1
theta=[0, pi/4, pi/2, 3*pi/4, pi, 5*pi/4]; %Creating a theta row vector
r=2;
%Since r is a constant, declaring it ahead allows me to use it as r
% instead of 2
x=r*cos(theta); %Calculating the row vector x
y=r*sin(theta); %Calculating the row v
Math 2660-125
Exam 4
10/27/2011
PRINT Name:
Each problem counts 5 points.
1. (1) What is the rank of a matrix (denition)?
1
2
2
3
2
5
4
8
(2) Find the rank of the matrix A =
1 3 2 5
0
2
0
4
.
Solution.
(1) The rank of a matrix is the dimension of its ro
Math 2660-125
Exam 3
10/13/2011
PRINT Name:
Each problem counts 5 points.
1. Determine whether the set
x1
S = x2 : x1 + x2 = x2 R3
3
x3
is a subspace of R3 . Explain your answer.
Recall that a nonempty subset S of a vector space V is a subspace of V if S
Math 2660-125
Exam 2
09/22/2011
PRINT Name:
Each problem counts 5 points.
101
1. Let A = 3 3 7 .
213
(a) Find A1 and verify your answer.
(b) Use A1 to solve Ax = b for b = (1, 2, 3) .
Solution.
(a) Applying elementary row operations on [A|I ] to get reduc
function LAB04ex1
t0 = 0; tf = 40; y0 = [1.5;5];
[t,Y] = ode45(@f,[t0,tf],y0,[]);
[t,Y]
y = Y(:,1); v = Y(:,2); % y in output has 2 columns corresponding to u1 and
u2
figure(1);
plot(t,y,'b-+',t,v,'ro-');
legend('y(t)','v(t)=y'(t)');
ylim([-1.5,1.5])
grid