Math 449 - Homework  3 - Solutions

1) Exercise 1 (iii), section 3.1, p. 108.

Given the vectors X=(4, -8, 1), Y=(1, -12, -11),

(a) X + Y = (5, -20, -10);

(b) X - Y = (3, 4, 12);

(c) 3*X = (12, -24, 3);

(d) ||X|| = (4^2+(-8)^2+1^2)^(1/2) = 9;

(e) 7*Y - 4*X = (7, -84, -77) - (16, -32, 4) = (-9, -52, -81);

(f) X.Y = 4*1 + (-8)*(-12) + 1*(-11) = 89;

(g) ||7*Y - 4*X|| = ((-9)^2 + (-52)^2 + (-81)^2)^(1/2) = (9346)^(1/2) 
                                                       (approx. = 96.675);

 
************************************************************************** 
 
2)  Exercise 6 of section 3.1, page 109.

(a) and (c) are symmetric, but (b) and (d) are not.

************************************************************************** 
 
3)  Exercise 8 of section 3.2, page 109.

The inverse matrix of A*B is the unique matrix D such that D*(A*B)=I.

So it suffices to show that B^(-1)*A^(-1) has this property. And, in fact,

            (B^(-1)*A^(-1))*(A*B) = B^(-1)*(A^(-1)*(A*B))
                                  = B^(-1)*((A^(-1)*A)*B)
                                  = B^(-1)*(I*B)
                                  = B^(-1)*B
                                  = I

 
************************************************************************** 

4) Exercise 14 of section 3.2, page 109.

The (i,j)-entry of the matrix (A*B)' is equal to the (j,i)-entry of A*B, 

which is  

    (A*B)(j,i) = A(j,1)*B(1,i) + A(j,2)*B(2,i) + ... + A(j,n)*B(n,i), 
    
whereas the (i,j)-entry of B'*A' is

(B'*A')(i,j) = B'(i,1)*A'(1,j) + B'(i,2)*A'(2,j) + ... + B'(i,n)*A'(n,j)

             = B(1,i)*A(j,1) + B(2,i)*A(j,2) + ... + B(n,i)*A(j,n)
             
             = A(j,1)*B(1,i) + A(j,2)*B(2,i) + ... + A(j,n)*B(n,i).
             
Since the two expressions agree for all (i,j), we must have (A*B)'=B'*A'.

************************************************************************** 

5) Do Algorithms and Programs 2 of section 3.2, page 120.

Let Rx(a), Ry(b), Rz(c) be the rotation matrices about the x, y, z-axes

by angles a, b, c, respectively, as shown on page 115. We wish to 

transform the cube under the rotation matrix R=Rz(pi/6)*Rx(pi/12). 

Thus we have

Rz=[cos(pi/6) -sin(pi/6)   0;
    sin(pi/6)  cos(pi/6)   0;
    0          0           1];
   
Rx=[1          0           0;
    0 cos(pi/12) -sin(pi/12);
    0 sin(pi/12)  cos(pi/12)];

R=Rz*Rx;

The vertices of the cube in initial position are

v0=[0 0 0];
v1=[0 0 1];
v2=[0 1 0];
v3=[0 1 1];
v4=[1 0 0];
v5=[1 0 1];
v6=[1 1 0];
v7=[1 1 1];

We can draw the cube by tracing the polygonal curve along the edges of the

cube in the order of the rows of the matrix A below. (Can you find a more 

efficient way? Note that I'm tracing a few edges more than once.)

A = [v0; v1; v3; v2; v6; v4; v5; v7; v6; v4; v0; v2; v3; v7; v5; v1];

You can plot the cube in initial position using the Matlab command:

>>plot3(A(:,1), A(:,2), A(:,3))
>>axis equal
>>grid

To plot the rotated cube by Rx, note that (Rx*A')'=A*Rx' gives the edge-path

of the cube in this new position. So, write 

>>A=A*Rx';

and use the same plot command above.

Similarly, the third position is given by the edge-path

>>A=A*Rz';


************************************************************************** 

6) Exercise 6 of section 3.3, page 124.

We want to solve the system

  5*x_1                       =-10
    x_1 + 3*x_2               =  4
  3*x_1 + 4*x_2 + 2*x_3       =  2
   -x_1 + 3*x_2 - 6*x_3 - x_4 =  5
   
    
 Let A be the lower triangular matrix of coefficients for the above system.
 
 It is immediate that 
 
                det(A)=5*3*2*(-1) = -30, 
                
 so there is a unique solution. The solution is:
 
 x_1 = -10/5                                             =-2
 x_2 = (4 - x_1)/4 = (4 + 2)/3                           = 2
 x_3 = (2 - 3*x_1 - 4*x_2)/2 = (2 + 6 - 8)/2             = 0
 x_4 = (5 + x_1 - 3*x_2 + 6*x_3)/(-1) = -(5 - 2 - 6 + 0) = 3.
   

************************************************************************** 

7) Algorithms and Programs 2 and 3 of section 3.3, page 125.




%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%
function X=forsub(A,B)

%Input  - A is a n x n lower-triangular nonsingular matrix
%       - B is a n x 1 matrix
%Output - X is the solution to the linear system AX = B
%

n=length(B);
X=zeros(n,1);

X(1)=B(1)/A(1,1);

for k=2:n
   X(k)=(B(k)-A(k,1:k-1)*X(1:n))/A(k,k);
end
%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%


We now apply this to the system AX=B, where A is the 20 x 20 matrix whose

(i,j)-entry is i+j if i>=j and 0 otherwise; and B is the 20 x 1 matrix
whose (i,1)-entry is i.

Thus,

A=zeros(20,20);
for i=1:20
   for j=1:i
       A(i,j)=i+j;
   end
end   

for i=1:20
   B(i)=i;
end


Setting X=forsub(A,B) we obtain

X=[
    0.5000;
    0.1250;
    0.0625;
    0.0391;
    0.0273;
    0.0205;
    0.0161;
    0.0131;
    0.0109;
    0.0093;
    0.0080;
    0.0070;
    0.0062;
    0.0055;
    0.0050;
    0.0045;
    0.0041;
    0.0038;
    0.0035;
    0.0032]

**************************************************************************