Download Solutions 9 - University of Bristol

Comp Maths Model Solutions 9 1
1. Can choose your own numbers, but here’s my example of how it works:
>> a = [1 2 3];
>> b = [3 2 1];
>> c = [2 3 1];
>> cross(a,cross(b,c))
ans =
13
-8
1
>> dot(a,c)*b-dot(a,b)*c
ans =
13
-8
1
>> dot(a,cross(b,c))*a
ans =
12
24
36
>> cross(cross(a,b),cross(a,c))
ans =
12
24
36
2. So set matrix and RHS vector and then invert
>> A = [3 -1 4 ; 0 1 -1; 2 6 -1];
>> b = [0 ; 1 ; 1]
>> x = A\b
x =
2.22222
-0.88889
-1.88889
3. (a) Here’s the code... a bit long but there are no obvious short-cuts.
function r = vecmap(x,alpha,beta,gamma)
Rx = zeros(3,3);
Ry = zeros(3,3);
Rz = zeros(3,3);
Rx(1,1) = 1;
Rx(2,2) = cos(alpha);
Rx(2,3) = -sin(alpha);
Rx(3,3) = cos(alpha);
Rx(3,2) = sin(alpha);
Ry(2,2) = 1;
Ry(1,1) = cos(beta);
1
c
University
of Bristol 2016
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1
Ry(1,3) = sin(beta);
Ry(3,3) = cos(beta);
Ry(3,1) =-sin(beta);
Rz(3,3) = 1;
Rz(1,1) = cos(gamma);
Rz(1,2) =-sin(gamma);
Rz(2,2) = cos(gamma);
Rz(2,1) = sin(gamma);
r = Rz*Ry*Rx*x;
end
(b) Set x = [1 ; 1 ; 1]; and call r = vecmap(x,pi,pi,pi) to find r = (1, 1, 1)T .
So mapping leaves position vector unchanged. Easy to see geometrically that rotation
by π around each axis will get you back to the starting position. Alternatively, it’s
easy to see from their definition that Rz (π)Ry (π)Rx (π) = I.
(c) The condition RT = R−1 is equivalent to RRT = I and for each of the three
matrices this is easy to confirm. Indeed, once you can done one of them the others
are basically the same.
If we have r = Rz (γ)Ry (β)Rx (α)x then it follows that
−1
−1
T
T
T
x = R−1
x (α)Ry (β)Rz (γ)r = Rx (α)Ry (β)Rz (γ)r
and this is the inverse mapping.
4. (a) Here’s the code
A = rand(2,2);
d = eig(A)
x = [1 ; 1];
inv(eye(2)-A)*x
y = x;
m = 200;
for j=1:m
y = x + A*y;
end
y
%
%
%
%
%
%
set A to be a random 2x2 matrix
print eigenvalues of A
define x to be the column vector (1,1)^T
print (I-A)^{-1}x
set y to first term the sequence, y_0
set m
% update y using the recurrence relation given
% print the value of the series
(b) Run the script: E.g. of output when eigenvalues (d)less than unity in modulus:
d =
0.923550826838525
-0.811043090074474
ans =
12.9566641630964
2
13.2144570358461
y =
12.9566626842223
13.2144555262378
5. det(A) = 0 if and only if det(A − 0I) = 0 and so there is a λ = 0.
6. (a) True. Do >> det(A) and >> d = eig(A); followed by >> d(1)*d(2)*d(3)*d(4)
(b) True. >> d = eig(A) and >> e = eig(A’) give the same answers.
(c) True. >> d = eig(A) and >> e = eig(inv(A)); followed by >> 1./e to reciprocate the output.
(d) False. >> d = eig(A*B) is not the same as >> e = eig(A).*eig(B).
(e) False. >> d = eig(A+B) is not the same as >> e = eig(A)+eig(B).
7. This:
0.4
0.2
0
-0.2
-0.4
-1
0
1
2
3
4
5
6
Figure 1: Output from the code given in Q7 with n = 120. Observe two distinct regions
of eigenvalues – one occupying a circular domain of fixed radius approximately 0.3 and a
set of real eigenvalues which vary in position with n.
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