Practice Set B: Problems 10-13

10.

(a)

A1 = [3 4 5; 2 -3 7; 1 -6 1]; b = [2; -1; 3]; �

format short; x = A1\b �

x =

�� 2.6196

�� -0.2283

�� -0.9891 �

A1*x �

ans =

�� 2.0000

�� -1.0000

�� 3.0000 �

(b)

A2 = [3 -9 8; 2 -3 7; 1 -6 1]; b = [2; -1; 3];

x = A2\b �

Warning: Matrix is close to singular or badly scaled.

�� Results may be inaccurate. RCOND = 4.189521e-018.

x =

�� -6.0000

�� -1.3333

�� 1.0000 �

The matrix A2 is singular.� In fact

det(A2) �

ans =

�� 0 �

(c)

A3 = [1 3 -2 4; -2 3 4 -1; -4 -3 1 2; 2 3 -4 1]; b3 = [1; 1; 1; 1];

x = A3\b3 �

x =

�� -0.5714

�� 0.3333

�� -0.2857

�� 0 �

�A3*x

ans =

�� 1.0000

�� 1.0000 �

(d)

syms a b c d x y u v;

A4 = [a b; c d]; A4\[u; v] �

ans =

[ -(-d*u+b*v)/(d*a-b*c)]

[� (-c*u+v*a)/(d*a-b*c)] �

det(A4) �

ans =

d*a-b*c �

The determinant of the coefficient matrix is the denominator in the answer.� So one gets an answer only if the coefficient matrix is non-singular.

11.

(a)

rank(A1)

ans =

�� 3 �

rank(A2) �

ans =

�� 2 �

rank(A3) �

ans =

�� 4 �

rank(A4) �

ans =

2 �

MATLAB implicitly assumes ad -� bc �� 0 here.

(b)

Only the second one computed is singular.

(c)

det(A1)

inv(A1) �

ans =

�� 92

ans =

�� 0.4239�� -0.3696�� 0.4674

�� 0.0543�� -0.0217�� -0.1196

�� -0.0978�� 0.2391�� -0.1848 �

det(A2) �

ans =

�� 0 �

The matrix A2 does not have an inverse.

det(A3)

inv(A3)

ans =

�� 294

ans =

�� 0.1837�� -0.1531�� -0.2857�� -0.3163

�� 0�� 0.1667�� 0�� 0.1667

�� 0.1633�� 0.0306 ��-0.1429�� -0.3367

�� 0.2857�� -0.0714�� 0�� -0.2143 �

det(A4)�

inv(A4) �

ans =

d*a-b*c

ans =

[� d/(d*a-b*c), -b/(d*a-b*c)]

[ -c/(d*a-b*c),� a/(d*a-b*c)] �

12.

(a)

[U1, R1] = eig(A1) �

U1 =

� Columns 1 through 2

� -0.9749�� 0.6036� ��

� -0.2003�� 0.0624 + 0.5401i

�� 0.0977�� -0.5522 + 0.1877i

� Column 3

�� 0.6036��

�� 0.0624 - 0.5401i

� -0.5522 - 0.1877i

R1 =

� Columns 1 through 2

�� 3.3206�� 0��

�� 0�� -1.1603 + 5.1342i

�� 0�� 0��

� Column 3

�� 0��

� -1.1603 - 5.1342i �

A1*U1 - U1*R1 �

ans =

� 1.0e-014 *

� Columns 1 through 2

�� 0.3109�� 0.2776 - 0.3997i

� -0.0444�� 0 - 0.0777i

� -0.0833�� 0.0999 - 0.1776i

� Column 3

�� 0.2776 + 0.3997i

�� 0 + 0.0777i

�� 0.0999 + 0.1776i �

This is zero.� Notice the "e-014".

[U2, R2] = eig(A2) �

U2 =

� Columns 1 through 2

�� 0.9669�� 0.7405��

�� 0.1240�� 0.4574 - 0.2848i

� -0.2231�� 0.2831 + 0.2848i

� Column 3

�� 0.7405��

�� 0.4574 + 0.2848i

�� 0.2831 - 0.2848i

R2 =

� Columns 1 through 2

� -0.0000�� 0��

�� 0�� 0.5000 + 6.5383i

�� 0�� 0��

� Column 3

�� 0��

�� 0.5000 - 6.5383i �

A2*U2 - U2*R2 �

ans =

� 1.0e-014 *

� Columns 1 through 2

� -0.2224�� -0.2554 - 0.2665i

� -0.1498�� 0.0888��

� -0.1156�� -0.0222 + 0.0666i

� Column 3

� -0.2554 + 0.2665i

�� 0.0888��

� -0.0222 - 0.0666i �

Same comment as in (a).

[U3, R3] = eig(A3) �

U3 =

� Columns 1 through 2

� -0.2446 - 0.4647i� -0.2446 + 0.4647i

�� 0.6254�� 0.6254��

�� 0.0025 + 0.3017i�� 0.0025 - 0.3017i

� -0.1736 - 0.4603i� -0.1736 + 0.4603i

� Columns 3 through 4

� -0.5621 + 0.1062i� -0.5621 - 0.1062i

�� 0.1982 + 0.0654i�� 0.1982 - 0.0654i

� -0.5833�� -0.5833��

� -0.2215 - 0.4898i� -0.2215 + 0.4898i

R3 =

��Columns 1 through 2

�� 4.0755 + 4.1517i�� 0��

�� 0�� 4.0755 - 4.1517i

�� 0�� 0��

� Columns 3 through 4

�� 0�� 0��

� -1.0755 + 2.7440i�� 0��

�� 0�� -1.0755 - 2.7440i �

A3*U3 - U3*R3 �

ans =

� 1.0e-014 *

� Columns 1 through 2

� -0.1554 - 0.2665i� -0.1554 + 0.2665i

�� 0.4441 + 0.0888i�� 0.4441 - 0.0888i

� -0.0888 + 0.2220i� -0.0888 - 0.2220i

�� 0.0888 - 0.4441i�� 0.0888 + 0.4441i

� Columns 3 through 4

� -0.1887 - 0.0444i� -0.1887 + 0.0444i

�� 0.3608 + 0.3331i�� 0.3608 - 0.3331i

�� 0.2442 - 0.0444i�� 0.2442 + 0.0444i

� -0.3775 - 0.0666i� -0.3775 + 0.0666i �

And ditto yet again.

[U4, R4] = eig(A4) �

U4 =

[�� 1,�� 1]

[ -(-1/2*d+1/2*a-1/2*(d^2-2*d*a+a^2+4*b*c)^(1/2))/b, -(-1/2*d+1/2*a+1/2*(d^2-2*d*a+a^2+4*b*c)^(1/2))/b]

R4 =

[ 1/2*d+1/2*a+1/2*(d^2-2*d*a+a^2+4*b*c)^(1/2),�� 0]

[�� 0, 1/2*d+1/2*a-1/2*(d^2-2*d*a+a^2+4*b*c)^(1/2)] �

A4*U4 - U4*R4 �

ans =

[�� 0,�� 0]

[ c-d*(-1/2*d+1/2*a-1/2*(d^2-2*d*a+a^2+4*b*c)^(1/2))/b+(-1/2*d+1/2*a-1/2*(d^2-2*d*a+a^2+4*b*c)^(1/2))/b*(1/2*d+1/2*a+1/2*(d^2-2*d*a+a^2+4*b*c)^(1/2)), c-d*(-1/2*d+1/2*a+1/2*(d^2-2*d*a+a^2+4*b*c)^(1/2))/b+(-1/2*d+1/2*a+1/2*(d^2-2*d*a+a^2+4*b*c)^(1/2))/b*(1/2*d+1/2*a-1/2*(d^2-2*d*a+a^2+4*b*c)^(1/2))] �

simplify(ans) �

ans =

[ 0, 0]

[ 0, 0] �

(b)

A = [1 0 2; -1 0 4; -1 -1 5];

clear U1 U2 R1 R2 �

[U1, R1] = eig(A) �

U1 =

�� -0.8165�� 0.5774�� 0.7071

�� -0.4082�� 0.5774�� -0.7071

�� -0.4082�� 0.5774�� 0.0000

R1 =

�� 2.0000�� 0�� 0

�� 0�� 3.0000�� 0

�� 0�� 0�� 1.0000 �

B = [5 2 -8; 3 6 -10; 3 3 -7]; �

[U2, R2] = eig(B) �

U2 =

�� 0.8165�� -0.5774�� 0.7071

�� 0.4082�� -0.5774�� -0.7071

�� 0.4082�� -0.5774�� 0.0000

R2 =

�� 2.0000�� 0�� 0

�� 0�� -1.0000�� 0

�� 0�� 0�� 3.0000 �

We observe that the columns of U1 are negatives of the corresponding columns of U2.� Finally,

A*B - B*A �

ans =

�� 0�� 0�� 0

�� 0�� 0�� 0 �

13.

(a)

If we set X_n to be the column matrix with entries x_n, y_n, z_n, and M the square matrix with entries 1, 1/4, 0; 0, 1/2, 0; 0, 1/4, 1 then X_n+₁= MX_n.

(b)

We have X_n= MX_n_-1= M²X_n_-2= � = MⁿX₀.

(c)

M = [1, 1/4, 0; 0, 1/2, 0; 0, 1/4, 1];

[U,R] = eig(M)

U =

�� 1.0000�� 0�� -0.4082

�� 0�� 0�� 0.8165

�� 0�� 1.0000�� -0.4082

R =

�� 1.0000�� 0�� 0

�� 0�� 1.0000�� 0

�� 0�� 0�� 0.5000 �

(d)

M should be URU ^-1.� Let's check:

M - U*R*inv(U)

ans =

�� 0�� 0�� 0

�� 0�� 0�� 0 �

It is totally evident that R_� is the diagonal matrix with entries 1, 1, 0.� Therefore M_� = UR_�U ^-1. So

Minf = U*diag([1, 1, 0])*inv(U)

Minf =

�� 1.0000�� 0.5000�� 0

�� 0�� 0�� 0

�� 0�� 0.5000�� 1.0000 �

(e)

syms x0 y0 z0; X0 = [x0; y0; z0]; Minf*X0

ans =

[ x0+1/2*y0]

[�� 0]

[ 1/2*y0+z0] �

Half of the mixed genotype migrates to the dominant genotype and the other half of the mixed genotype migrates to the recessive genotype.� These are added to the two original pure types, whose proportions are preserved.

(f)

M^5*X0�

ans =

[ x0+31/64*y0]

[�� 1/32*y0]

[ 31/64*y0+z0] �

M^10*X0 �

ans =

[ x0+1023/2048*y0]

[�� 1/1024*y0]

[ 1023/2048*y0+z0] �

(g)

If you use the suggested alternate model, then only the first three columns of the table are relevant, the transition matrix M becomes M = [1 1/2 0; 0 1/2 1; 0 0 0], and we leave it to you to compute that the eventual population distribution is [1 0 0], independent of the initial population.