MATH1025 AF

FW00F

Section 1.4

15. (a) If A and B are n by n matrices the entry in row i and column j of the matrix product AB is ai1b1j + ai2b2j + ... + ainbnj. Thus if row i of A is all zero this entry in the product is also zero.

This is true for any value of j which means that row i of AB is zero.

Now if B were the inverse of A, AB would be the unit matrix which has no zero rows. Hence we have a contradiction and A cannot have an inverse if it has a zero row.

(b) Similarly, the entry in row i and column j of BA is bi a1j + bi2a2j + ... + binanj. Thus if column j of A is zero, this entry is zero for any value of i. Thus BA has a zero column and A cannot have an inverse.

17. If A is invertible then A-1 exists. Multiply AB = 0 on the left by this inverse to get A-1AB = A-10 or B = 0.

18. Multiply the partitioned matrices together to get

Now AA-1 = I so that if this product is to be the unit matrix, then BA-1 + AC must be the zero matrix, i.e. AC = -BA-1. Multiply both sides by A-1 we get A-1AC = IC = C = -A-1BA-1.

21. Since (AB)T = BTAT then (BBT)T = (BT)TBT = BBT and hence BBT is symmetric.

Since (A + B)T = AT + BT then (B + BT)T = BT + (BT)T = BT + B = B + BT and hence B + BT is symmetric.

Similarly, (B - BT)T = BT - (BT)T = BT - B = -(B - BT) and hence B - BT is skew-symmetric.

Section 1.5

6. (a) The augmented matrix for the inverse is

R1 -> 1/3 R1; R2 -> R2 - R1; R3 -> R3 - 2 R1

R3-> 2/5 R3

R2 -> -3/4 R2; R3 -> R3 - 7/3 R2

R1-> R1-4/3R2

R1 -> R1 + 1/3 R3; R2 -> R2 + 5/2 R3

Thus the inverse is

(b) Augmented matrix is:

R1 -> -R1; R2 -> R2 -2 R1; R3 -> R3 - 4 R1

R2 -> 1/10 R2; R3 -> R3 + 10 R2

Since there is now a row over zeros in the left side of the third row, there is no solution and the matrix has no inverse.

(c) Augmented matrix is:

R3 -> R3 - R2

R3 -> R3 - R1

R3 -> (-1/2) R3

R1 -> R1 - R3; R2 -> R2 - R3

Thus the inverse is

8. (a) The inverse is

(b) Interchange the first and fourth rows and the second and third rows and then divide by the diagonal elements on the left to get the inverse

(c) The augmented matrix is

If we divide the first row by k and then subtract it from the second row; then divide the second row by k and subtract it from the third row; then divide the third row by k and subtract it from the fourth row; then finally divide the fourth row by k we get the inverse

Section 1.6

12. The augmented matrix is

R1 -> -R1; R2 -> R2 - R1; R3 -> R3 -6R1

R2 -> (1/13)R2; R3 -> R3 -28 R2

R3 -> (13/2)R3; R2 -> R2 + (1/13)R3; R1 -> R1 + R3

R1 -> R1 + 4 R2

Thus the solution to the first system is x1 = -18, x2 = -1, x3 = -14 and to the second system is

x1 = -421/2, x2 = -25/2, x3 = -327/2.

Section 1.7

7. By Theorem 1.7.1 a triangular matrix is not invertible if at least one of its diagonal elements is zero. Thus for A and B both to be not invertible we must have a + b - 1 = 0 and 2a - 3b - 7 = 0 which yields the system

a + b = 1

2a -3b = 7

which has the solution a = 2 and b = -1.

18. If ATA = A then taking the transpose of each side we get (ATA)T = AT(AT)T = ATA = AT. Thus A = AT and A is symmetric. Also from the original equation we get A2 = A.

22. (a)Since (A-1)T = (AT)-1 by Theorem 1.4.10

= (-A)-1 since A is skew-symmetric

= -A-1 by Theorem 1.4.8

then A-1 is skew-symmetric if A is invertible and skew-symmetric.

(b) Let B=AT. Then BT =A. But AT = -A so B = -BT and therefore AT is skew-symmetric if A is.

(A + B)T = AT + BT = -A - B = -(A + B) so A + B is skew-symmetric if A and B are.

(A - B)T = AT - BT = -A - (-B) = -(A - B) so A - B is skew-symmetric if A and B are.

(kA)T = kAT = k(-A) - -(kA) so kA is skew-symmetric if A is.

(c) Since (A + AT)T is symmetric for any square matrix A and A - AT is skew-symmetric for any square matrix A (question 21 of section 1.4) we can write

then any square matrix can be written as the sum of a symmetric and a skew-symmetric matrix.

24. (a) Solving Ly = b where L =and b =

we get y1 = 1, y2 = (-2 + 2y1)/3 = 0 and y3 = -2y1 - 4y2 = -2. Now solve Ux = y where

U =we get x3 = y3/4 = -1/2, x2 = y2 - 2x3 = 1 and x1 = (y1 + x2 - 3x3)/2 = 7/4.

(b) In a similar manner we get y1 = 4/2 = 2, y2 = -5 - 4y1 = -13 and y3 = (2 + 3y1 + 2y2)/3 = -6. Then x3 = y3/2 = -3, x2 = (y2 - x3)/4 = -5/2 and x1 = (y1 + 5x2 - 2x3)/3 = -3/2.

Section 11.6

2. (a) x(1) = Px(0) =, x(2) = Px(1) =, x(3) = Px(2) =.

(b) P is regular since all of its entries are positive. Let the steady-state vector be q. Then Pq = q or (P - I)q = 0. The coefficient matrix is

R1 -> 1/(-0.8)R1; R2 -> R2 - (0.6)R1; R3 -> R3 - (0.2)R1

R2 -> 1/(-0.525)R2; R3 -> R3 - (0.525)R2; R1 -> R1 + (0.125)R2

Thus the solution is q1 = 22t, q2 = 29t and q3 = 21t. But q1 + q2 + q3 = 1 or 72t = 1. Thus

q1 = 22/72, q2 = 29/72 and q3 = 21/72.

8. The transition matrix has elements pij which represents the proportion of people moving from region j to region i each year. Thus P =. We solve for the steady state vector as in question 2. Thus P - I =

R1 -> 1/(-0.1)R1; R2 -> R2 - (0.05)R1; R3 -> R3 - (0.05)R1

R2 -> 1/(-0.175)R2; R3 -> R3 - (0.175)R2; R1 -> R1 + (1.5)R2

Thus q1 = 13t, q2 = 4t and q3 = 7t. But q1 + q2 + q3 = 1 or 24t = 1.

Thus q1 = 13/24, q2 = 4/24 and q3 = 7/24 and after a long time 13/24 of the population lives in region 1, 1/6 in region 2 and 7/24 in region 3.