Section 9.2: Matrices Definition: A matrix A consists of a rectangular array of numbers, or elements, arranged in m rows and n columns. That is, a 11 a 12 a 1n a 21 a 22 a 2n A =...... a m1 a m2 a mn A matrix with m rows and n columns is called an m n matrix. The numbers m and n are called the dimensions of the matrix. The element lying in the ith row and jth column is denoted by a ij. The matrix A can be denoted using the shorthand A = (a ij ). The sum and difference of two m n matrices A = (a ij ) and B = (b ij ) are defined componentwise: A + B = (a ij ) + (b ij ) = (a ij + b ij ), A B = (a ij ) (b ij ) = (a ij b ij ). Similarly, if α is a scalar (real or complex), then the product αa is defined componentwise: This is called scalar multiplication. αa = α(a ij ) = (αa ij ). Example: Consider the matrices Find A B + 2C. A = [ ] 1 2 0 3 B = [ ] 0 1 2 4 C = [ ] 1 2 1 1 1
Definition: If A = (a ij ) is an m n matrix and B = (b ij ) is an n r matrix, then the product AB = C = (c ij ) is the m r matrix with entries c ij = n a ik b kj = a i1 b 1j + a i2 b 2j + + a in b nj k=1 This is known as matrix multiplication. Example: Consider the matrices A = [ ] 1 2 3 4 B = [ ] 2 2 1 3 Find the matrix products AB and BA. Example: Consider the matrices 3 2 A = 2 4 B = 1 3 Compute the matrix products AB and BA. [ 2 1 ] 3 4 1 6 Note: In general, matrix multiplication is not commutative. That is, AB BA. 2
Notation: If A is a square (n n) matrix, it can be multiplied by itself a number of times. If k is a positive integer, then A k = AA A }{{} k factors Example: Find an expression for A n if A = [ ] 1 1 1 1 Definition: If A = (a ij ) is an m n matrix, then the transpose of A, denoted by A T or A, is the n m matrix obtained by interchanging the rows and columns of A. That is, A T = (a ji ) Example: Consider the matrices A = [ ] 1 2 3 4 and B = [ ] 2 1 5 3 Verify that (AB) T = B T A T. 3
Definition: The identity matrix is a square (n n) matrix 1 0 0 0 1 0 I =...... 0 0 1 That is, I is the matrix with 1 s on the main diagonal and zeros elsewhere. The identity matrix satisfies AI = IA = A for any square matrix A. Definition: A square matrix A is called nonsingular if there exists a matrix B such that AB = BA = I, where I is the identity matrix. In this case, the matrix B is called the inverse of A and denoted by A 1. Thus, AA 1 = A 1 A = I. Matrices that do not have an inverse are called singular. Example: Consider the matrices 1 2 3 1 2 5 A = 0 1 4 and B = 0 1 4 0 0 1 0 0 1 Verify that A and B are inverses of each other. 4
For any square matrix, there is an associated scalar which tells us whether the matrix is singular or nonsingular. Definition: The determinant of a 2 2 matrix [ ] a11 a A = 12 a 21 a 22 is the scalar defined by det A = a 11 a 12 a 21 a 22 = a 11a 22 a 12 a 21 The determinant of A determines whether it is singular or nonsingular. In particular, A is nonsingular if and only if det A 0. Theorem: (Inverse of a 2 2 Matrix) If A is the 2 2 matrix then its inverse is [ a11 a A = 12 a 21 a 22 A 1 = 1 det A ], [ ] a22 a 12 a 21 a 11 Example: Determine whether each matrix is nonsingular and find its inverse, if it exists. [ ] 3 5 (a) A = 2 4 (b) B = [ ] 2 4 3 6 5
In general, the inverse of a square matrix A can be calculated by augmenting A with the identity matrix and performing Gaussian elimination. Example: Show that A = [ ] 2 5 is nonsingular and find its inverse. 1 3 Example: Find the inverse of the nonsingular matrix 1 1 1 A = 2 1 1 1 1 1 6
Solving Linear Systems Consider a system of n linear equations in n variables a 11 x 1 + a 12 x 2 + + a 1n x n = b 1, a 21 x 1 + a 22 x 2 + + a 2n x n = b 2, a n1 x 1 + a n2 x 2 + + a nn x n = b n. This system can be expressed in matrix form as Ax = b, where A = (a ij ), x = (x i ), and b = (b i ). If b = 0, the system is called homogeneous. Otherwise, it is called nonhomogeneous. If the coefficient matrix A is nonsingular, then the system has a unique solution given by x = A 1 b. In particular, the homogeneous system Ax = 0 has only the trivial solution x = 0. On the other hand, if A is singular, then A 1 does not exist and the system Ax = 0 has infinitely many nonzero solutions in addition to the trivial solution.. Example: Consider the linear system 2x + y = 2 4x + 6y = 8 (a) Write this system in matrix form. (b) Solve this system using an inverse matrix. 7
Example: Determine whether A = [ ] 1 1 2 1 is nonsingular. If so, compute its inverse. In either case, solve Ax = 0. Example: Determine whether B = [ ] 1 3 1 3 is nonsingular. If so, compute its inverse. In either case, solve Bx = 0. 8
The Leslie Matrix Suppose that we would like to model the population size of some species with discrete breeding seasons and age-dependent reproductive fitness (e.g. perennial plants, cicadas, salmon). Here, we consider an age-structured discrete time model proposed by Patrick Leslie in 1965. It is assumed that the population is closed to migration and, since only females produce offspring, only the female population is modeled. Suppose that the population is divided into m + 1 distinct age classes. Let N i (t) denote the number of females of age i = 0, 1, 2,..., m at time t = 0, 1, 2,.... Then the vector N 0 (t) N 1 (t) N(t) = N 2 (t). N m (t) describes the number of females in each age class at time t = 0, 1, 2,.... Let P i [0, 1] denote the fraction of females of age i that survive to age i + 1. Then for i = 0, 1,..., m 1. N i+1 (t + 1) = P i N i (t) Let F i 0 denote the average number of female offspring produced by one female of age i which survive through the time interval in which they were born. Then N 0 (t + 1) = F 0 N 0 (t) + F 1 N 1 (t) + F 2 N 2 (t) + + F m N m (t). These equations can be written in matrix form as where N(t + 1) = LN(t), F 0 F 1 F 2 F m 1 F m P 0 0 0 0 0 0 P 1 0 0 0 L = 0 0 P 2 0 0....... 0 0 0 P m 1 0 The matrix L is called the Leslie matrix. In general, a Leslie matrix has the fertilities on the first row and the survival probabilities on the subdiagonal. All other entries are zero. One can project forward in time by repeated multiplication by the Leslie matrix, N(1) = LN(0), N(2) = LN(1) = L 2 N(0), and in general, N(t) = L t N(0). 9
The relationships among the age groups in the Leslie matrix can be represented by a directed graph, or digraph, with nodes representing each age group and directed edges representing a relation between the two groups. The Leslie matrix model is illustrated by a digraph below. Example: Assume that a population is divided into three age classes and that 80% of the females of age zero and 10% of the females of age one survive until the end of the next breeding season. Moreover, assume that females of age one have an average of 1.6 female offspring and females of age 2 have an average of 3.9 female offspring. If the initial population consists of 1000 females of age zero, 100 females of age one, and 20 females of age two, find the Leslie matrix and age distribution at time t = 3. 10
Example: Consider the Leslie matrix 0 5 0 L = 0.8 0 0. 0 0.3 0 Determine the number of age classes in the population, the fraction of one-year-olds that survive until the end of the next breeding season, and the average number of female offspring of a two-year-old female. Example: Consider the Leslie matrix 2 3 2 1 L = 0.4 0 0 0 0 0.6 0 0. 0 0 0.8 0 Determine the number of age classes in the population, the fraction of two-year-olds that survive until the end of the next breeding season, and the average number of female offspring of a one-year-old female. 11
Stable Age Distribution Example: Suppose that a population is divided into two age classes with Leslie matrix L = and suppose that N 0 (0) = 100 and N 1 (0) = 100. [ ] 1.5 2, 0.08 0 (a) Find the age distributions for t = 0, 1, 2,..., 7. The following table summarizes the age distributions for t = 0, 1, 2,..., 7. Time, t N 0 (t) N 1 (t) 0 100 100 1 350 8 2 541 28 3 868 43 4 1388 69 5 2221 111 6 3553 178 7 5685 284 12
(b) Compute the successive ratios q 0 (t) = N 0(t) N 0 (t 1) and q 1 (t) = N 1(t) N 1 (t 1) for t = 1, 2,..., 7, and determine lim t q 0 (t) and lim t q 1 (t). The following table summarizes the successive ratios for t = 1, 2,..., 7. Therefore, we suspect that Time, t q 0 (t) q 1 (t) 1 3.5 0.08 2 1.55 3.5 3 1.6 1.536 4 1.5991 1.605 5 1.6001 1.609 6 1.5997 1.604 7 1.6001 1.596 lim q 0(t) = lim q 1 (t) = 1.6. t t In the long run, the population size is increasing by a factor of 1.6 each generation. 13
(c) Compute the fraction of females which are age 0, p(t) = for t = 0, 1, 2,..., 7, and determine lim t p(t). N 0 (t) N 0 (t) + N 1 (t) The following table summarizes the fraction of females which are age 0 for t = 0, 1, 2,..., 7. Therefore, we suspect that Time, t p(t) 0 0.5 1 0.9777 2 0.9508 3 0.9528 4 0.9526 5 0.9524 6 0.9523 7 0.9524 lim p(t) = 0.9524. t In the long run, approximately 95.24% of the population is age 0. 14
Although the population size is increasing (by a constant factor of 1.6 each generation), the fraction of females in age class 0 and 1 is convergent (to 95.24% and 14.76%, respectively). This constant distribution of the population is called the stable age distribution. If we start the population in the stable age distribution, the fraction of females in age class 0 will remain the same, about 95.24%, and the population will increase by a constant factor of 1.6 each generation. A stable age distribution for this population is [ ] 2000 N(0) =. 100 It follows that N(1) = LN(0) = [ ] [ ] 1.5 2 2000 0.08 0 100 The fraction of females in age class 0 remains the same: = [ ] 3200. 160 N 0 (t) N 0 (t) + N 1 (t) = 3200 3200 + 160 = 2000 2000 + 100 = 0.9524. Similarly, yields the same result. 1.6 [ ] 2000 = 100 [ ] 3200 160 Therefore, if N = [N 0, N 1 ] T denotes a stable age distribution, then LN = λn, where λ = 1.6, and N 0 N 0 + N 1 95.24%. In Section 9.3, we obtain a much simpler method to compute the stable age distribution. 15