Section 9.2: Matrices.. a m1 a m2 a mn

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1 Section 9.2: Matrices Definition: A matrix is a rectangular array of numbers: a 11 a 12 a 1n a 21 a 22 a 2n A = a m1 a m2 a mn In general, a ij denotes the (i, j) entry of A. That is, the entry in the ith row and jth column. The matrix A can be denoted using the shorthand A = (a ij ). Definition: If A = (a ij ) is an m n matrix and c is a scalar, then ca is the m n matrix whose (i, j) entry is ca ij. This is known as scalar multiplication. Definition: If A = (a ij ) and B = (b ij ) are both m n matrices, then the sum A + B is the m n matrix whose (i, j) entry is a ij + b ij for each ordered pair (i, j). Example: Consider the matrices (a) Find A B + 2C. A = [ ] B = [ ] C = [ ] (b) Find a matrix D such that A + C = 2A B + D. 1

2 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 = [ ] and B = [ ] Compute the matrix products AB and BA. Example: Consider the matrices 3 2 A = 2 4 and B = 1 3 Compute the matrix products AB and BA. [ 2 1 ] Note: In general, matrix multiplication is not commutative. That is, AB BA. 2

3 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 = [ ] Definition: The transpose of an m n matrix A is the n m matrix B with entries b ij = a ji. That is, transposition interchanges the rows and columns of a matrix. The transpose of A is denoted by A T or A. Example: Consider the matrices A = [ ] and B = [ ] Verify that (AB) T = B T A T. 3

4 Just as the number 1 acts as an identity for multiplication of real numbers, there is a special matrix I that acts as an identity for matrix multiplication. For any matrix, A, IA = AI = A Definition: The identity matrix is a square (n n) matrix I = That is, I is the matrix with 1 s on the main diagonal and zeros elsewhere. 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 A = and B = Verify that A and B are inverses of each other. 4

5 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 = [ ] (c) C = [ ]

6 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 A =

7 Solving Linear Systems Consider a system of n linear equations in n variables a 11 x 1 + a 12 x a 1n x n = b 1, a 21 x 1 + a 22 x a 2n x n = b 2,. a n1 x 1 + a n2 x a nn x n = b n. This system can be expressed in matrix form as A x = b, where a 11 a 12 a 1n x 1 b 1 a 21 a 22 a 2n A =......, x = x 2., b 2 b =. a n1 a n2 a nn If the coefficient matrix A is nonsingular, then the system has a unique solution given by x = A 1 b x n b n 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

8 A linear system of the form A x = 0 is called homogeneous. Clearly x = 0 is a solution of any homogeneous system. Therefore, it is called the trivial solution. If A is nonsingular, the homogeneous system A x = 0 has only the trivial solution. Indeed x = A 1 0 = 0. However, if A is singular, then A 1 does not exist and the system A x = 0 has infinitely many nontrivial solutions. Example: Determine whether A = [ ] is nonsingular. If so, compute its inverse. In either case, solve A x = 0. Example: Determine whether B = [ ] is nonsingular. If so, compute its inverse. In either case, solve B x = 0. 8

9 The Leslie Matrix Suppose that we would like to model the population size of a species with discrete breeding seasons (e.g. perennial plants, cicadas, salmon). Several models assume that all individuals have the same reproductive fitness. However, for many species, reproductive fitness is age-dependent. Here, we consider an age-structured, discrete-time model introduced by Patrick Leslie. Since only females produce offspring, we will model only the female population. The number of individuals is determined at the end of each breeding season. Suppose that the population is divided into m+1 distinct age classes and let N x (t) denote the number of females of age x = 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 j [0, 1] denote the fraction of females of age j at time t that survive to time t + 1. Then N 1 (t + 1) = P 0 N 0 (t) N 2 (t + 1) = P 1 N 1 (t). N m (t + 1) = P m 1 N m 1 (t) Let F j 0 denote the average number of female offspring per female of age j which survive to the end of the next breeding season. 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 The matrix L is called the Leslie matrix. N(t + 1) = LN(t), F 0 F 1 F 2 F m 1 F m P P L = 0 0 P P m 1 0 9

10 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

11 Example: Assume that a population is divided into four age classes and that 65% of the females of age zero, 40% of the females of age one, and 30% of the females of age two survive until the end of the next breeding season. Moreover, assume that females of age one have an average of 2.8 female offspring, females of age 2 have an average of 7.2 female offspring, and females of age 3 have an average of 3.7 female offspring. If the initial population consists of 1500 females of age zero, 500 females of age one, 250 females of age two, and 50 females of age three, find the Leslie matrix and age distribution at time t = 3. 11

12 Example: Consider the Leslie matrix L = 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 L = 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. 12

13 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, (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)

14 (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) 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. 14

15 (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) lim p(t) = t In the long run, approximately 95.24% of the population is age 0. 15

16 Although the population size is increasing (by a constant factor of 1.6 each generation), the percentage of females in age class 0 is convergent (to 95.24%). This constant percentage 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) = [ ] [ ] The fraction of females in age class 0 remains the same: = [ ] N 0 (t) N 0 (t) + N 1 (t) = = = Similarly, yields the same result. 1.6 [ ] 2000 = 100 [ ] Therefore, if N = [N 0, N 1 ] T denotes a stable age distribution, then where λ = 1.6, and LN = λn, N 0 N 0 + N %. In Section 9.3, we obtain a much easier method to compute the stable age distribution. 16

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