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Basic Matrix Operations Matrix – a rectangular array of numbers, variables, and both Order (dimensions) – describes the number of rows and columns in a.

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Presentation on theme: "Basic Matrix Operations Matrix – a rectangular array of numbers, variables, and both Order (dimensions) – describes the number of rows and columns in a."— Presentation transcript:

1 Basic Matrix Operations Matrix – a rectangular array of numbers, variables, and both Order (dimensions) – describes the number of rows and columns in a matrix 2 456 37810 Row 1 Row 2 Column 1 Column 2 Column 3 Column 4 This is a 2 x 4 matrix. RowsColumns

2 VinsTonisSals VinsTonisSals Pizza$10.10$10.86$10.65 A = Drinks$1.09$0.89$1.05 Salad$3.69$3.89$3.85 Interpret the values for A 12, A 21, and A 32. A 12 says the price of a pizza at Tonis is $10.86 Element – each individual entry notation – A mn : A is the name of the matrix, m is the row number and n is column number. interpretation – what the number represents, describe in words A 21 says the price of drinks at Vins is $1.09 A 32 says the price of salad at Tonis is $3.89

3 Square matrix – a matrix, where the rows and columns are equal Column matrix – a matrix made up of just one column and any number of rows Row matrix – a matrix made up of just one row and any number of columns 23 65 12-52 12 14 62 121145324365

4 Adding and subtracting matrices – (matrices of the same dimensions can be added or subtracted) add or subtract the element of the first matrix by the element of the second matrix in the same position and the answer goes in that same position in the matrix that is representing the answer. 234-5 -6789 += 2 + 43 + -5 -6 + 87 + 9 6-2 216 =

5 Main diagonal – the elements whose row number and column number are the same. Zero matrix – a matrix of any dimensions made up of all zeros 1234 5678 9101112 13141516 000000000000

6 Scalar Multiplication – (think Distributive Property for Matrices) the element of the matrix are multiplied by a value outside of the matrix 823 -1116 5 40 115 -5580 =

7 Matrix Multiplication – In order to multiply two matrices the columns of the first matrix must be equal to the rows of the second matrix. [A] x [B] (m x k) (k x n) When doing word problems, make sure the labels of the rows and columns you are multiplying match as well as the numbers.

8 A group of friends are going out for pizza. Given are the ordering options and the prices for pizza and salad at the two pizza places. Option 1 Option 2 Pizza42 Salad68 Vin’s Toni’s Pizza 11.25 11.95 Salad 3.69 2.85 Find the cost of each option at each of the two pizza places. = [A] = [B]

9 = [A]= [B] First take [A][B]. Why is this wrong? Option 1 Option 2 Pizza42 Salad68 Vin’s Toni’s Pizza11.25 11.95 Salad3.69 2.85 42684268 11.25 11.95 3.69 2.85 ( 4 P Opt1)(11.25 P @V) + (2 P Opt 2)(3.69 S @V) The labels in the second product do not match each other How do we fix this?

10 42684268 11.25 11.95 3.69 2.85 We need to Transpose the first matrix. 11.25 11.95 3.69 2.85 46284628 And now multiply the two matrices together ( 4 P Opt1)(11.25 P @V) + (6 S Opt 1)(3.69 S @V) Now all of the labels match in the product for Row 1 Column 1

11 Multiplication of a Column by a Row A = 4 5 8 B = 2 6 9 Now, lets multiply A x B. 4 5 8269269 4 (2) + 5 (6) + 8(9) = 110 1 x 3 3 x 1=

12 Transpose – ( A T ) of the matrix A is the matrix obtained by interchanging the rows and columns of matrix A. 123456123456 A = A T = 142536142536 The transpose function is in the Matrix catalog, MATH column and the second choice down.

13 If the matrices pass the first test of the columns in the first being equal to the rows in the second use the following steps to multiply. A B 2345678923456789 Take the numbers in the first row of matrix A times the numbers in each of the columns in matrix B. Then do the same for the numbers in the second row of matrix A times the numbers in each of the columns in matrix B. 2(4) + 3(8)2(5) + 3(9)3237 6(4) + 7(8)6(5) + 7(9)8093

14 Inverse – we use the inverse of a matrix when we need to divide it is denoted as A -1 Some of you may remember using a formula to find the inverse in Algebra 2 but in this class we will use the inverse button on the calculator (fourth button down on left hand side of calculator) Identity Matrix – this acts like the number one for matrices, denoted with a capital I. This matrix must be a square matrix and have the number one in the main diagonal and 0’s everywhere else. Example:10or100 01010 001

15 Leslie Growth Model The Leslie Growth Model is used to predict population levels at different age distributions. Needed Information: The length of each cycle (how long an animal is in each age group) The length of each cycle (how long an animal is in each age group) The birthrate in each of the cycles The birthrate in each of the cycles The survival rate of each of the cycles The survival rate of each of the cycles The initial distribution in each of the cycles. The initial distribution in each of the cycles. (birthrate and survival rates will remain constant in the problems and only the female population is tracted)

16 Population Growth Suppose that a certain animal has the following birthrate, survival rate, and population distribution. Age (yrs)BirthrateSurvivalDistribution 0-2 0 0.8 25 2-4 0.7 0.7 32 4-6 1.5 0.7 15 6-8 1.5 0.6 12 8-10 0.4 0 10 a.) Find the population distribution for the next two cycles. b.) Find the growth rate for each cycle.

17 Finding the population distribution after one cycle Step 1: Multiply the distribution in each cycle times the birthrate for the corresponding cycle and add all the products 25(0) + 32(.7) + 15(1.5) + 12(1.5) + 10(.4) = 66.90 (round all numbers to 2 decimal places) This number (66.90) will be the number of newborns (age 0-2) after one cycle.

18 Age (yrs)BirthrateSurvivalDistribution Cycle 1 0-2 0 0.8 25 66.9 0-4 0.7 0.7 32 4-6 1.5 0.7 15 6-8 1.5 0.6 12 8-10 0.4 0 10 Now, we need to find out many of the animals in each distribution survived and move to the next highest age group.

19 Find the number of animals that survive to live another day. Step 2: Multiply the numbers of animals that start in the first age group times the survival rate in that group and place this number in the next highest age group (one row down the chart) 25 (.8) = 20 (these are the animals that are now 2-4 years old) 32 (.7) = 22.4 (these are the animals that are now 4-6)

20 Age (yrs)BirthrateSurvivalDistribution Cycle 1 0-2 0 0.8 25 66.9 0-4 0.7 0.7 32 20 4-6 1.5 0.7 15 22.4 6-8 1.5 0.6 12 8-10 0.4 0 10 Now, fill in the rest of the chart in the same manner. x x

21 15 (.7) = 10.5 12 (.6) = 7.2 10 ( 0 ) = 0 (the last one will always be zero because this is the life expectancy of the animals)

22 Age (yrs) Birthrate Survival Distribution Cycle 1 0-2 0 0.8 25 66.90 0-4 0.7 0.7 32 20 4-6 1.5 0.7 15 22.4 6-8 1.5 0.6 12 10.5 8-10 0.4 0 10 7.2

23 Find the Growth Rate New – Old Old = Growth Rate Distribution Cycle 1 25 66.90 32 20 15 22.4 12 10.5 10 7.2 94 127 127 – 94 94 Cycle 1=.35 or 35%

24 Now, use the Cycle 1 numbers and the given birth and survival rates to find the age distribution for Cycle 2 and the Growth Rate.

25 Age (yrs)BR SR Dist Cycle 1 Cycle 2 0-2 0 0.8 25 66.90 66.23 2-4 0.7 0.7 32 20 53.52 4-6 1.5 0.7 15 22.4 14 6-8 1.5 0.6 12 10.5 15.68 8-10 0.4 0 10 7.2 6.3

26 Population Growth Suppose that a certain animal has the following birthrate, survival rate, and population distribution. Age (yrs)BirthrateSurvivalDistribution 0-2 0 0.7 20 2-4 0.8 0.8 15 4-6 0.9 0.9 23 6-8 1.2 0.5 17 8-10 0.5 0.4 8 10-12 0.2 0.3 5 10-12 0.2 0.3 5 12-14 0 0 5 12-14 0 0 5 a.) Find the distribution for the next three cycles. b.) Find the growth rate for the next three cycles. c.) After how many cycles will the population be over 1,000,000?

27 Leslie Matrices in the calculator The population distribution is a row matrix [A] The birth and survival rates make up the Leslie matrix. (will always be a square matrix) The birthrates are the first column and the survival rates and zeros make up the other columns. The survival rate go in the super diagonal (one above the main diagonal) and zeros are all other elements of the matrix.

28 [A] = [ 20 15 23 17 8 5 5 ] [B] = 0 0.7 0 0 0 0 0 0.8 0 0.8 0 0 0 0 0.8 0 0.8 0 0 0 0 0.9 0 0 0.9 0 0 0 0.9 0 0 0.9 0 0 0 1.2 0 0 0 0.5 0 0 1.2 0 0 0 0.5 0 0 0.5 0 0 0 0 0.4 0 0.5 0 0 0 0 0.4 0 0.2 0 0 0 0 0 0.3 0.2 0 0 0 0 0 0.3 0 0 0 0 0 0 0 0 0 0 0 0 0 0 Now, [A][B] = Cycle 1 [A][B] 2 = Cycle 2

29 Total Population Make another matrix [C], this is a column matrix with all 1’s for elements. This matrix will have as many rows as the distribution matrix had columns. For this example: [C] = 1 111111 [A][B][C] = Total Population for Cycle 1 [A][B] 2 [C] = Total Population Cycle 2


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