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Objectives

Objectives. Use matrices to display mathematical and real-world data. Find sums, differences, and scalar products of matrices.

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Objectives

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  1. Objectives Use matrices to display mathematical and real-world data. Find sums, differences, and scalar products of matrices. The table shows the top scores for girls in barrel racing at the 2004 National High School Rodeo finals. The data can be presented in a table or a spreadsheet as rows and columns of numbers. You can also use a matrix to show table data. A matrix is a rectangular array of numbers enclosed in brackets.

  2. Matrix A has two rows and three columns. A matrix with m rows and n columns has dimensionsm n, read “m by n,” and is called an mn matrix. A has dimensions 2  3. Each value in a matrix is called an entry of the matrix.

  3. The address of an entry is its location in a matrix, expressed by using the lower case matrix letter with row and column number as subscripts. The score 16.206 is located in row 2 column 1, so a21 is 16.206.

  4. 3.95 5.95 3.75 5.60 3.50 5.25 P = The prices for different sandwiches are presented at right. A. Display the data in matrix form. B. What are the dimensions of P? P has three rows and two columns, so it is a 3  2 matrix.

  5. The prices for different sandwiches are presented at right. C. What is entry P32? What does is represent? The entry at P32, in row 3 column 2, is 5.25. It is the price of a 9 in. tuna sandwich. D. What is the address of the entry 5.95? The entry 5.95 is at P12.

  6. Use matrix M to answer the questions below. a. What are the dimensions of M? 3  4 b. What is the entry at m32? 11 c. The entry 0 appears at what two addresses? m14 and m23

  7. You can add or subtract two matrices only if they have the same dimensions.

  8. 3 –2 1 0 3 + 1–2 + 4 1 + (–2)0 + 3 4 2 –1 3 + = = 1 4 –2 3 Add or subtract, if possible. 3 –2 1 0 4 7 2 5 1 –1 1 4 –2 3 2 –2 3 1 0 4 W = , X = , Y = , Z = W + Y Add each corresponding entry. W + Y =

  9. 2 9 –1 4 1 –5 2 –2 3 1 0 4 4 7 2 5 1 –1 – = Add or subtract, if possible. 3 –2 1 0 4 7 2 5 1 –1 1 4 –2 3 2 –2 3 1 0 4 W = , X = , Y = , Z = X – Z Subtract each corresponding entry. X – Z =

  10. Add or subtract, if possible. 3 –2 1 0 4 7 2 5 1 –1 1 4 –2 3 2 –2 3 1 0 4 W = , X = , Y = , Z = X + Y X is a 2  3 matrix, and Y is a 2  2 matrix. Because X and Y do not have the same dimensions, they cannot be added.

  11. Add or subtract if possible. 4 –2 –3 10 2 6 3 2 0 –9 –5 14 4 –1 –5 3 2 8 0 1 –3 3 0 10 A = , C = , D = B = , B + D B – A D – B

  12. You can multiply a matrix by a number, called a scalar. To find the product of a scalar and a matrix, or the scalar product, multiply each entry by the scalar. 3 –2 1 0 2 –1 1 4 –2 3 0 4 4 7 2 5 1 –1 P = R = Q= Evaluate 2P Evaluate -4R

  13. 6.75 13.50 7.20 15.75 8.10 18.00 9.00 20.25 7.5 15 8 17.5 9 20 10 22.5 7.5 15 8 17.5 9 20 10 22.5 7.5 15 8 17.5 9 20 10 22.5 0.75 1.5 0.8 1.75 0.9 2 1 2.25 – Use a scalar product to find the prices if a 10% discount is applied to the prices above. You can multiply by 0.1 and subtract from the original numbers. – 0.1 =

  14. 30 17.5 25 14 40 22.5 150 87.5 125 70 200 112.5 150 87.5 125 70 200 112.5 – 120 70 100 56 160 90 Check It Out! Example 3 Use a scalar product to find the prices if a 20% discount is applied to the ticket service prices. You can multiply by 0.8. WHY??? 0.8 =

  15. 3 –2 1 0 2 –1 1 4 –2 3 0 4 4 7 2 5 1 –1 P = R = Q= Evaluate 3P — Q, if possible. P and Q do not have the same dimensions; they cannot be subtracted after the scalar products are found.

  16. 3 12 –6 9 0 12 1 4 –2 3 0 4 3 –2 1 0 2 –1 3 –2 1 0 2 –1 3(1) 3(4) 3(–2) 3(3) 3(0) 3(4) 3 –2 1 0 2 –1 = = 3 – – – 0 14 –7 9 –2 13 3 –2 1 0 2 –1 1 4 –2 3 0 4 4 7 2 5 1 –1 P = Q= R = Evaluate 3R — P, if possible. =

  17. 4 –2 –3 10 4 –1 –5 3 2 8 3 2 0 –9 D = [6 –3 8] A = B = C = Evaluate 3B + 2C, if possible. B and C do not have the same dimensions; they cannot be added after the scalar products are found.

  18. 4 –2 –3 10 3 2 0 –9 = 2 –3 8 –4 –6 20 –9 –6 0 27 –1 –10 –6 47 2(4) 2(–2) 2(–3) 2(10) –3(3) –3(2) –3(0) –3(–9) = + = = + 4 –2 –3 10 4 –1 –5 3 2 8 3 2 0 –9 D = [6 –3 8] A = B = C = Evaluate 2A – 3C, if possible.

  19. Lesson Quiz 1. What are the dimensions of A? 2. What is entry D12? Evaluate if possible. 3. 2A — C 4.C + 2D 5.10(2B + D) 3  2 –2 Not possible

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