FeatureLesson Algebra 2 Lesson Main Lesson 4-1 Use the table at the right. (For help, go to Skills Handbook page 842.) Organizing Data Into Matrices How.

FeatureLesson Algebra 2 Lesson Main Lesson 4-1 Use the table at the right. (For help, go to Skills Handbook page 842.) Organizing Data Into Matrices How many units were imported to the United States in 1996? How many were imported in 2000? U.S. Passenger Vehicles and Light Trucks Imports And Exports (millions) 1996 1998 2000 Imports4.6785.1856.964 Exports1.2951.3311.402 Source: U.S. Department of Commerce. Check Skills You’ll Need 4-1 4.678 million units 6.964 million units

FeatureLesson Algebra 2 Lesson Main Write the dimensions of each matrix. Lesson 4-1 Organizing Data Into Matrices a. 7 –4 12 9 The matrix has 2 rows and 2 columns and is therefore a 2  2 matrix. b. The matrix has 1 row and 3 columns and is therefore a 1  3 matrix. 0 6 15 Quick Check Additional Examples 4-1

FeatureLesson Algebra 2 Lesson Main Identify each matrix element. K = Lesson 4-1 Organizing Data Into Matrices 3 –1 –8 5 1 8 4 9 8 –4 7 –5 a. k 12 b. k 32 c. k 23 d. k 34 Element k 12 is –1.Element k 32 is –4. a. K = k 12 is the element in the first row and second column. 3 –1 –8 5 1 8 4 9 8 –4 7 –5 b. K = k 32 is the element in the third row and second column. 3 –1 –8 5 1 8 4 9 8 –4 7 –5 Additional Examples 4-1

FeatureLesson Algebra 2 Lesson Main (continued) K = Lesson 4-1 Organizing Data Into Matrices 3 –1 –8 5 1 8 4 9 8 –4 7 –5 a. k 12 b. k 32 c. k 23 d. k 34 Element k 23 is 4.Element k 34 is –5. c. K = k 23 is the element in the second row and third column. 3 –1 –8 5 1 8 4 9 8 –4 7 –5 d. K = k 34 is the element in the third row and the fourth column. 3 –1 –8 5 1 8 4 9 8 –4 7 –5 Quick Check Additional Examples 4-1

FeatureLesson Algebra 2 Lesson Main Three students kept track of the games they won and lost in a chess competition. They showed their results in a chart. Write a 2  3 matrix to show the data. Let each row represent the number of wins and losses and each column represent a student. Lesson 4-1 Organizing Data Into Matrices = Win X = Loss Ed X X Jo X Lew X X X X 5 6 3 2 1 4 Wins Losses Ed Jo Lew Quick Check Additional Examples 4-1

FeatureLesson Algebra 2 Lesson Main Refer to the table. a.Write a matrix N to represent the information. Lesson 4-1 Organizing Data Into Matrices U.S. Passenger Car Imports And Exports (millions) 1996 1998 2000 Imports4.6785.1856.964 Exports1.2951.3311.402 Source: U.S. Department of Commerce. Use 2  3 matrix. 4.678 5.185 6.964 1.295 1.331 1.402 N = Import Exports 1996 1998 2000 Each column represents a different year. Each row represents imports and exports. Additional Examples 4-1

FeatureLesson Algebra 2 Lesson Main (continued) b.Which element represents exports for 2000? Lesson 4-1 Organizing Data Into Matrices U.S. Passenger Car Imports And Exports (millions) 1996 1998 2000 Imports4.6785.1856.964 Exports1.2951.3311.402 Source: U.S. Department of Commerce. Exports are in the second row. The year 2000 is in the third column. Element n 23 represents the number of exports for 2000. Quick Check Additional Examples 4-1

FeatureLesson Algebra 2 Lesson Main 1.Write the dimensions of the matrix. M = Lesson 4-1 Organizing Data Into Matrices 8 4 0 1 9 3 –5 0 –1 2 6 1 2.Identify the elements m 24, m 32, and m 13 of the matrix M in question 1. 3.The table shows the amounts of the deposits and withdrawals for the checking accounts of four bank customers. Show the data in a 2  4 matrix. Label the rows and columns. Deposits Withdrawals A$450$370 B$475$289 C$364$118 D$420$400 3  4 0, 2, 0 450 475 364 420 370 289 118 400 A B C D Deposits Withdrawals Lesson Quiz 4-1

FeatureLesson Algebra 2 Lesson Main Simplify the elements of each matrix. 1.2. 3.4. 5.6. Lesson 4-2 Adding and Subtracting Matrices (For help, go to Skills Handbook page 845.) 10 + 4 0 + 4 –2 + 4 –5 + 4 5 – 2 3 – 2 –1 – 2 0 – 2 –2 + 3 0 – 3 1 – 3 –5 + 3 3 + 1 4 + 9 –2 + 0 5 + 7 8 – 4 –5 – 1 9 – 1 6 – 9 2 + 4 6 – 8 4 – 3 5 + 2 Check Skills You’ll Need 4-2

FeatureLesson Algebra 2 Lesson Main 1. = 2. = 3. = 4. = 5. = 6. = Solutions Lesson 4-2 Adding and Subtracting Matrices 10 + 4 0 + 4 –2 + 4 –5 + 4 14 4 2 –1 5 – 2 3 – 2 –1 – 2 0 – 2 3 1 –3 –2 –2 + 3 0 – 3 1 – 3 –5 + 3 1 –3 –2 3 + 1 4 + 9 –2 + 0 5 + 7 4 13 –2 12 8 – 4 –5 – 1 9 – 1 6 – 9 4 –6 8 –3 2 + 4 6 – 8 4 – 3 5 + 2 6 –2 1 7 Check Skills You’ll Need 4-2

FeatureLesson Algebra 2 Lesson Main The table shows information on ticket sales for a new movie that is showing at two theaters. Sales are for children (C) and adults (A). Lesson 4-2 Adding and Subtracting Matrices Theater C A C A 1 198 350 54 439 2 201 375 58 386 Matinee Evening a. Write two 2  2 matrices to represent matinee and evening sales. Theater 1 198 350 Theater 2 201 375 Matinee C A Theater 1 54 439 Theater 2 58 386 Evening C A Additional Examples 4-2

FeatureLesson Algebra 2 Lesson Main (continued) Lesson 4-2 Adding and Subtracting Matrices b. Find the combined sales for the two showings. 198 350 201 375 + 54 439 58 386 = 198 + 54 350 + 439 201 + 58 375 + 386 = Theater 1 252 789 Theater 2 259 761 C A Quick Check Additional Examples 4-2

FeatureLesson Algebra 2 Lesson Main Find each sum. Lesson 4-2 Adding and Subtracting Matrices a.b. 9 0 –4 6 + 0 3 –8 –5 1 + –3 8 5 –1 = 9 + 0 0 + 0 –4 + 0 6 + 0 = 3 + (–3) –8 + 8 –5 + 5 1 + (–1) = 9 0 –4 6 = 0 Quick Check Additional Examples 4-2

FeatureLesson Algebra 2 Lesson Main A = and B =. Find A – B. Method 1: Use additive inverses. Lesson 4-2 Adding and Subtracting Matrices 4 8 –2 0 7 –9 4 5 A – B = A + (–B) = + 4 8 –2 0 –7 9 –4 –5 Write the additive inverses of the elements of the second matrix. 4 + (–7) 8 + 9 –2 + (–4) 0 + (–5) Add corresponding elements = –3 17 –6 –5 = Simplify. Additional Examples 4-2

FeatureLesson Algebra 2 Lesson Main (continued) Method 2: Use subtraction. Lesson 4-2 Adding and Subtracting Matrices A – B = – 4 8 –2 0 7 –9 4 5 4 – 7 8 – (–9) –2 – 4 0 – 5 Subtract corresponding elements = –3 17 –6 –5 = Simplify. Quick Check Additional Examples 4-2

FeatureLesson Algebra 2 Lesson Main Solve X – = for the matrix X. Lesson 4-2 Adding and Subtracting Matrices 2 5 3 –1 8 0 10 –3 –4 9 6 –9 X – + = + 2 5 3 –1 8 0 10 –3 –4 9 6 –9 2 5 3 –1 8 0 2 5 3 –1 8 0 2 5 3 –1 8 0 Add to each side of the equation. X – = 10 –3 –4 9 6 –9 2 5 3 –1 8 0 12 2 –1 8 14 –9 X = Simplify. Quick Check Additional Examples 4-2

FeatureLesson Algebra 2 Lesson Main Determine whether the matrices in each pair are equal. Lesson 4-2 Adding and Subtracting Matrices M = ; N = 8 + 9 5 –6 –1 0 0.7 17 5 4 – 10 –2 + 1 0 – 7979 Both M and N have three rows and two columns, but – 0.7. M and N are not equal matrices. 7979 =/ a. M = ; N = 8 + 9 5 –6 –1 0 0.7 17 5 4 – 10 –2 + 1 0 – 7979 Additional Examples 4-2

FeatureLesson Algebra 2 Lesson Main (continued) Lesson 4-2 Adding and Subtracting Matrices b. P = ; Q = Both P and Q have two rows and two columns, and their corresponding elements are equal. P and Q are equal matrices. 3 –4 40 –3 – 27 9 16 4 12 4 8 0.2 P = ; Q = 3 –4 40 –3 – 27 9 16 4 12 4 8 0.2 Quick Check Additional Examples 4-2

FeatureLesson Algebra 2 Lesson Main Solve the equation Lesson 4-2 Adding and Subtracting Matrices Since the two matrices are equal, their corresponding elements are equal. 2m – n –3 8 –4m + 2n = for m and n. 15 m + n 8 –30 2m – n = 15–3 = m + n–4m + 2n = –30 2m – n –3 8 –4m + 2n = 15 m + n 8 –30 Additional Examples 4-2

FeatureLesson Algebra 2 Lesson Main (continued) Lesson 4-2 Adding and Subtracting Matrices The solutions are m = 4 and n = –7. Solve for m and n. 2m – n = 15 m + n = –3 3m = 12Add the equations. m = 4Solve for m. 4 + n = –3Substitute 4 for m. n = –7Solve for n. Quick Check Additional Examples 4-2

FeatureLesson Algebra 2 Lesson Main Find each sum or difference. 1. +2. – 3.What is the additive identity for 2  4 matrices? 4.Solve the equation for x and y. 5.Are the following matrices equal? 6.Solve X – = for the matrix X. Lesson 4-2 Adding and Subtracting Matrices 2 8 0 –12 –9 6 6 –5 –3 1 4 8 –5 4 –3 –2 9 5 4 –6 –2x –1 5 x + y 18 –3x + 4y x – 2y –16 = 3 0.5 – ; 0.50 0.4 –0.6 6262 2525 2323 4 3 1 5 2 7 0 6 –7 14 6 –17 0 3 –5 3 –9 10 0 0 x = –9, y = –7 6 10 1 11 no, – –0.6 2323 =/ Lesson Quiz 4-2

FeatureLesson Algebra 2 Lesson Main (For help, go to Lesson 4-2.) Lesson 4-3 Matrix Multiplication Find each sum. 1. + + 3 5 2 8 3 5 2 8 3 5 2 8 2. + + + + –4 7 –4 7 –4 7 –4 7 –4 7 3. + + + –1 3 4 0 –2 –5 –1 3 4 0 –2 –5 –1 3 4 0 –2 –5 –1 3 4 0 –2 –5 Check Skills You’ll Need 4-3

FeatureLesson Algebra 2 Lesson Main Solutions Lesson 4-3 Matrix Multiplication 1. + + = = 3 5 2 8 3 5 2 8 3 5 2 8 3 + 3 + 3 5 + 5 + 5 2 + 2 + 2 8 + 8 + 8 9 15 6 24 2. + + + + = –4 7 –4 7 –4 7 –4 7 –4 7 –4 + (–4) + (–4) + (–4) + (–4) 7 + 7 + 7 + 7 + 7 = –20 35 3. + + + –1 3 4 0 –2 –5 –1 3 4 0 –2 –5 –1 3 4 0 –2 –5 –1 3 4 0 –2 –5 = 4(–1) 4(3) 4(4) 4(0) 4(–2) 4(–5) –4 12 16 0 –8 –20 Check Skills You’ll Need 4-3

FeatureLesson Algebra 2 Lesson Main The table shows the salaries of the three managers (M1, M2, M3) in each of the two branches (A and B) of a retail clothing company. The president of the company has decided to give each manager an 8% raise. Show the new salaries in a matrix. Lesson 4-3 Matrix Multiplication Store M1 M2 M3 A $38,500 $40,000 $44,600 B $39,000 $37,800 $43,700 1.08 38500 40000 44600 39000 37800 43700 = Multiply each element by 1.08. 1.08(38500) 1.08(40000) 1.08(44600) 1.08(39000) 1.08(37800) 1.08(43700) Additional Examples 4-3

FeatureLesson Algebra 2 Lesson Main (continued) Lesson 4-3 Matrix Multiplication The new salaries at branch A are $41,580, $43,200, and $48,168. = A B 41580 43200 48168 42120 40824 47196 M1 M2 M3 The new salaries at branch B are $42,120, $40,824, and $47,196. Quick Check Additional Examples 4-3

FeatureLesson Algebra 2 Lesson Main Find the sum of –3M + 7N for M = and N =. Lesson 4-3 Matrix Multiplication 2 –3 0 6 –5 –1 2 9 –3M + 7N = –3 + 7 2 –3 0 6 –5 –1 2 9 = + –6 9 0 –18 –35 –7 14 63 = –41 2 14 45 Quick Check Additional Examples 4-3

FeatureLesson Algebra 2 Lesson Main Solve the equation –3Y + 2 =. Lesson 4-3 Matrix Multiplication 6 9 –12 15 27 –18 30 6 –3Y + 2 = 6 9 –12 15 27 –18 30 6 –3Y + = Scalar multiplication. 12 18 –24 30 27 –18 30 6 –3Y = – Subtract from each side. 27 –18 30 6 12 18 –24 30 12 18 –24 30 –3Y = Simplify. 15 –36 54 –24 Y = – = 15 –36 54 –24 1313 –5 12 –18 8 Multiply each side by – and simplify. 1313 Additional Examples 4-3

FeatureLesson Algebra 2 Lesson Main (continued) Lesson 4-3 Matrix Multiplication –3Y + 2 = 6 9 –12 15 27 –18 30 6 –3 + 2 Substitute. 6 9 –12 15 27 –18 30 6 –5 12 –18 8 + Multiply. 12 18 –24 30 27 –18 30 6 15 –36 54 –24 = Simplify. 27 –18 30 6 27 –18 30 6 Check: Quick Check Additional Examples 4-3

FeatureLesson Algebra 2 Lesson Main Find the product of and. Multiply a 11 and b 11. Then multiply a 12 and b 21. Add the products. Lesson 4-3 Matrix Multiplication –2 5 3 –1 4 –4 2 6 –2 5 3 –1 4 –4 2 6 = (–2)(4) + (5)(2) = 2 The result is the element in the first row and first column. Repeat with the rest of the rows and columns. –2 5 3 –1 4 –4 2 6 = (–2)(4) + (5)(6) = 38 2 –2 5 3 –1 4 –4 2 6 = (3)(4) + (–1)(2) = 10 2 38 Additional Examples 4-3

FeatureLesson Algebra 2 Lesson Main (continued) Lesson 4-3 Matrix Multiplication 2 38 10 –2 5 3 –1 4 –4 2 6 = (3)(–4) + (–1)(6) = –18The product of and is. –2 5 3 –1 4 –4 2 6 2 38 10 –18 Quick Check Additional Examples 4-3

FeatureLesson Algebra 2 Lesson Main Matrix A gives the prices of shirts and jeans on sale at a discount store. Matrix B gives the number of items sold on one day. Find the income for the day from the sales of the shirts and jeans. A = $18 $22 B = Lesson 4-3 Matrix Multiplication Prices Number of Items Sold Shirts Jeans Shirts 109 Jeans 76 Multiply each price by the number of items sold and add the products. 18 22 = (18)(109) + (22)(76) = 3634 109 76 The store’s income for the day from the sales of shirts and jeans was $3634. Quick Check Additional Examples 4-3

FeatureLesson Algebra 2 Lesson Main Use matrices P = and Q =. Determine whether products PQ and QP are defined or undefined. Lesson 4-3 Matrix Multiplication 3 –1 2 5 9 0 0 1 8 6 5 7 0 2 0 3 1 1 –1 5 2 Find the dimensions of each product matrix. (3  3) (3  4) 3  4 PQ product equal matrix (3  4) (3  3) QP not equal Product PQ is defined and is a 3  4 matrix. Product PQ is undefined, because the number of columns of Q is not equal to the number of rows in P. Quick Check Additional Examples 4-3

FeatureLesson Algebra 2 Lesson Main Use matrices A, B, C, and D. Lesson 4-3 Matrix Multiplication A = B = C = D = 2 3 –1 0 –5 4 –7 1 0 2 6 –6 294294 –3 2 –1 1. Find 8A.2.Find AC.3.Find CD. 4.Is BD defined or undefined? 5.What are the dimensions of (BC)D? 16 24 –8 0 –40 32 27 –29 –6 4 –2 –27 18 –9 –12 8 –4 undefined 2  3 Lesson Quiz 4-3

FeatureLesson Algebra 2 Lesson Main (For help, go to Lesson 2-6.) Lesson 4-4 Geometric Transformations with Matrices 1.y = x + 2; left 4 units2.ƒ(x) = x + 2; up 5 units 3.g(x) = |x|; right 3 units4.y = x; down 2 units 5.y = |x – 3|; down 2 units6.ƒ(x) = –2|x|; right 2 units Without using graphing technology, graph each function and its translation. Write the new function. 1212 1313 Check Skills You’ll Need 4-4

FeatureLesson Algebra 2 Lesson Main 1.y = x + 2; left 4 units: y = x + 6 2.ƒ(x) = x + 2; up 5 units: ƒ(x) = x + 7; Solutions Lesson 4-4 Geometric Transformations with Matrices 3.g(x) = |x| right 3 units: g(x) = |x – 3| 4.y = x; down 2 units: y = x – 2 1212 1212 Check Skills You’ll Need 4-4

FeatureLesson Algebra 2 Lesson Main 5.y = |x – 3|; down 2 units: y = |x – 3| – 2 6.ƒ(x) = –2|x| right 2 units: ƒ(x) = –2|x + 4| Solutions (continued) Lesson 4-4 Geometric Transformations with Matrices 1313 1313 Check Skills You’ll Need 4-4

FeatureLesson Algebra 2 Lesson Main Triangle ABC has vertices A(1, –2), B(3, 1) and C(2, 3). Use a matrix to find the vertices of the image translated 3 units left and 1 unit up. Graph ABC and its image ABC. Lesson 4-4 Geometric Transformations with Matrices The coordinates of the vertices of the image are A (–2, –1), B (0, 2), C (–1, 4). Vertices ofTranslationVertices of the Triangle Matrixthe image 1 3 2 –2 1 3 + = –3 –3 –3 1 1 1 –2 0 –1 –1 2 4 Subtract 3 from each x-coordinate. Add 1 to each y-coordinate. A B C Quick Check Additional Examples 4-4

FeatureLesson Algebra 2 Lesson Main The figure in the diagram is to be reduced by a factor of. Find the coordinates of the vertices of the reduced figure. Lesson 4-4 Geometric Transformations with Matrices 2323 Quick Check Write a matrix to represent the coordinates of the vertices. A B C D E A B C D E 2323 0 2 3 –1 –2 3 2 –2 –3 0 = 0 2 – – 2 – –2 0 4343 4343 4343 4343 2323 Multiply. Additional Examples 4-4 The new coordinates are A (0, 2), B (, ), C (2, – ), D (–, –2), and E (–, 0). 4343 4343 4343 2323 4343

FeatureLesson Algebra 2 Lesson Main Lesson 4-4 Geometric Transformations with Matrices Additional Examples 4-4 1 0 0 –1 2 3 4 –1 0 –2 = 2 3 4 1 0 2 Reflect the triangle with coordinates A(2, –1), B(3, 0), and C(4, –2) in each line. Graph triangle ABC and each image on the same coordinate plane. a. x-axis b. y-axis c. y = x –1 0 0 1 2 3 4 –1 0 –2 = –2 –3 –4 –1 0 –2 0 1 1 0 2 3 4 –1 0 –2 = 2 3 4 0 –1 –1 0 2 3 4 –1 0 –2 = 1 0 2 –2 –3 –4 d. y = –x

FeatureLesson Algebra 2 Lesson Main (continued) Lesson 4-4 Geometric Transformations with Matrices a. x-axis 1 0 0 –1 2 3 4 –1 0 –2 = 2 3 4 1 0 2 b. y-axis –1 0 0 1 2 3 4 –1 0 –2 = –2 –3 –4 –1 0 –2 c. y = x 1 0 0 1 2 3 4 –1 0 –2 = 2 3 4 0 –1 –1 0 2 3 4 –1 3 –2 = 1 0 2 –2 –3 –4 d. y = –x Quick Check Additional Examples 4-4

FeatureLesson Algebra 2 Lesson Main Lesson 4-4 Geometric Transformations with Matrices Additional Examples 4-4 Rotate the triangle from Additional Example 3 as indicated. Graph the triangle ABC and each image on the same coordinate plane. a. 90  c. 270  b. 180  d. 360  0 –1 1 0 2 3 4 –1 0 –2 = 1 0 2 2 3 4 –1 0 0 –1 2 3 4 –1 0 –2 = –2 –3 –4 1 0 2 0 1 –1 0 2 3 4 –1 0 –2 = –2 –3 –4 1 0 0 1 2 3 4 –1 0 –2 = 2 3 4 –1 0 –2 Quick Check

FeatureLesson Algebra 2 Lesson Main 1.Write a matrix equation that represents a translation of triangle ABC 7 units left and 3 units up. 2.Write a matrix equation that represents a dilation of triangle ABC with a scale factor of 5. 3.Use matrix multiplication to reflect triangle ABC in the line y = –x. Then draw the preimage and image on the same coordinate plane. For these questions, use triangle ABC with vertices A(–1, 1), B(2, 2), and C(1, –2). Lesson 4-4 Geometric Transformations with Matrices –1 2 1 1 2 –2 + = –7 –7 –7 3 3 3 –8 –5 –6 4 5 1 –1 2 1 1 2 –2 5 = –5 10 5 5 10 –10 0 –1 –1 0 + = –1 2 1 1 2 –2 –1 –2 2 1 –2 –1 Lesson Quiz 4-4

FeatureLesson Algebra 2 Lesson Main 1a.3(4)b.2(6)c.3(4) – 2(6) 2a.3(–4)b.2(–6)c.3(–4) – 2(–6) 3a.–3(–4)b.2(–6)c.–3(–4) – 2(–6) 4a.–3(4)b.–2(–6)c.–3(4) – (–2)(–6) Lesson 4-5 (For help, go to Skills Handbook page 845.) 2 X 2 Matrices, Determinants, and Inverses Simplify each group of expressions. Check Skills You’ll Need 4-5

FeatureLesson Algebra 2 Lesson Main 1a.3(4) = 12 1b.2(6) = 12 1c.3(4) – 2(6) = 12 – 12 = 0 2a.3(–4) = –12 2b.2(–6) = –12 2c.3(–4) – 2(–6) = –12 – (–12) = –12 + 12 = 0 3a.–3(–4) = 12 3b.2(–6) = –12 3c.–3(–4) – 2(–6) = 12 – (–12) = 12 + 12 = 24 4a.–3(4) = –12 4b.–2(–6) = 12 4c.–3(4) – (–2)(–6) = –12 – 12 = –24 Solutions Lesson 4-5 2 X 2 Matrices, Determinants, and Inverses Check Skills You’ll Need 4-5

FeatureLesson Algebra 2 Lesson Main Show that matrices A and B are multiplicative inverses. A = B = Lesson 4-5 2 X 2 Matrices, Determinants, and Inverses 3 –1 7 1 0.1 0.1 –0.7 0.3 AB = 3 –1 7 1 0.1 0.1 –0.7 0.3 = (3)(0.1) + (–1)(–0.7) (3)(0.1) + (–1)(0.3) (7)(0.1) + (1)(–0.7) (7)(0.1) + (1)(0.3) = 1 0 0 1 AB = I, so B is the multiplicative inverse of A. Quick Check Additional Examples 4-5

FeatureLesson Algebra 2 Lesson Main Evaluate each determinant. a. det b. det c. det Lesson 4-5 2 X 2 Matrices, Determinants, and Inverses 7 8 –5 –9 4 –3 5 6 a –b b a = = (7)(–9) – (8)(–5) = –23 7 8 –5 –9 = = (4)(6) – (–3)(5) = 39 4 –3 5 6 = = (a)(a) – (–b)(b) = a 2 + b 2 a –b b a Quick Check Additional Examples 4-5

FeatureLesson Algebra 2 Lesson Main Determine whether each matrix has an inverse. If it does, find it. Lesson 4-5 2 X 2 Matrices, Determinants, and Inverses Find det X. ad – bc = (12)(3) – (4)(9) Simplify. = 0 12 4 9 3 Since det X = 0, the inverse of X does not exist. Find det Y. ad – bc = (6)(20) – (5)(25) Simplify. = –5 6 5 25 20 Since the determinant 0, the inverse of Y exists.=/ a. X = b. Y = Additional Examples 4-5

FeatureLesson Algebra 2 Lesson Main (continued) Lesson 4-5 2 X 2 Matrices, Determinants, and Inverses = – 20 –5 Substitute –5 for the –25 6 determinant. 1515 = Multiply. –4 1 5 –1.2 Y –1 = 20 –5 Change signs. –25 6 Switch positions. 1 det Y Quick Check 20 –5 Use the determinant to –25 6 write the inverse. = 1 det Y Additional Examples 4-5

FeatureLesson Algebra 2 Lesson Main Solve X = for the matrix X. The matrix equation has the form AX = B. First find A –1. Lesson 4-5 2 X 2 Matrices, Determinants, and Inverses 9 25 4 11 3 –7 A –1 = 1 ad – bc d –b –c a Use the definition of inverse. = 1 (9)(11) – (25)(4) 11 –25 –4 9 Substitute. = –11 25 4 –9 Simplify. Use the equation X = A –1 B. X = –11 25 4 –9 Substitute. 3 –7 Additional Examples 4-5

FeatureLesson Algebra 2 Lesson Main (continued) Lesson 4-5 2 X 2 Matrices, Determinants, and Inverses = = (–11)(3) + (25)(–7) (4)(3) + (–9)(–7) Multiply and simplify. –208 75 Check: X = 9 25 4 11 Use the original equation. 3 –7 9 25 4 11 Substitute. 3 –7 –208 75 Multiply and simplify. 3 –7 9(–208) + 25(75) 4(–208) + 11(75) = 3 –7 3 –7 Quick Check Additional Examples 4-5

FeatureLesson Algebra 2 Lesson Main In a city with a stable group of 45,000 households, 25,000 households use long distance carrier A, and 20,000 use long distance carrier B. Records show that over a 1-year period, 84% of the households remain with carrier A, while 16% switch to B. 93% of the households using B stay with B, while 7% switch to A. a.Write a matrix to represent the changes in long distance carriers. Lesson 4-5 2 X 2 Matrices, Determinants, and Inverses 0.84 0.07 0.16 0.93 To A To B From A B Write the percents as decimals. Additional Examples 4-5

FeatureLesson Algebra 2 Lesson Main (continued) b.Predict the number of households that will be using distance carrier B next year. Lesson 4-5 2 X 2 Matrices, Determinants, and Inverses 25000 20000 Use A Use B Write the information in a matrix. 25000 20000 0.84 0.07 0.16 0.93 22,400 22,600 = 22,600 households will use carrier B. Additional Examples 4-5

FeatureLesson Algebra 2 Lesson Main (continued) Lesson 4-5 2 X 2 Matrices, Determinants, and Inverses First find the determinant of. 0.84 0.07 0.16 0.93 0.84 0.07 0.16 0.93 = 0.77 About 28,400 households used carrier A. Multiply the inverse matrix by the information matrix in part (b). Use a calculator and the exact inverse. 28,377 16,623 25,000 20,000 0.93–0.07 –.016 0.84 1 0.77 c.Use the inverse of the matrix from part (a) to find, to the nearest hundred households, the number of households that used carrier A last year. Quick Check Additional Examples 4-5

FeatureLesson Algebra 2 Lesson Main 1.Is the inverse of ? How do you know? 2.Find the determinant of. 3.Find the inverse of. 4.Solve the equation X = for X. Lesson 4-5 2 X 2 Matrices, Determinants, and Inverses 5 2 –2 1 1 2 2 5 –12 5 –16 4 –2 4 3 –7 20 35 1 2 –3 7 no; Answers may vary. Sample is, which is not the 2  2 identity matrix. 1 2 2 5 5 2 –2 –1 1 0 0 –1 32 –3.5 –2 –1.5 –1 –50.2 28.6 Lesson Quiz 4-5

FeatureLesson Algebra 2 Lesson Main Find the product of the circled elements in each matrix. 1.2.3. 4.5.6. Lesson 4-6 3 X 3 Matrices, Determinants, and Inverses (For help, go to Skills Handbook page 845.) 2 3 0 –1 3 –2 4 –3 –4 2 3 0 –1 3 –2 4 –3 –4 2 3 0 –1 3 –2 4 –3 –4 2 3 0 –1 3 –2 4 –3 –4 2 3 0 –1 3 –2 4 –3 –4 2 3 0 –1 3 –2 4 –3 –4 Check Skills You’ll Need 4-6

FeatureLesson Algebra 2 Lesson Main 1.(2)(3)(–4) = 6(–4) = –24 2.(0)(–1)(–3) = 0(–3) = 0 3.(3)(–2)(4) = (–6)(4) = –24 4.(2)(–2)(–3) = (–4)(–3) = 12 5.(3)(–1)(–4) = (–3)(–4) = 12 6.(0)(3)(4) = (0)(4) = 0 Solutions Lesson 4-6 3 X 3 Matrices, Determinants, and Inverses Check Skills You’ll Need 4-6

FeatureLesson Algebra 2 Lesson Main Evaluate the determinant of X =. Lesson 4-6 3 X 3 Matrices, Determinants, and Inverses 8 –4 3 –2 9 5 1 6 0 = [(8)(9)(0) + (–2)(6)(3) + (1)(–4)(5)]Use the – [(8)(6)(5) + (–2)(–4)(0) + (1)(9)(3)]definition. 8 –4 3 –2 9 5 1 6 0 = [0 + (–36) + (–20)] – [240 + 0 + 27]Multiply. = –56 – 267 = –323.Simplify. The determinant of X is –323. Quick Check Additional Examples 4-6

FeatureLesson Algebra 2 Lesson Main Enter matrix T into your graphing calculator. Use the matrix submenus to evaluate the determinant of the matrix. The determinant of the matrix is –65. Lesson 4-6 3 X 3 Matrices, Determinants, and Inverses T = 4 2 3 –2 –1 5 1 3 6 Quick Check Additional Examples 4-6

FeatureLesson Algebra 2 Lesson Main Determine whether the matrices are multiplicative inverses. a. C =, D = Lesson 4-6 3 X 3 Matrices, Determinants, and Inverses 0.5 0 0 0 0 0.5 0 1 1 2 0 0 0 2 1 0 2 0 0.5 0 0 0 0 0.5 0 1 1 2 0 0 0 2 1 0 2 0 1 0 0 0 1 0 0 4 1 = Since CD I, C and D are not multiplicative inverses. =/ Additional Examples 4-6

FeatureLesson Algebra 2 Lesson Main (continued) Lesson 4-6 3 X 3 Matrices, Determinants, and Inverses b. A =, B = 0 0 1 0 1 0 1 0 –1 1 0 1 0 1 0 1 0 0 0 1 0 0 0 1 = 0 1 0 1 0 –1 1 0 1 0 1 0 1 0 0 Since AB = I, A and B are multiplicative inverses. Quick Check Additional Examples 4-6

FeatureLesson Algebra 2 Lesson Main Solve the equation. Lesson 4-6 3 X 3 Matrices, Determinants, and Inverses 2 0 1 0 1 4 1 0 0 –1 8 –2 X = Let A =. 2 0 1 0 1 4 1 0 0 Find A –1. X = 0 0 1 –4 1 8 1 0 –2 –1 8 –2 Use the equation X = A –1 C. Multiply. X = –2 –4 3 Quick Check Additional Examples 4-6

FeatureLesson Algebra 2 Lesson Main Use the alphabet table and the encoding matrix. matrix K =. Lesson 4-6 3 X 3 Matrices, Determinants, and Inverses 0.5 0.25 0.25 0.25 –0.5 0.5 0.5 1 –1 a. Find the decoding matrix K –1. K –1 =Use a graphing calculator. 0 2 1 2 –2.5 –0.75 2 –1.5 –1.25 Additional Examples 4-6

FeatureLesson Algebra 2 Lesson Main (continued) Lesson 4-6 3 X 3 Matrices, Determinants, and Inverses b. Decode.Zero indicates a space holder. 11.25 16.75 24.5 5.75 17 5.5 1.5 –12 15 = 0 2 1 2 –2.5 –0.75 2 –1.5 –1.25 11.25 16.75 24.5 5.75 17 5.5 1.5 –12 15 13 22 26 7 024 12 23 22 Use the decoding matrix from part (a). Multiply. The numbers 13 22 26 7 0 24 12 23 22 correspond to the letters NEAT CODE. Quick Check Additional Examples 4-6

FeatureLesson Algebra 2 Lesson Main 1.Use pencil and paper to evaluate the determinant of 2.Determine whether the matrices are multiplicative inverses. 3.Solve the equation for M. Lesson 4-6 3 X 3 Matrices, Determinants, and Inverses –2 –4 2 3 1 0 5 –6 –2. 1 1 –1 –1 0 1 0 –1 1 ; 1 0 1 1 1 0 1 1 1 –1 –1 1 1 2 –1 0 –1 1 M = –1 –4 3 –66 yes 4 –5 –2 Lesson Quiz 4-6

FeatureLesson Algebra 2 Lesson Main Lesson 4-7 Inverse Matrices and Systems (For help, go to Lesson 3-6.) 1. 2. 3. 5x + y = 14 4x + 3y = 20 x – y – z = –9 3x + y + 2z = 12 x = y – 2z –x + 2y + z = 0 y = –2x + 3 z = 3x Solve each system. Check Skills You’ll Need 4-7

FeatureLesson Algebra 2 Lesson Main 1.2. Solutions Lesson 4-7 Inverse Matrices and Systems 5x + y = 14 4x + 3y = 20 Solve the first equation for y: y = –5x + 14 Substitute this into the second equation: 4x + 3(–5x + 14)= 20 4x – 15x + 42= 20 –11x + 42= 20 –11x= –22 x= 2 Use the first equation with x = 2: 5(2) + y= 14 10 + y= 14 y= 4 The solution is (2, 4). x – y – z = –9 3x + y + 2z = 12 x = y – 2z Use the first equation with x = y – 2z (third equation): (y – 2z) – y – z = –9 –3z = –9 z = 3 Use the second equation with x = y – 2z (third equation) and z = 3: 3(y – 2z) + y + 2z = 12 3(y – 2(3) + y + 2(3) = 12 3y – 18 + y + 6 = 12 4y – 12 = 12 4y = 24 y = 6 Use the third equation with y = 6 And z = 3: x = 6 – 2(3) = 6 – 6 = 0 The solution is (0, 6, 3). Check Skills You’ll Need 4-7

FeatureLesson Algebra 2 Lesson Main Solutions (continued) Lesson 4-7 Inverse Matrices and Systems Use the first equation with y = –2x + 3 (second equation) and z = 3x (third equation): –x + 2(–2x + 3) + 3x= 0 –x – 4x + 6 + 3x= 0 2x + 6= 0 –2x= –6 x= 3 –x + 2y + z = 0 y = –2x + 3 z = 3x Use the second equation with x = 3: y = –2(3) + 3 = –6 + 3 = –3 Use the third equation with x = 3: z = 3(3) = 9 The solution is (3, –3, 9). 3. Check Skills You’ll Need 4-7

FeatureLesson Algebra 2 Lesson Main Write the system as a matrix equation. Lesson 4-7 Inverse Matrices and Systems –3x – 4y + 5z = 11 –2x + 7y = –6 –5x + y – z = 20 Then identify the coefficient matrix, the variable matrix, and the constant matrix. Matrix equation: = –3 –4 5 –2 7 0 –5 1 –1 xyzxyz 11 –6 20 Coefficient matrix –3 –4 5 –2 7 0 –5 1 –1 xyzxyz Variable matrix 11 –6 20 Constant matrix Quick Check Additional Examples 4-7

FeatureLesson Algebra 2 Lesson Main Solve the system. Lesson 4-7 Inverse Matrices and Systems 2x + 3y = –1 x – y = 12 2 3 1 –1 xyxy –1 12 = Write the system as a matrix equation. A –1 = Find A –1. 1515 3535 1515 2525 – = A –1 B = = Solve for the variable matrix. xyxy 1515 3535 1515 2525 – –1 12 7 –5 Additional Examples 4-7

FeatureLesson Algebra 2 Lesson Main (continued) Lesson 4-7 Inverse Matrices and Systems The solution of the system is (7, –5). Check: 2x + 3y = –1 x – y = 12Use the original equations. 2(7) + 3(–5) –1 (7) – (–5) 12Substitute. 14 – 15 = –1 7 + 5 = 12Simplify. Quick Check Additional Examples 4-7

FeatureLesson Algebra 2 Lesson Main Solve the system. Step 1:Write the system as a matrix equation. Lesson 4-7 Inverse Matrices and Systems 7x + 3y + 2z = 13 –2x + y – 8z = 26 x – 4y +10z = –13 Step 2:Store the coefficient matrix as matrix A and the constant matrix as matrix B. 7 3 2 –2 1 –8 1 –4 10 13 26 –13 xyzxyz = The solution is (9, –12, –7). Quick Check Additional Examples 4-7

FeatureLesson Algebra 2 Lesson Main A linen shop has several tables of sheets and towels on special sale. The sheets are all priced the same, and so are the towels. Mario bought 3 sheets and 5 towels at a cost of $137.50. Marco bought 4 sheets and 2 towels at a cost of $118.00. Find the price of each item. Lesson 4-7 Inverse Matrices and Systems Define: Let x = the price of one sheet. Let y = the price of one towel. Write: = 3 5 4 2 xyxy 137.50 118.00 Use a graphing calculator. Store the coefficient matrix as matrix A and the constant matrix as matrix B. Relate: 3 sheets and 5 towels cost $137.50. 4 sheets and 2 towels cost $118.00. The price of a sheet is $22.50. The price of a towel is $14.00. Quick Check Additional Examples 4-7

FeatureLesson Algebra 2 Lesson Main Write the coefficient matrix for each system. Use it to determine whether the system has exactly a unique solution. Lesson 4-7 Inverse Matrices and Systems a. 4x – 2y = 7 –6x + 3y = 5 A = ; det A = = 4(3) – (–2)(–6) = 0 4 –2 –6 3 4 –2 –6 3 Since det A = 0, the matrix does not have an inverse and the system does not have a unique solution. Additional Examples 4-7

FeatureLesson Algebra 2 Lesson Main (continued) Lesson 4-7 Inverse Matrices and Systems b. 12x + 8y = –3 3x – 7y = 50 A = ; det A = = 12(–7) – 8(–3) = –60 12 8 3 –7 12 8 3 –7 Since det A 0, the matrix has an inverse and the system has a unique solution. =/ Quick Check Additional Examples 4-7

FeatureLesson Algebra 2 Lesson Main 1. 2. Determine whether each system has a unique solution. 3. 4. Lesson 4-7 Inverse Matrices and Systems yes 3x + 2y = –6 2x – 3y = 61 2x + 4y + 5z = –3 7x + 9y + 4z = 19 –3x + 2y + 8z = 0 7x – 2y = 15 –28x + 8y = 7 20x + 5y = 33 –32x + 8y = 47 = ; (8, –15) 3 2 2 –3 –6 61 xyxy 2 4 5 7 9 4 –3 2 8 –3 19 0 xyzxyz = ; (–10, 13, –7) no Write each system as a matrix equation. Then solve the system. Lesson Quiz 4-7

FeatureLesson Algebra 2 Lesson Main 1.2.3. Lesson 4-8 Augmented Matrices and Systems (For help, go to Lessons 4-5 and 4-6.) 4.5.6. –1 2 0 3 0 1 –1 3 2 1 –1 5 0 1 –3 4 5 –1 –1 0 1 3 4 5 –1 2 0 0 –1 1 0 2 –1 3 4 0 –2 –1 5 Evaluate the determinant of each matrix. Check Skills You’ll Need 4-8

FeatureLesson Algebra 2 Lesson Main 1.det = (–1)(3) – (2)(0) = –3 – 0 = –3 2.det = (0)(3) – (1)(–1) = 0 – (–1) = 1 3.det = (2)(5) – (1)(–1) = 10 – (–1) = 11 4.det = [(0)(5)(–1) + (4)(0)(–3) + (–1)(1)(–1)] – [(0)(0)(–1) + (4)(1)(1) + (–1)(5)(–3)] = [0 + 0 + 1] – [0 + 4 + 15] = 1 – 19 = –18 Solutions Lesson 4-8 Augmented Matrices and Systems –1 2 0 3 0 1 –1 3 2 1 –1 5 0 1 –3 4 5 –1 –1 0 1 Check Skills You’ll Need 4-8

FeatureLesson Algebra 2 Lesson Main 5.det = [(3)(2)(1) + (–1)(–1)(5) + (0)(4)(0)] – [(3)(–1)(0) + (–1)(4)(1) + (0)(2)(5)] = [6 + 5 + 0] – [0 + (–4) + 0] = 11 – (–4) = 15 6.det =[(0)(4)(5) + (3)(–1)(–1) + (–2)(2)(0)] – [(0)(–1)(0) + (3)(2)(5) + (–2)(4)(–1)] = [0 + 3 + 0] – [0 + 30 + 8] = 3 – 38 = –35 Solutions (continued) Lesson 4-8 Augmented Matrices and Systems 3 4 5 –1 2 0 0 –1 1 0 2 –1 3 4 0 –2 –1 5 Check Skills You’ll Need 4-8

FeatureLesson Algebra 2 Lesson Main Use Cramer’s rule to solve the system. Lesson 4-8 Augmented Matrices and Systems Evaluate three determinants. Then find x and y. 7x – 4y = 15 3x + 6y = 8 D = = 54 7 –4 3 6 D x = = 122 15 –4 8 6 D y = = 11 7 15 3 8 x = = DxDDxD 61 27 11 54 DyDDyD y = = The solution of the system is,. 61 27 11 54 Quick Check Additional Examples 4-8

FeatureLesson Algebra 2 Lesson Main Find the y-coordinate of the solution of the system. Lesson 4-8 Augmented Matrices and Systems –2x + 8y + 2z = –3 –6x + 2z = 1 –7x – 5y + z = 2 D = = –24Evaluate the determinant. –2 8 2 –6 0 2 –7 –5 1 D y = = 20Replace the y-coefficients with the constants and evaluate again. –2 –3 2 –6 1 2 –7 2 1 y = = – = –Find y. 20 24 DyDDyD 5656 The y-coordinate of the solution is –. 5656 Quick Check Additional Examples 4-8

FeatureLesson Algebra 2 Lesson Main Write an augmented matrix to represent the system Lesson 4-8 Augmented Matrices and Systems –7x + 4y = –3 x + 8y = 9 System of equations –7x + 4y = –3 x + 8y = 9 x-coefficientsy-coefficientsconstants Augmented matrix –7 4 –3 1 8 9 Draw a vertical bar to separate the coefficients from constants. Quick Check Additional Examples 4-8

FeatureLesson Algebra 2 Lesson Main Write a system of equations for the augmented matrix. Lesson 4-8 Augmented Matrices and Systems 9 –7 –1 2 5 –6 Augmented matrix 9 –7 –1 2 5 –6 x-coefficientsy-coefficientsconstants System of equations 9x – 7y = –1 2x + 5y = –6 Quick Check Additional Examples 4-8

FeatureLesson Algebra 2 Lesson Main Use an augmented matrix to solve the system Lesson 4-8 Augmented Matrices and Systems x – 3y = –17 4x + 2y = 2 1 –3 –17 4 2 2 Write an augmented matrix. Multiply Row 1 by –4 and add it to Row 2. Write the new augmented matrix. 1 –3 –17 0 14 70 –4(1 –3 –17) 4 2 2 0 14 70 1 14 1 –3 –17 0 1 5 Multiply Row 2 by. Write the new augmented matrix. (0 14 70) 0 1 5 1 14 Additional Examples 4-8

FeatureLesson Algebra 2 Lesson Main (continued) Lesson 4-8 Augmented Matrices and Systems 1 14 1 –3 –17 0 1 5 (0 14 70) 0 1 5 1 0 –2 0 1 5 1 –3 –17 3(0 1 5) 1 0 –2 Multiply Row 2 by 3 and add it to Row 1. Write the final augmented matrix. The solution to the system is (–2, 5). Check: x – 3y = –17 4x + 2y = 2Use the original equations. (–2) – 3(5) –17 4(–2) + 2(5) 2 Substitute. –2 – 15 –17 –8 + 10 2Multiply. –17 = –17 2 = 2 Quick Check Additional Examples 4-8

FeatureLesson Algebra 2 Lesson Main Use the rref feature on a graphing calculator to solve the system Lesson 4-8 Augmented Matrices and Systems 4x + 3y + z = –1 –2x – 2y + 7z = –10. 3x + y + 5z = 2 Step 1:Enter the augmented matrix as matrix A. Step 2:Use the rref feature of your graphing calculator. The solution is (7, –9, –2). Additional Examples 4-8

FeatureLesson Algebra 2 Lesson Main (continued) Lesson 4-8 Augmented Matrices and Systems Partial Check: 4x + 3y + z = –1Use the original equation. 4(7) + 3(–9) + (–2) –1Substitute. 28 – 27 – 2 –1Multiply. –1 = –1Simplify. Quick Check Additional Examples 4-8

FeatureLesson Algebra 2 Lesson Main 1.Use Cramer’s Rule to solve the system. 2.Suppose you want to use Cramer’s Rule to find the value of z in the following system. Write the determinants you would need to evaluate. 3.Solve the system by using an augmented matrix. 4.Solve the system by using an augmented matrix. Lesson 4-8 Augmented Matrices and Systems (–1, 8, –3) 3x + 2y = –2 5x + 4y = 8 –7x + 3y + 9z = 12 5x + 3z = 8 4x – 6y + z = –2 4x + y – z = 7 –2x + 2y + 5z = 3 7x – 3y – 9z = –4 5x + y = 1 3x – 2y = 24 (–12, 17) D =, D z = –7 3 9 5 0 3 4 –6 1 –7 3 12 5 0 8 4 –6 –2 (2, –9) Lesson Quiz 4-8

FeatureLesson Algebra 2 Lesson Main Lesson 4-1 Use the table at the right. (For help, go to Skills Handbook page 842.) Organizing Data Into Matrices How.

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FeatureLesson Algebra 2 Lesson Main Lesson 4-1 Use the table at the right. (For help, go to Skills Handbook page 842.) Organizing Data Into Matrices How.

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Presentation on theme: "FeatureLesson Algebra 2 Lesson Main Lesson 4-1 Use the table at the right. (For help, go to Skills Handbook page 842.) Organizing Data Into Matrices How."— Presentation transcript:

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