1. 4-5 – 2x2 Matrices, Determinants, & Inverses

2. Objectives Evaluating Determinants of 2x2 Matrices Using Inverse Matrices to Solve Equations

3. Vocabulary A square matrix is a matrix with the same number of columns as rows. Multiplicative Identity Matrix is an n x n square matrix with 1’s along the main diagonal and 0’s elsewhere.

4. Vocabulary Multiplicative Inverse of a Matrix If A and B are n x n matrices, and AB = BA = 1, then B is the multiplicative inverse of A, written A AA = A A = 1

5. Show that matrices A and B are multiplicative inverses. A = B = 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. Verifying Inverses

6. Vocabulary The determinant of a 2 x 2 matrix is ad – bc. det A

7. Evaluate each determinant. a. det b. det c. det 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 Evaluating the Determinant of a 2x2 Matrix

8. Vocabulary Let A =. If det A = 0, then A has no inverse. If det A ≠ 0, then A =

9. Determine whether each matrix has an inverse. If it does, find it. 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 = Finding an Inverse Matrix

10. (continued) = – 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 20 –5 Use the determinant to –25 6 write the inverse. = 1 det Y Continued

11. Solve X = for the matrix X. The matrix equation has the form AX = B. First find A –1. 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 Solving a Matrix Equation

12. (continued) = = (–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 Continued

13. 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. 0.84 0.07 0.16 0.93 To A To B From A B Write the percents as decimals. Real World Example

14. (continued) b.Predict the number of households that will be using distance carrier B next year. 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. Continued

15. (continued) 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. Continued

16. Homework Pg 203 # 1,4,5,14,18, 22