1. Copyright © Cengage Learning. All rights reserved. 8 Matrices and Determinants

2. 8.3 Copyright © Cengage Learning. All rights reserved. THE INVERSE OF A SQUARE MATRIX

3. 3 Verify that two matrices are inverses of each other. Use Gauss-Jordan elimination to find the inverses of matrices. Use a formula to find the inverses of 2  2 matrices. Use inverse matrices to solve systems of linear equations. What You Should Learn

4. 4 The Inverse of a Matrix

5. 5 This section further develops the algebra of matrices. To begin, consider the real number equation ax = b. To solve this equation for x, multiply each side of the equation by a –1 (provided that a  0 ). ax = b (a –1 a)x = a –1 b (1)x = a –1 b x = a –1 b The number a –1 is called the multiplicative inverse of a because a –1 a = 1.

6. 6 The Inverse of a Matrix The definition of the multiplicative inverse of a matrix is similar.

7. 7 Example 1 – The Inverse of a Matrix Show that B is the inverse of A, where and Solution: To show that B is the inverse of A, show that AB = I = BA, as follows.

8. 8 Example 1 – Solution As you can see, AB = I = BA. This is an example of a square matrix that has an inverse. Note that not all square matrices have inverses. cont’d

9. 9 Finding Inverse Matrices

10. 10 Finding Inverse Matrices If a matrix A has an inverse, A is called invertible (or nonsingular); otherwise, A is called singular. A nonsquare matrix cannot have an inverse. To see this, note that if A is of order m  n and B is order of n  m (where m  n), the products AB and BA are of different orders and so cannot be equal to each other. Not all square matrices have inverses. If, however, a matrix does have an inverse, that inverse is unique. Example 2 shows how to use a system of equations to find the inverse of a matrix.

11. 11 Example 2 – Finding the Inverse of a Matrix Find the inverse of. Solution: To find the inverse of A, try to solve the matrix equation AX = I for x. A X I

12. 12 Example 2 – Solution Equating corresponding entries, you obtain two systems of linear equations. x 11 + 4x 21 = 1 –x 11 – 3x 21 = 0 x 12 + 4x 22 = 0 –x 12 – 3x 22 = 1 Solve the first system using elementary row operations to determine that x 11 = –3 and x 21 = 1. cont’d Linear system with two variables, x 11 and x 21. Linear system with two variables, x 12 and x 22.

13. 13 Example 2 – Solution From the second system you can determine that x 12 = –4 and x 22 = 1. Therefore, the inverse of A is X = A –1 You can use matrix multiplication to check this result. Check: cont’d

14. 14 Example 2 – Solution cont’d

15. 15 Finding Inverse Matrices In Example 2, note that the two systems of linear equations have the same coefficient matrix A. Rather than solve the two systems represented by and

16. 16 Finding Inverse Matrices Separately, you can solve them simultaneously by adjoining the identity matrix to the coefficient matrix to obtain. This “doubly augmented” matrix can be represented as [A I ]. AI …

17. 17 Finding Inverse Matrices By applying Gauss-Jordan elimination to this matrix, you can solve both systems with a single elimination process. R 1 + R 2 → –4R 2 + R 1 →

18. 18 Finding Inverse Matrices So, from the “doubly augmented” matrix [A I ], you obtain the matrix [I A –1 ]. This procedure (or algorithm) works for any square matrix that has an inverse. … … AII A –1

19. 19 Finding Inverse Matrices

20. 20 The Inverse of a 2  2 Matrix

21. 21 The Inverse of a 2  2 Matrix Using Gauss-Jordan elimination to find the inverse of a matrix works well (even as a computer technique) for matrices of order 3  3 or greater. For 2  2 matrices, however, many people prefer to use a formula for the inverse rather than Gauss-Jordan elimination.

22. 22 The Inverse of a 2  2 Matrix This simple formula, which works only for 2  2 matrices, is explained as follows. If A is a 2  2 matrix given by then A is invertible if and only if ad – bc  0. Moreover, if ad – bc  0, the inverse is given by. The denominator ad – bc is called the determinant of the 2  2 matrix A. Formula for inverse of matrix A

23. 23 Example 4 – Finding the Inverse of a 2  2 Matrix If possible, find the inverse of each matrix. a. b. Solution: a. For the matrix A, apply the formula for the inverse of a 2  2 matrix to obtain ad – bc = (3)(2) – (–1)(–2) = 4.

24. 24 Example 4 – Solution Because this quantity is not zero, the inverse is formed by interchanging the entries on the main diagonal, changing the signs of the other two entries, and multiplying by the scalar as, follows. Substitute for a, b, c, d, and the determinant. Multiply by the scalar. cont’d

25. 25 Example 4 – Solution b. For the matrix B, you have ad – bc = (3)(2) – (–1)(–6) = 0 which means that B is not invertible. cont’d

26. 26 Systems of Linear Equations

27. 27 Systems of Linear Equations You know that a system of linear equations can have exactly one solution, infinitely many solutions, or no solution. If the coefficient matrix A of a square system (a system that has the same number of equations as variables) is invertible, the system has a unique solution, which is defined as follows.

28. 28 Example 5 – Solving a System Using an Inverse Matrix You are going to invest $10,000 in AAA-rated bonds, AA-rated bonds, and B-rated bonds and want an annual return of $730. The average yields are 6% on AAA bonds, 7.5% on AA bonds, and 9.5% on B bonds. You will invest twice as much in AAA bonds as in B bonds. Your investment can be represented as x + y + z = 10,000 0.06x + 0.075y + 0.095z = 730 x – 2z = 0 where x, y, and z represent the amounts invested in AAA, AA, and B bonds, respectively. Use an inverse matrix to solve the system.

29. 29 Example 5 – Solution Begin by writing the system in the matrix form AX = B. Then, use Gauss-Jordan elimination to find A –1.

30. 30 Example 5 – Solution Finally, multiply B by A –1 on the left to obtain the solution. X = A –1 B The solution of the system is x = 4000, y = 4000, and z = 2000. So, you will invest $4000 in AAA bonds, $4000 in AA bonds, and $2000 in B bonds. cont’d