Copyright © Cengage Learning. All rights reserved. 2 SYSTEMS OF LINEAR EQUATIONS AND MATRICES Read pp Stop at “Inverse of a Matrix” box

Copyright © Cengage Learning. All rights reserved. 2.6 The Inverse of a Square Matrix

3

4 If a is a nonzero real number, then there exists a unique real number a –1 (that is, ) such that The use of the (multiplicative) inverse of a real number enables us to solve algebraic equations of the form ax = b (13)

5 The Inverse of a Square Matrix Multiplying both sides of (13) by a –1, we have For example, since the inverse of 2 is 2 –1 =, we can solve the equation 2x = 5

6 The Inverse of a Square Matrix By multiplying both sides of the equation by 2 –1 =, giving 2 –1 (2x) = 2 –1  5 We can use a similar procedure to solve the matrix equation AX = B where A, X, and B are matrices of the proper sizes.

7 The Inverse of a Square Matrix To do this we need the matrix equivalent of the inverse of a real number. Such a matrix, whenever it exists, is called the inverse of a matrix. Not every square matrix has an inverse. A square matrix that has an inverse is said to be nonsingular. A matrix that does not have an inverse is said to be singular.

8 The Inverse of a Square Matrix An example of a singular matrix is given by If B had an inverse given by where a, b, c, and d are some appropriate numbers, then by the definition of an inverse.

9 The Inverse of a Square Matrix we would have BB –1 = I; that is, which implies that 0 = 1—an impossibility! This contradiction shows that B does not have an inverse.

10 Solving Systems of Equations with Inverses

11 Solving Systems of Equations with Inverses We now show how the inverse of a matrix may be used to solve certain systems of linear equations in which the number of equations in the system is equal to the number of variables. For simplicity, let’s illustrate the process for a system of three linear equations in three variables: a 11 x 1 + a 12 x 2 + a 13 x 3 = b 1 a 21 x 1 + a 22 x 2 + a 23 x 3 = b 2 a 31 x 1 + a 32 x 2 + a 33 x 3 = b 3 (15)

12 Solving Systems of Equations with Inverses Let’s write You should verify that System (15) of linear equations may be written in the form of the matrix equation AX = B (16)

13 Solving Systems of Equations with Inverses If A is nonsingular, then the method of this section may be used to compute A –1. Next, multiplying both sides of Equation (16) by A –1 (on the left), we obtain A –1 AX = A –1 BorIX = A –1 Bor X = A –1 B the desired solution to the problem. In the case of a system of n equations with n unknowns, we have the following more general result.

14 Solving Systems of Equations with Inverses In the case of a system of n equations with n unknowns, we have the following more general result.

15 Example 4 Solve the following systems of linear equations: a. 2x + y + z = 1 3x + 2y + z = 2 2x + y + 2z = –1 b. 2x + y + z = 2 3x + 2y + z = –3 2x + y + 2z = 1

16 Example 4 – Solution We may write the given systems of equations in the form AX = B and AX = C respectively, where cont’d

17 Example 4 – Solution The inverse of the matrix A, was found in Example 1. Using this result, we find that the solution of the first system (a) is cont’d

18 Example 4 – Solution or x = 2, y = –1, and z = –2. cont’d

19 Example 4 – Solution The solution of the second system (b) is X = A –1 C or x = 8, y = –13, and z = –1. cont’d

20 Practice p. 137 Self-Check Exercises #1-3