1. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 1 5 Systems and Matrices Copyright © 2013, 2009, 2005 Pearson Education, Inc.

2. 2 5.8 Matrix Inverses Identity Matrices Multiplicative Inverses Solving Systems Using Inverse Matrices

3. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 3 Identity Matrices By the identity property for real numbers, for any real number a. If there is to be a multiplicative identity matrix I, such that for any matrix A, then A and I must be square matrices of the same dimension.

4. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 4 2  2 Identity Matrix If I 2 represents the 2 x 2 identity matrix, then

5. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 5 Identity Matrices To verify that I 2 is the 2  2 identity matrix, we must show that AI = A and IA = A for any matrix A. Let Then

6. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 6 Identity Matrices and

7. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 7 n  n Identity Matrix The n x n identity matrix is I n, where The element a ij = 1 when i = j (the diagonal elements) and a ij = 0 otherwise.

8. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 8 Example 1 VERIFYING THE IDENTITY PROPERTYOF I 3 Let Give the 3  3 identity matrix I 3 and show that AI 3 = A. Solution The 3  3 identity matrix is

9. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 9 Example 1 VERIFYING THE IDENTITY PROPERTYOF I 3 By the definition of matrix multiplication,

10. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 10 Multiplicative Inverses For every nonzero real number a, there is a multiplicative inverse that satisfies both of the following. (Recall: is also written a – 1.) In a similar way, if A is an n  n matrix, then its multiplicative inverse, written A –1, must satisfy both of the following. This means that only a square matrix can have a multiplicative inverse.

11. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 11 Caution Although a -1 = for any nonzero real number a, if A is a matrix, then We do not use the symbol, since 1 is a number and A is a matrix.

12. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 12 Multiplicative Inverses To find the matrix A –1 we use row transformations, introduced earlier in this chapter. As an example, we find the inverse of Let the unknown inverse matrix be symbolized as follows.

13. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 13 Multiplicative Inverses By the definition of matrix inverse, AA –1 = I 2. Use matrix multiplication for the product above.

14. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 14 Multiplicative Inverses Set the corresponding elements equal to obtain a system of equations (1) (2) (3) (4)

15. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 15 Since equations (1) and (3) involve only x and z, while equations (2) and (4) involve only y and w, these four equations lead to two systems of equations. Multiplicative Inverses Write the two systems as augmented matrices. and

16. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 16 Multiplicative Inverses andyields Each of these systems can be solved by the Gauss-Jordan method. However, since the elements to the left of the vertical bar are identical, the two systems can be combined into one matrix.

17. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 17 Multiplicative Inverses Interchange R1 and R2 to get 1 in the upper left-hand corner. – 2R1 + R2 We can solve simultaneously using matrix row transformations. We need to change the numbers on the left of the vertical bar to the 2 x 2 identity matrix.

18. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 18 Multiplicative Inverses R2 + R1

19. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 19 Multiplicative Inverses The numbers in the first column to the right of the vertical bar in the final matrix give the values of x and z. The second column gives the values of y and w. That is,

20. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 20 Multiplicative Inverses so that

21. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 21 Multiplicative Inverses To check, multiply A by A –1. The result should be I 2.

22. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 22 Multiplicative Inverses Thus,

23. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 23 Finding an Inverse Matrix To obtain A – 1 for any n  n matrix A for which A – 1 exists, follow these steps. Step 1 Form the augmented matrix [A  I n ] where I n is the n  n identity matrix. Step 2 Perform row transformations on [A  I n ] to obtain a matrix of the form [I n  B]. Step 3 Matrix B is A – 1.

24. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 24 Note To confirm that two n  n matrices A and B are inverses of each other, it is sufficient to show that AB = I n. It is not necessary to show also that BA = I n.

25. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 25 Example 2 FINDING THE INVERSE OF A 3  3 MATRIX Find A –1 if Solution Use row transformations as follows. Step 1 Write the augmented matrix [A  I 3 ].

26. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 26 Example 2 FINDING THE INVERSE OF A 3  3 MATRIX Step 2 Since 1 is already in the upper left-hand corner as desired, begin by using the row transformation that will result in 0 for the first element in the second row. Multiply the elements of the first row by – 2 and add the result to the second row. – 2R1 + R2

27. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 27 Example 2 FINDING THE INVERSE OF A 3  3 MATRIX – 3R1 + R3 To get 0 for the first element in the third row, multiply the elements of the first row by – 3 and add to the third row.

28. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 28 Example 2 FINDING THE INVERSE OF A 3  3 MATRIX – ½ R2 To get 1 for the second element in the second row, multiply the elements of the second row by – ½.

29. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 29 Example 2 FINDING THE INVERSE OF A 3  3 MATRIX – ⅓ R2 To get 1 for the third element in the third row, multiply the elements of the third row by – ⅓.

30. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 30 Example 2 FINDING THE INVERSE OF A 3  3 MATRIX – 1R3 + R1 To get 0 for the third element in the first row, multiply the elements of the third row by – 1 and add to the first row

31. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 31 Example 2 FINDING THE INVERSE OF A 3  3 MATRIX – 3/2 R3 + R2 To get 0 for the third element in the second row, multiply the elements of the third row by – 3/2 and add to the second row.

32. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 32 Example 2 FINDING THE INVERSE OF A 3  3 MATRIX Step 3 The last transformation shows that the inverse is

33. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 33 Example 3 INDENTIFYING A MATRIX WITH NO INVERSE Find A –1, if possible, given that Solution Using row transformations to change the first column of the augmented matrix results in the following matrices.

34. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 34 Example 3 INDENTIFYING A MATRIX WITH NO INVERSE and (We multiplied the elements in row one by in the first step, and in the second step we added the negative of row one to row two.) At this point, the matrix should be changed so that the second row, second element will be 1.

35. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 35 Example 3 INDENTIFYING A MATRIX WITH NO INVERSE Since that element is now 0, there is no way to complete the desired transformation, so A –1 does not exist for this matrix A. Just as there is no multiplicative inverse for the real number 0, not every matrix has a multiplicative inverse. Matrix A is an example of such a matrix. and

36. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 36 Solving Systems Using Inverse Matrices Matrix inverses can be used to solve square linear systems of equations. (A square system has the same number of equations as variables.) For example, consider the following linear system of three equations with three variables.

37. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 37 Solving Systems Using Inverse Matrices The definition of matrix multiplication can be used to rewrite the system using matrices (1) (To see this, multiply the matrices on the left.)

38. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 38 Solving Systems Using Inverse Matrices then the system given in (1) becomes AX = B.

39. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 39 Solving Systems Using Inverse Matrices If A – 1 exists, then both sides of AX = B can be multiplied on the left as shown. Associative property Inverse property Identity property Matrix A – 1 B gives the solution of the system.

40. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 40 Solution of the Matrix Equation AX = B If A is an n  n matrix with inverse A – 1, X is an n  1 matrix of variables, and B is an n  1 matrix, then the matrix equation has the solution

41. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 41 Example 4 SOLVING SYSTEMS OF EQUATIONS USING MATRIX INVERSES Use the inverse of the coefficient matrix to solve each system. Solution (a) The system can be written in matrix form as

42. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 42 Example 4 SOLVING SYSTEMS OF EQUATIONS USING MATRIX INVERSES where An equivalent matrix equation is AX = B with solution X = A – 1 B. Use the methods described in this section to determine that

43. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 43 Example 4 SOLVING SYSTEMS OF EQUATIONS USING MATRIX INVERSES Now, find A – 1 B. Since X = A – 1 B The final matrix shows that the solution set of the system is {(2, 0)}.

44. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 44 Example 4 SOLVING SYSTEMS OF EQUATIONS USING MATRIX INVERSES Use the inverse of the coefficient matrix to solve each system. Solution (b) The coefficient matrix A for this system is and its inverse A –1 was found in Example 2.

45. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 45 Example 4 SOLVING SYSTEMS OF EQUATIONS USING MATRIX INVERSES Let

46. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 46 Example 4 SOLVING SYSTEMS OF EQUATIONS USING MATRIX INVERSES Since we have X = A – 1 B, we have A – 1 from Example 2 The solution set is {(2,1, –3)}.