1. 5.8 - 1 10 TH EDITION LIAL HORNSBY SCHNEIDER COLLEGE ALGEBRA

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

3. 5.8 - 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 size.

4. 5.8 - 4 2  2 Identity Matrix If I 2 represents the identity matrix, then

5. 5.8 - 5 Identity Matrices To verify that I 2 is the 2  2 identity matrix, we must show that I = A and IA + A for any matrix A. Let Then

6. 5.8 - 6 Identity Matrices and

7. 5.8 - 7 n  n Identity Matrix The identity matrix is where The element a ij = 1 when i = j (the diagonal elements) and a ij = 0 otherwise.

8. 5.8 - 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 identity matrix is

9. 5.8 - 9 Example 1 VERIFYING THE IDENTITY PROPERTYOF I 3 By the definition of matrix multiplication,

10. 5.8 - 10 Multiplicative Inverses For every nonzero real number a, there is a multiplicative inverse such that (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 This means that only a square matrix can have a multiplicative inverse.

11. 5.8 - 11 Caution Although a -1 = for any nonzero real number a, if A is a matrix, In fact, has no meaning, since 1 is a number and A is a matrix.

12. 5.8 - 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

13. 5.8 - 13 Multiplicative Inverses By the definition of matrix inverse, AA – 1 = I 2, or By matrix multiplication,

14. 5.8 - 14 Multiplicative Inverses Setting corresponding elements equal gives the system of equations (1) (2) (3) (4)

15. 5.8 - 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 Writing the two systems as augmented matrices gives and

16. 5.8 - 16 Multiplicative Inverses and yields

17. 5.8 - 17 Multiplicative Inverses Interchange R1 and R2 to get 1 in the upper left corner. – 2 R1 + R2

18. 5.8 - 18 Multiplicative Inverses R2 + R1

19. 5.8 - 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. 5.8 - 20 Multiplicative Inverses so that

21. 5.8 - 21 Multiplicative Inverses Check:

22. 5.8 - 22 Multiplicative Inverses Thus,

23. 5.8 - 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. 5.8 - 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. 5.8 - 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. 5.8 - 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. 5.8 - 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. 5.8 - 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. 5.8 - 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. 5.8 - 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. 5.8 - 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. 5.8 - 32 Example 2 FINDING THE INVERSE OF A 3  3 MATRIX Step 3 The last transformation shows that the inverse is

33. 5.8 - 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

34. 5.8 - 34 Example 3 INDENTIFYING A MATRIX WITH NO INVERSE results in the following matrices: 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. 5.8 - 35 Example 3 INDENTIFYING A MATRIX WITH NO INVERSE results in the following matrices: and 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.

36. 5.8 - 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, given the linear system

37. 5.8 - 37 Solving Systems Using Inverse Matrices the definition of matrix multiplication can be used to rewrite the system as (1)

38. 5.8 - 38 Solving Systems Using Inverse Matrices (To see this, multiply the matrices on the left.)

39. 5.8 - 39 Solving Systems Using Inverse Matrices then the system given in (1) becomes AX = B. If A – 1 exists, then both sides of AX = B can be multiplied on the left to get Associative property Inverse property Identity property Matrix A – 1 B gives the solution of the system.

40. 5.8 - 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. 5.8 - 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. 5.8 - 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. 5.8 - 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. 5.8 - 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

45. 5.8 - 45 Example 4 SOLVING SYSTEMS OF EQUATIONS USING MATRIX INVERSES and its inverse A – 1 was found in Example 2. Let

46. 5.8 - 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)}.