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# Matrix multiplication, inverse

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### Transcript of "Matrix multiplication, inverse"

1. 1. Announcements Quiz 2 on Wednesday Jan 27 on sections 1.4, 1.5, 1.7 and 1.8 If you have any grading issues with quiz 1, please discuss with me asap. Test 1 will be on Feb 1, Monday in class. More details later.
2. 2. Last Class.. Two matrices are equal if they have the same size the corresponding entries are all equal
3. 3. Last Class.. Two matrices are equal if they have the same size the corresponding entries are all equal If A and B are m × n matrices, the sum A + B is also an m × n matrix
4. 4. Last Class.. Two matrices are equal if they have the same size the corresponding entries are all equal If A and B are m × n matrices, the sum A + B is also an m × n matrix The columns of A + B is the sum of the corresponding columns of A and B .
5. 5. Last Class.. Two matrices are equal if they have the same size the corresponding entries are all equal If A and B are m × n matrices, the sum A + B is also an m × n matrix The columns of A + B is the sum of the corresponding columns of A and B . A + B is dened only if A and B are of the same size.
6. 6. Matrix Multiplication Denition Suppose A is an m × n matrix and let B be another matrix of size n × p with columns b1 , b2 , . . . , bp , the product AB is the m × p matrix whose columns are Ab1 , Ab2 , . . . , Abp In other words, AB = A b1 b2 . . . bp = Ab1 Ab2 . . . Abp
7. 7. Matrix Multiplication For multiplication of 2 matrices to be possible, The number of columns in the rst matrix (multiplicant) must be same as the number of rows in the second matrix (multiplier).
8. 8. Matrix Multiplication For multiplication of 2 matrices to be possible, The number of columns in the rst matrix (multiplicant) must be same as the number of rows in the second matrix (multiplier). If both matrices are square matrices, you can always multiply them either way (that is AB or BA)
9. 9. Matrix Multiplication For multiplication of 2 matrices to be possible, The number of columns in the rst matrix (multiplicant) must be same as the number of rows in the second matrix (multiplier). If both matrices are square matrices, you can always multiply them either way (that is AB or BA) For arbitrarily sized matrices, it is possible that you can nd AB but not nd BA. Or you could nd BA but not AB . Sometimes, both AB and BA will not exist.
10. 10. Matrix Multiplication For multiplication of 2 matrices to be possible, The number of columns in the rst matrix (multiplicant) must be same as the number of rows in the second matrix (multiplier). If both matrices are square matrices, you can always multiply them either way (that is AB or BA) For arbitrarily sized matrices, it is possible that you can nd AB but not nd BA. Or you could nd BA but not AB . Sometimes, both AB and BA will not exist. Thus in general AB = BA
11. 11. Questions 1. Suppose A is a 6 × 4 matrix and B is a 3 × 4 matrix, does AB exist?
12. 12. Questions 1. Suppose A is a 6 × 4 matrix and B is a 3 × 4 matrix, does AB exist? NO 2. What about BA?
13. 13. Questions 1. Suppose A is a 6 × 4 matrix and B is a 3 × 4 matrix, does AB exist? NO 2. What about BA?NO 3. Suppose A is a 6 × 4 matrix and B is a 4 × 3 matrix, does AB exist?
14. 14. Questions 1. Suppose A is a 6 × 4 matrix and B is a 3 × 4 matrix, does AB exist? NO 2. What about BA?NO 3. Suppose A is a 6 × 4 matrix and B is a 4 × 3 matrix, does AB exist? Yes 4. What about BA?
15. 15. Questions 1. Suppose A is a 6 × 4 matrix and B is a 3 × 4 matrix, does AB exist? NO 2. What about BA?NO 3. Suppose A is a 6 × 4 matrix and B is a 4 × 3 matrix, does AB exist? Yes 4. What about BA? NO
16. 16. Questions Suppose A is a 2 × 5 matrix and B is a 5 × 3 matrix, what are the sizes of AB and BA? B A AB   ∗ ∗ ∗  ∗ ∗ ∗  ∗ ∗ ∗ ∗ ∗   ∗ ∗ ∗ ∗ ∗ ∗ =   ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗     ∗ ∗ ∗  ∗ ∗ ∗ 2×5 5×3 The product BA is not dened here.
17. 17. Questions Suppose A is a 2 × 5 matrix and B is a 5 × 3 matrix, what are the sizes of AB and BA? B A AB   ∗ ∗ ∗  ∗ ∗ ∗  ∗ ∗ ∗ ∗ ∗   ∗ ∗ ∗ ∗ ∗ ∗ =   ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗     ∗ ∗ ∗  ∗ ∗ ∗ 2×5 5×3 Match The product BA is not dened here.
18. 18. Questions Suppose A is a 2 × 5 matrix and B is a 5 × 3 matrix, what are the sizes of AB and BA? B A AB   ∗ ∗ ∗  ∗ ∗ ∗  ∗ ∗ ∗ ∗ ∗   ∗ ∗ ∗ ∗ ∗ ∗ =   ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗     ∗ ∗ ∗  ∗ ∗ ∗ 2×5 5×3 Match Size of AB The product BA is not dened here.
19. 19. Questions Suppose A is a 2 × 5 matrix and B is a 5 × 3 matrix, what are the sizes of AB and BA? B A AB   ∗ ∗ ∗  ∗ ∗ ∗  ∗ ∗ ∗ ∗ ∗   ∗ ∗ ∗ ∗ ∗ ∗ =   ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗     ∗ ∗ ∗  ∗ ∗ ∗ 2×5 5×3 2×3 Match Size of AB The product BA is not dened here.
20. 20. The Row-Column rule The following is the more ecient and easy way for computing AB . If the product AB is dened, the entry in row i and column j of AB is the sum of the products of corresponding entries from row i of A and column j of B For example, The entry in the rst row, second column in AB is the sum of the products of the entries from the rst row of A and the second column of B .
21. 21. Example 0 2   2 3 −1 Find AB where A = and B =  −2 3 0 1 2  4 2
22. 22. B : 3 rows 2 columns     0 2       0 −2 3   2×       4 2           2 3 −1     −10 ∗           0 1 2      ∗ ∗  A : 2 rows 3 columns
23. 23. B : 3 rows 2 columns     0 2       0 −2 3   2×     2)   (− 4 2     3×       2 3 −1     −10 ∗           0 1 2      ∗ ∗  A : 2 rows 3 columns
24. 24. B : 3 rows 2 columns     0 2       0 −2 3   2×     2)   (− 4 2   4   3× 1 )× (−       2 3 −1     −10 ∗           0 1 2      ∗ ∗  A : 2 rows 3 columns
25. 25. B : 3 rows 2 columns     0 2       0 −2 3   2×     2) +   (− 4 2   4   3× + 1 )× (−       2 3 −1     −10 ∗           0 1 2      ∗ ∗  A : 2 rows 3 columns
26. 26. B : 3 rows 2 columns     0 2     2   −2 3   2×       4 2           2 3 −1     10 11            0 1 2   ∗ ∗  A : 2 rows 3 columns
27. 27. B : 3 rows 2 columns     0 2     2   −2 3   2×     3   3× 4 2           2 3 −1     10 11            0 1 2   ∗ ∗  A : 2 rows 3 columns
28. 28. B : 3 rows 2 columns     0 2     2   −2 3   2×     3   3× 4 2   2   (− 1) ×       2 3 −1     10 11            0 1 2   ∗ ∗  A : 2 rows 3 columns
29. 29. B : 3 rows 2 columns     0 2     2   −2 3   2×     + 3   3 ×+ 4 2   2   (− 1) ×       2 3 −1     10 11            0 1 2   ∗ ∗  A : 2 rows 3 columns
30. 30. B : 3 rows 2 columns     0 2       −2 3     0     0× 4 2           2 3 −1     10 11         6   0 1 2      ∗  A : 2 rows 3 columns
31. 31. B : 3 rows 2 columns     0 2       −2 3     0     0× 4 2   2)   (− 1×       2 3 −1     10 11         6   0 1 2      ∗  A : 2 rows 3 columns
32. 32. B : 3 rows 2 columns     0 2       −2 3     0     0× 4 2   2)   (− 1× 4 2×       2 3 −1     10 11         6   0 1 2      ∗  A : 2 rows 3 columns
33. 33. B : 3 rows 2 columns     0 2       −2 3     0     0× 4 2   2) +   (− 1× + 4 2×       2 3 −1     10 11         6   0 1 2      ∗  A : 2 rows 3 columns
34. 34. B : 3 rows 2 columns     0 2       −2 3     2 0×     4 2           2 3 −1     −10 11           0 1 2 6 7       A : 2 rows 3 columns
35. 35. B : 3 rows 2 columns     0 2       −2 3     2 0×     3 4 2     1×       2 3 −1     −10 11           0 1 2 6 7       A : 2 rows 3 columns
36. 36. B : 3 rows 2 columns     0 2       −2 3     2 0×     3 4 2     1× 2 2×       2 3 −1     −10 11           0 1 2 6 7       A : 2 rows 3 columns
37. 37. B : 3 rows 2 columns     0 2       −2 3     2 0×     3 4 2 +     1× 2 + 2×       2 3 −1     −10 11           0 1 2 6 7       A : 2 rows 3 columns C = A × B : 2 rows 2 columns
38. 38. Identity Matrix A diagonal matrix with 1's on the diagonals. (This automatically means that it is a square matrix). Use the notation In for the identity matrix of order n × n. A matrix A of suitable order multiplied with the identity matrix gives back A. Example 1 0 0 1 I2 1 0 0    0 1 0  0 0 1 I3
39. 39. Problem 2, sec 2.1 Parts 3 and 4 7 −5 1 1 2 −5 If B = ,C= and E = . Find CB 1 −4 −3 −2 1 3 and EB .
40. 40. Problem 2, sec 2.1 Parts 3 and 4 7 −5 1 1 2 −5 If B = ,C= and E = . Find CB 1 −4 −3 −2 1 3 and EB . Obviously the product EB cannot be found (E has 1 column and B has 2 rows).
41. 41. Problem 2, sec 2.1 Parts 3 and 4 7 −5 1 1 2 −5 If B = ,C= and E = . Find CB 1 −4 −3 −2 1 3 and EB . Obviously the product EB cannot be found (E has 1 column and B has 2 rows). 1 2 7 −5 1 CB = −2 1 1 −4 −3
42. 42. Problem 2, sec 2.1 Parts 3 and 4 7 −5 1 1 2 −5 If B = ,C= and E = . Find CB 1 −4 −3 −2 1 3 and EB . Obviously the product EB cannot be found (E has 1 column and B has 2 rows). 1 2 7 −5 1 CB = −2 1 1 −4 −3 1.7 + 2.1 1.(−5) + 2.(−4) 1.1 + 2.(−3) = (−2).7 + 1.1 (−2).(−5) + 1.(−4) (−2).1 + 1.(−3) 9 −13 −5 = −13 6 −5
43. 43. Problem 4, sec 2.1 9 −1 3   If A =  −8 7 −6 , nd A − 5I3 and (5I3 )A. −4 1 8
44. 44. Problem 4, sec 2.1 9 −1 3   If A =  −8 7 −6 , nd A − 5I3 and (5I3 )A. −4 1 8 1 0 0 5 0 0     Since I3 =  0 1 0 , 5I3 =  0 5 0 , 0 0 1 0 0 5 9 −1 3 5 0 0 4 −1 3       A − 5I3 =  −8 7 −6  −  0 5 0  =  −8 2 −6  −4 1 8 0 0 5 −4 1 3
45. 45. Problem 4, sec 2.1 5 0 0 9 −1 3    (5I3 )A =  0 5 0   −8 7 −6  = 0 0 5 −4 1 8 5.9 + 0.(−8) + 0.(−4) 5.(−1) + 0.7 + 0.1 5.3 + 0.(−6) + 0.8    0.9 + 5.(−8) + 0.(−4) 0.(−1) + 5.7 + 0.1 0.3 + 5.(−6) + 0.8  0.9 + 0.(−8) + 5.(−4) 0.(−1) + 0.7 + 5.1 0.3 + 0.(−6) + 5.8 45 −5 15   = −40 35 −30  −20 5 40
46. 46. Problem 10, sec 2.1 2 −3 8 4 5 −2 A= , B= and C = . Verify that −4 6 5 5 3 1 AB = AC but B = C .
47. 47. Problem 10, sec 2.1 2 −3 8 4 5 −2 A= , B= and C = . Verify that −4 6 5 5 3 1 AB = AC but B = C . Solution: This example shows that the usual cancellation rule does not apply to matrices in general. (Usually, if ab = ac , you can cancel a on both sides (if a = 0 ) and this will give b = c .) 2 −3 8 4 1 −7 AB = = −4 6 5 5 −2 14 2 −3 5 −2 1 −7 AC = = −4 6 3 1 −2 14 Thus AB = AC but clearly, B = C .
48. 48. Transpose of a Matrix If A is an m × n matrix, then the transpose of A is denoted by AT .
49. 49. Transpose of a Matrix If A is an m × n matrix, then the transpose of A is denoted by AT . AT is an n × m matrix whose columns are the rows of A. (Or swap the rows and columns of A).
50. 50. Transpose of a Matrix If A is an m × n matrix, then the transpose of A is denoted by AT . AT is an n × m matrix whose columns are the rows of A. (Or swap the rows and columns of A). If A = AT , then A is called a symmetric matrix.
51. 51. Transpose of a Matrix If A is an m × n matrix, then the transpose of A is denoted by AT . AT is an n × m matrix whose columns are the rows of A. (Or swap the rows and columns of A). If A = AT , then A is called a symmetric matrix. (AT )T = A.
52. 52. Transpose of a Matrix If A is an m × n matrix, then the transpose of A is denoted by AT . AT is an n × m matrix whose columns are the rows of A. (Or swap the rows and columns of A). If A = AT , then A is called a symmetric matrix. (AT )T = A. Example 1 2   1 3 5 If A =  3 4 , then AT = 2 4 6 5 6
53. 53. Powers of a Matrix If A is an n × n matrix, then Ak is A multiplied k times
54. 54. Powers of a Matrix If A is an n × n matrix, then Ak is A multiplied k times Example 1 2 1 2 1 2 7 10 If A = , then A2 = = 3 4 3 4 3 4 15 22
55. 55. Powers of a Matrix If A is an n × n matrix, then Ak is A multiplied k times Example 1 2 1 2 1 2 7 10 If A = , then A2 = = 3 4 3 4 3 4 15 22 7 10 1 2 37 54 Also, A3 = A2 A = = 15 22 3 4 81 118
56. 56. Powers of a Matrix If A is an n × n matrix, then Ak is A multiplied k times Example 1 2 1 2 1 2 7 10 If A = , then A2 = = 3 4 3 4 3 4 15 22 7 10 1 2 37 54 Also, A3 = A2 A = = 15 22 3 4 81 118 Please do this and convince yourself that this is true.
57. 57. Section 2.2 Inverse of a Matrix Recall that for numbers the inverse (or the multilpicative inverse) is its reciprocal. The inverse of 8 is 1/8 or 8−1 .
58. 58. Section 2.2 Inverse of a Matrix Recall that for numbers the inverse (or the multilpicative inverse) is its reciprocal. The inverse of 8 is 1/8 or 8−1 . Satises 8−1 .8 = 1 and 8.8−1 = 1
59. 59. Section 2.2 Inverse of a Matrix Recall that for numbers the inverse (or the multilpicative inverse) is its reciprocal. The inverse of 8 is 1/8 or 8−1 . Satises 8−1 .8 = 1 and 8.8−1 = 1 Can generalize the concept of inverse to a matrix
60. 60. Section 2.2 Inverse of a Matrix Recall that for numbers the inverse (or the multilpicative inverse) is its reciprocal. The inverse of 8 is 1/8 or 8−1 . Satises 8−1 .8 = 1 and 8.8−1 = 1 Can generalize the concept of inverse to a matrix The matrix involved must be a square matrix No slanted line notation for matrix inverses.
61. 61. Section 2.2 Inverse of a Matrix Given an n × n matrix A, we want to nd another n × n matrix C such that AC = In and CA = In where In is the identity matrix of size n × n. Such a matrix C is called the inverse of A.
62. 62. Section 2.2 Inverse of a Matrix Given an n × n matrix A, we want to nd another n × n matrix C such that AC = In and CA = In where In is the identity matrix of size n × n. Such a matrix C is called the inverse of A. The inverse of a matrix is unique.
63. 63. Section 2.2 Inverse of a Matrix Given an n × n matrix A, we want to nd another n × n matrix C such that AC = In and CA = In where In is the identity matrix of size n × n. Such a matrix C is called the inverse of A. The inverse of a matrix is unique. We denote the inverse of A by A−1
64. 64. Section 2.2 Inverse of a Matrix Given an n × n matrix A, we want to nd another n × n matrix C such that AC = In and CA = In where In is the identity matrix of size n × n. Such a matrix C is called the inverse of A. The inverse of a matrix is unique. We denote the inverse of A by A−1 Thus if A is an n × n matrix, AA−1 = A−1 A = In
65. 65. Section 2.2 Inverse of a Matrix Not every n × n matrix has an inverse If A−1 exists, we say that A is invertible.
66. 66. Section 2.2 Inverse of a Matrix Not every n × n matrix has an inverse If A−1 exists, we say that A is invertible. A matrix that is not invertible is also called a singular matrix. Invertible matrices are also called non-singular matrices.
67. 67. Section 2.2 Inverse of a Matrix Not every n × n matrix has an inverse If A−1 exists, we say that A is invertible. A matrix that is not invertible is also called a singular matrix. Invertible matrices are also called non-singular matrices. Denition a b Given a matrix A = , the quantity ad − bc is called the c d determinant of A.
68. 68. Finding Inverse of a 2 ×2 Matrix a b Let A = . If ad − bc = 0 then A is invertible and c d 1 d −b A−1 = ad − bc −c a .
69. 69. Finding Inverse of a 2 ×2 Matrix a b Let A = . If ad − bc = 0 then A is invertible and c d 1 d −b A−1 = ad − bc −c a . So if the determinant of A (or det A) is equal to 0, A−1 does not exist.
70. 70. Finding Inverse of a 2 ×2 Matrix Steps for a 2 × 2 matrix A 1. First check whether det A=0. If so, stop. A is not invertible.
71. 71. Finding Inverse of a 2 ×2 Matrix Steps for a 2 × 2 matrix A 1. First check whether det A=0. If so, stop. A is not invertible. 2. If det A = 0, swap the main diagonal elements of A.
72. 72. Finding Inverse of a 2 ×2 Matrix Steps for a 2 × 2 matrix A 1. First check whether det A=0. If so, stop. A is not invertible. 2. If det A = 0, swap the main diagonal elements of A. 3. Then change the sign of both o diagonal elements (don't swap these)
73. 73. Finding Inverse of a 2 ×2 Matrix Steps for a 2 × 2 matrix A 1. First check whether det A=0. If so, stop. A is not invertible. 2. If det A = 0, swap the main diagonal elements of A. 3. Then change the sign of both o diagonal elements (don't swap these) 4. Divide this matrix (after steps 2 and 3) by detA to give A−1 . (This divides each element of the resultant matrix.) 5. If you want to check your answer, you can see whether 1 0 AA−1 = 0 1 6. This method will not work for 3 × 3 or bigger matrices.
74. 74. Finding Inverse of a 2 ×2 Matrix Find the inverse of 1 2 A= 3 4 Solution: Here detA = (1)(4) − (2)(3) = −2 = 0. So we can nd A−1 .
75. 75. Finding Inverse of a 2 ×2 Matrix Find the inverse of 1 2 A= 3 4 Solution: Here detA = (1)(4) − (2)(3) = −2 = 0. So we can nd A−1 . Interchange the positions of 1 and 4. Change the signs of 2 and 3. Then we get 4 −2 −3 1 .
76. 76. Finding Inverse of a 2 ×2 Matrix Find the inverse of 1 2 A= 3 4 Solution: Here detA = (1)(4) − (2)(3) = −2 = 0. So we can nd A−1 . Interchange the positions of 1 and 4. Change the signs of 2 and 3. Then we get 4 −2 −3 1 . Divide each element of the matrix by detA which is -2. This gives −2 1 A−1 = 3/2 −1/2
77. 77. What's the use? Remember solving the matrix equation Ax = b for suitable A and x ?
78. 78. What's the use? Remember solving the matrix equation Ax = b for suitable A and x ? Theorem If A is an n × n invertible matrix, then for each vector b in Rn , the equation Ax = b has a unique solution x = A−1 b So, nd the inverse and multiply with the vector b to get the vector x.
79. 79. Example Use the inverse of the previous example to solve x1 + 2x2 = 2 3x1 + 4x2 = 4 Solution: Based on the previous theorem, x1 −2 1 2 0 x= = = x2 3/2 −1/2 4 1 A−1 b
80. 80. Example Theorem 1. If A is an invertible matrix, its inverse A−1 is also an invertible matrix and (A−1 )−1 = A
81. 81. Example Theorem 1. If A is an invertible matrix, its inverse A−1 is also an invertible matrix and (A−1 )−1 = A 2. If A is an invertible matrix, its transpose AT is also an invertible matrix and (AT )−1 = (A−1 )T
82. 82. Example Theorem 1. If A is an invertible matrix, its inverse A−1 is also an invertible matrix and (A−1 )−1 = A 2. If A is an invertible matrix, its transpose AT is also an invertible matrix and (AT )−1 = (A−1 )T 3. If A and B are invertible matrices, their product AB is also invertible
83. 83. Example Theorem 1. If A is an invertible matrix, its inverse A−1 is also an invertible matrix and (A−1 )−1 = A 2. If A is an invertible matrix, its transpose AT is also an invertible matrix and (AT )−1 = (A−1 )T 3. If A and B are invertible matrices, their product AB is also invertible and (AB )−1 = B −1 A−1 . This is called the shoes-socks principle. (Remember the order in which you put your socks and shoes on and the order in which you remove them?)
84. 84. Example Denition Elementary Matrix: A matrix obtained by doing one row operation on an identity matrix. Example 1 0 0   For I3 =  0 1 0  the following are elementary matrices. 0 0 1
85. 85. Example Denition Elementary Matrix: A matrix obtained by doing one row operation on an identity matrix. Example 1 0 0   For I3 =  0 1 0  the following are elementary matrices. 0 0 1 0 0 1   E1 =  0 1 0 , 1 0 0 R 1←→R 3
86. 86. Example Denition Elementary Matrix: A matrix obtained by doing one row operation on an identity matrix. Example 1 0 0   For I3 =  0 1 0  the following are elementary matrices. 0 0 1 0 0 1 4 0 0     E1 =  0 1 0 , E2 =  0 1 0 , 1 0 0 0 0 1 R 1←→R 3 4R 1
87. 87. Example Denition Elementary Matrix: A matrix obtained by doing one row operation on an identity matrix. Example 1 0 0   For I3 =  0 1 0  the following are elementary matrices. 0 0 1 0 0 1 4 0 0 1 0 0       E1 =  0 1 0 , E2 =  0 1 0 , E3 =  0 1 0  1 0 0 0 0 1 0 1 1 R 1←→R 3 4R 1 R 2+R 3
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