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# 01.03 squared matrices_and_other_issues

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A Introduction to Linear Algebra

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### 01.03 squared matrices_and_other_issues

1. 1. Mathematics for Artiﬁcial Intelligence Square Matrices and Related Matters Andres Mendez-Vazquez March 2, 2020 1 / 42
2. 2. Outline 1 Square Matrices Introduction The Inverse Solution to Ax = y Algorithm for the Inverse of a Matrix 2 Determinants Introduction Complexity Increases Reducing the Complexity Some Consequences of the deﬁnition Special Determinants 2 / 42
3. 3. Outline 1 Square Matrices Introduction The Inverse Solution to Ax = y Algorithm for the Inverse of a Matrix 2 Determinants Introduction Complexity Increases Reducing the Complexity Some Consequences of the deﬁnition Special Determinants 3 / 42
4. 4. Square Matrices Observation Square matrices are the only matrices that can have inverses. Further In a system of linear algebraic equations: 1 If the number of equations equals the number of unknowns 2 Then the associated coeﬃcient matrix A is square. Now, use the Gauss-Jordan What can happen? 4 / 42
5. 5. Square Matrices Observation Square matrices are the only matrices that can have inverses. Further In a system of linear algebraic equations: 1 If the number of equations equals the number of unknowns 2 Then the associated coeﬃcient matrix A is square. Now, use the Gauss-Jordan What can happen? 4 / 42
6. 6. Square Matrices Observation Square matrices are the only matrices that can have inverses. Further In a system of linear algebraic equations: 1 If the number of equations equals the number of unknowns 2 Then the associated coeﬃcient matrix A is square. Now, use the Gauss-Jordan What can happen? 4 / 42
7. 7. Outline 1 Square Matrices Introduction The Inverse Solution to Ax = y Algorithm for the Inverse of a Matrix 2 Determinants Introduction Complexity Increases Reducing the Complexity Some Consequences of the deﬁnition Special Determinants 5 / 42
8. 8. We have two possibilities First case The Gauss-Jordan form for An×n is the n × n identity matrix In Second case The Gauss-Jordan form for A has at least one row of zeros 6 / 42
9. 9. We have two possibilities First case The Gauss-Jordan form for An×n is the n × n identity matrix In Second case The Gauss-Jordan form for A has at least one row of zeros 6 / 42
10. 10. Explanation In the ﬁrst case We can show that A is invertible. How? Do you remember? EkEk−1...E2E1A = I, Setting B = EkEk−1...E2E1 We have BA = I therefore B = A−1 7 / 42
11. 11. Explanation In the ﬁrst case We can show that A is invertible. How? Do you remember? EkEk−1...E2E1A = I, Setting B = EkEk−1...E2E1 We have BA = I therefore B = A−1 7 / 42
12. 12. Explanation In the ﬁrst case We can show that A is invertible. How? Do you remember? EkEk−1...E2E1A = I, Setting B = EkEk−1...E2E1 We have BA = I therefore B = A−1 7 / 42
13. 13. Furthermore We can build the following matrix E−1 1 E−1 2 · · · E−1 k Then E−1 1 E−1 2 · · · E−1 k BA = E−1 1 E−1 2 · · · E−1 k I Thus E−1 1 E−1 2 · · · E−1 k (EkEk−1...E2E1) A = E−1 1 E−1 2 · · · E−1 k 8 / 42
14. 14. Furthermore We can build the following matrix E−1 1 E−1 2 · · · E−1 k Then E−1 1 E−1 2 · · · E−1 k BA = E−1 1 E−1 2 · · · E−1 k I Thus E−1 1 E−1 2 · · · E−1 k (EkEk−1...E2E1) A = E−1 1 E−1 2 · · · E−1 k 8 / 42
15. 15. Furthermore We can build the following matrix E−1 1 E−1 2 · · · E−1 k Then E−1 1 E−1 2 · · · E−1 k BA = E−1 1 E−1 2 · · · E−1 k I Thus E−1 1 E−1 2 · · · E−1 k (EkEk−1...E2E1) A = E−1 1 E−1 2 · · · E−1 k 8 / 42
16. 16. Therefore We have A = E−1 1 E−1 2 · · · E−1 k Theorem The following are equivalent: 1 The square matrix A is invertible. 2 The Gauss-Jordan or reduced echelon form of A is the identity matrix. 3 Acan be written as a product of elementary matrices. 9 / 42
17. 17. Therefore We have A = E−1 1 E−1 2 · · · E−1 k Theorem The following are equivalent: 1 The square matrix A is invertible. 2 The Gauss-Jordan or reduced echelon form of A is the identity matrix. 3 Acan be written as a product of elementary matrices. 9 / 42
18. 18. The Second Case The Gauss-Jordan form of An×n It can only have at most n leading entries. If the Gauss-Jordan form of A is not I We have something quite diﬀerent Then the GJ form has n − 1 or fewer leading entries Therefore, it has at least one row of zeros. 10 / 42
19. 19. The Second Case The Gauss-Jordan form of An×n It can only have at most n leading entries. If the Gauss-Jordan form of A is not I We have something quite diﬀerent Then the GJ form has n − 1 or fewer leading entries Therefore, it has at least one row of zeros. 10 / 42
20. 20. The Second Case The Gauss-Jordan form of An×n It can only have at most n leading entries. If the Gauss-Jordan form of A is not I We have something quite diﬀerent Then the GJ form has n − 1 or fewer leading entries Therefore, it has at least one row of zeros. 10 / 42
21. 21. Example Given A = 2 1 1 2 What do we need to do? Look at the blackboard... Therefore A−1 = E4E3E2E1 11 / 42
22. 22. Example Given A = 2 1 1 2 What do we need to do? Look at the blackboard... Therefore A−1 = E4E3E2E1 11 / 42
23. 23. Example Given A = 2 1 1 2 What do we need to do? Look at the blackboard... Therefore A−1 = E4E3E2E1 11 / 42
24. 24. Outline 1 Square Matrices Introduction The Inverse Solution to Ax = y Algorithm for the Inverse of a Matrix 2 Determinants Introduction Complexity Increases Reducing the Complexity Some Consequences of the deﬁnition Special Determinants 12 / 42
25. 25. If A is invertible a.k.a. full rank Equation Ax = y has the unique solution Ax = y ⇔ A−1 Ax = A−1 y ⇔ x = A−1 y If A is not invertible Then there is at least one free variable. There are non-trivial solutions to Ax = 0. If y = 0 Either Ax = y is inconsistent. Solutions to the system exist, but there are inﬁnitely many. 13 / 42
26. 26. If A is invertible a.k.a. full rank Equation Ax = y has the unique solution Ax = y ⇔ A−1 Ax = A−1 y ⇔ x = A−1 y If A is not invertible Then there is at least one free variable. There are non-trivial solutions to Ax = 0. If y = 0 Either Ax = y is inconsistent. Solutions to the system exist, but there are inﬁnitely many. 13 / 42
27. 27. If A is invertible a.k.a. full rank Equation Ax = y has the unique solution Ax = y ⇔ A−1 Ax = A−1 y ⇔ x = A−1 y If A is not invertible Then there is at least one free variable. There are non-trivial solutions to Ax = 0. If y = 0 Either Ax = y is inconsistent. Solutions to the system exist, but there are inﬁnitely many. 13 / 42
28. 28. If A is invertible a.k.a. full rank Equation Ax = y has the unique solution Ax = y ⇔ A−1 Ax = A−1 y ⇔ x = A−1 y If A is not invertible Then there is at least one free variable. There are non-trivial solutions to Ax = 0. If y = 0 Either Ax = y is inconsistent. Solutions to the system exist, but there are inﬁnitely many. 13 / 42
29. 29. If A is invertible a.k.a. full rank Equation Ax = y has the unique solution Ax = y ⇔ A−1 Ax = A−1 y ⇔ x = A−1 y If A is not invertible Then there is at least one free variable. There are non-trivial solutions to Ax = 0. If y = 0 Either Ax = y is inconsistent. Solutions to the system exist, but there are inﬁnitely many. 13 / 42
30. 30. Outline 1 Square Matrices Introduction The Inverse Solution to Ax = y Algorithm for the Inverse of a Matrix 2 Determinants Introduction Complexity Increases Reducing the Complexity Some Consequences of the deﬁnition Special Determinants 14 / 42
31. 31. Something Quite Important We have done something important It leads immediately to an algorithm for constructing the inverse of A. Observation Suppose Bn×p is another matrix with the same number of rows as An×n Then C = (A|B)n×(n+p) 15 / 42
32. 32. Something Quite Important We have done something important It leads immediately to an algorithm for constructing the inverse of A. Observation Suppose Bn×p is another matrix with the same number of rows as An×n Then C = (A|B)n×(n+p) 15 / 42
33. 33. Something Quite Important We have done something important It leads immediately to an algorithm for constructing the inverse of A. Observation Suppose Bn×p is another matrix with the same number of rows as An×n Then C = (A|B)n×(n+p) 15 / 42
34. 34. Then It is obvious EC = (EA|EB)n×(n+p) Where EA is a n × n matrix. EB is a n × p matrix 16 / 42
35. 35. Then It is obvious EC = (EA|EB)n×(n+p) Where EA is a n × n matrix. EB is a n × p matrix 16 / 42
36. 36. Algorithm Form the partitioned matrix C = (A|I) Apply the Gauss-Jordan reduction EkEk−1 · · · E1 (A|I) = (EkEk−1 · · · E1A|EkEk−1 · · · E1I) Therefore, if A is invertible (EkEk−1 · · · E1A|EkEk−1 · · · E1I) = I|A−1 17 / 42
37. 37. Algorithm Form the partitioned matrix C = (A|I) Apply the Gauss-Jordan reduction EkEk−1 · · · E1 (A|I) = (EkEk−1 · · · E1A|EkEk−1 · · · E1I) Therefore, if A is invertible (EkEk−1 · · · E1A|EkEk−1 · · · E1I) = I|A−1 17 / 42
38. 38. Algorithm Form the partitioned matrix C = (A|I) Apply the Gauss-Jordan reduction EkEk−1 · · · E1 (A|I) = (EkEk−1 · · · E1A|EkEk−1 · · · E1I) Therefore, if A is invertible (EkEk−1 · · · E1A|EkEk−1 · · · E1I) = I|A−1 17 / 42
39. 39. Remark The individual factors in the product of A−1 are not unique They depend on how we do the row reduction. 18 / 42
40. 40. Outline 1 Square Matrices Introduction The Inverse Solution to Ax = y Algorithm for the Inverse of a Matrix 2 Determinants Introduction Complexity Increases Reducing the Complexity Some Consequences of the deﬁnition Special Determinants 19 / 42
41. 41. Before the Matrix we had the Determinant You have seen determinants in your classes long ago A = a b c d =⇒ det (A) = ad − bc What about    a11 a12 a13 a21 a22 a23 a31 a32 a33    Then det (A) =a11a22a33 + a12a21a32 + a12a23a31 − ... a12a21a33 − a11a23a32 − a13a22a31 20 / 42
42. 42. Before the Matrix we had the Determinant You have seen determinants in your classes long ago A = a b c d =⇒ det (A) = ad − bc What about    a11 a12 a13 a21 a22 a23 a31 a32 a33    Then det (A) =a11a22a33 + a12a21a32 + a12a23a31 − ... a12a21a33 − a11a23a32 − a13a22a31 20 / 42
43. 43. Before the Matrix we had the Determinant You have seen determinants in your classes long ago A = a b c d =⇒ det (A) = ad − bc What about    a11 a12 a13 a21 a22 a23 a31 a32 a33    Then det (A) =a11a22a33 + a12a21a32 + a12a23a31 − ... a12a21a33 − a11a23a32 − a13a22a31 20 / 42
44. 44. Recursive Deﬁnition Deﬁnition Let A be a n × n matrix. Then the determinant of A is deﬁned as follow: det (A) = a11 if n = 1 n i=1 ai1Ai1 if n > 1 Where Aij is the (i, j) −cofactor where Aij = (−1)i+j det (Mij) Here Mij is the (n − 1) × (n − 1) matrix obtained from A by removing its ith row and jth column. 21 / 42
45. 45. Recursive Deﬁnition Deﬁnition Let A be a n × n matrix. Then the determinant of A is deﬁned as follow: det (A) = a11 if n = 1 n i=1 ai1Ai1 if n > 1 Where Aij is the (i, j) −cofactor where Aij = (−1)i+j det (Mij) Here Mij is the (n − 1) × (n − 1) matrix obtained from A by removing its ith row and jth column. 21 / 42
46. 46. Recursive Deﬁnition Deﬁnition Let A be a n × n matrix. Then the determinant of A is deﬁned as follow: det (A) = a11 if n = 1 n i=1 ai1Ai1 if n > 1 Where Aij is the (i, j) −cofactor where Aij = (−1)i+j det (Mij) Here Mij is the (n − 1) × (n − 1) matrix obtained from A by removing its ith row and jth column. 21 / 42
47. 47. Example We have    1 1 1 0 2 1 0 0 3    22 / 42
48. 48. Outline 1 Square Matrices Introduction The Inverse Solution to Ax = y Algorithm for the Inverse of a Matrix 2 Determinants Introduction Complexity Increases Reducing the Complexity Some Consequences of the deﬁnition Special Determinants 23 / 42
49. 49. Problems Will Robinson!!! We have that for a An×n det (A) has n factorials (n!)terms Problems The fastest computer of the world will take forever to ﬁnish 24 / 42
50. 50. Problems Will Robinson!!! We have that for a An×n det (A) has n factorials (n!)terms Problems The fastest computer of the world will take forever to ﬁnish 24 / 42
51. 51. Thus, we have some problems with that Floating point arithmetic It is not at all the same thing as working with real numbers. Representation x = (d1d2d3 · · · dn) × 2a1a2···am (1) The problem is at the round oﬀ When we do a calculation on a computer, we almost never get the right answer. 25 / 42
52. 52. Thus, we have some problems with that Floating point arithmetic It is not at all the same thing as working with real numbers. Representation x = (d1d2d3 · · · dn) × 2a1a2···am (1) The problem is at the round oﬀ When we do a calculation on a computer, we almost never get the right answer. 25 / 42
53. 53. Thus, we have some problems with that Floating point arithmetic It is not at all the same thing as working with real numbers. Representation x = (d1d2d3 · · · dn) × 2a1a2···am (1) The problem is at the round oﬀ When we do a calculation on a computer, we almost never get the right answer. 25 / 42
54. 54. Another Problems We would love the ﬂoating points to be represented uniformly The ﬂoating point numbers are distributed logarithmically which is quite diﬀerent from the even distribution of the rationals. Computations do not scale well 2 multiplications and 1 addition to compute the 2 × 2 determinant 12 multiplications and 5 additions to compute the 3 × 3 determinant 72 multiplications and 23 additions to 26 / 42
55. 55. Another Problems We would love the ﬂoating points to be represented uniformly The ﬂoating point numbers are distributed logarithmically which is quite diﬀerent from the even distribution of the rationals. Computations do not scale well 2 multiplications and 1 addition to compute the 2 × 2 determinant 12 multiplications and 5 additions to compute the 3 × 3 determinant 72 multiplications and 23 additions to 26 / 42
56. 56. Another Problems We would love the ﬂoating points to be represented uniformly The ﬂoating point numbers are distributed logarithmically which is quite diﬀerent from the even distribution of the rationals. Computations do not scale well 2 multiplications and 1 addition to compute the 2 × 2 determinant 12 multiplications and 5 additions to compute the 3 × 3 determinant 72 multiplications and 23 additions to 26 / 42
57. 57. Another Problems We would love the ﬂoating points to be represented uniformly The ﬂoating point numbers are distributed logarithmically which is quite diﬀerent from the even distribution of the rationals. Computations do not scale well 2 multiplications and 1 addition to compute the 2 × 2 determinant 12 multiplications and 5 additions to compute the 3 × 3 determinant 72 multiplications and 23 additions to 26 / 42
58. 58. Outline 1 Square Matrices Introduction The Inverse Solution to Ax = y Algorithm for the Inverse of a Matrix 2 Determinants Introduction Complexity Increases Reducing the Complexity Some Consequences of the deﬁnition Special Determinants 27 / 42
59. 59. Therefore How do we avoid to get us into problems? We need to deﬁne our determinant as a diﬀerent structure.... Deﬁnition The determinant of A is a real-valued function of the rows of A which we write as det (A) = det (r1, r2, ..., rn) 28 / 42
60. 60. Therefore How do we avoid to get us into problems? We need to deﬁne our determinant as a diﬀerent structure.... Deﬁnition The determinant of A is a real-valued function of the rows of A which we write as det (A) = det (r1, r2, ..., rn) 28 / 42
61. 61. Properties Multiplying a row by the constant c multiplies the determinant by c det (r1, r2, ..., cri, ..., rn) = cdet (r1, r2, ..., ri, ..., rn) If row i is the sum of the two row vectors x and y det (r1, r2, ..., x + y, ..., rn) =det (r1, r2, ..., x, ..., rn) + ... det (r1, r2, ..., y, ..., rn) Meaning The determinant is a linear function of each row. 29 / 42
62. 62. Properties Multiplying a row by the constant c multiplies the determinant by c det (r1, r2, ..., cri, ..., rn) = cdet (r1, r2, ..., ri, ..., rn) If row i is the sum of the two row vectors x and y det (r1, r2, ..., x + y, ..., rn) =det (r1, r2, ..., x, ..., rn) + ... det (r1, r2, ..., y, ..., rn) Meaning The determinant is a linear function of each row. 29 / 42
63. 63. Properties Multiplying a row by the constant c multiplies the determinant by c det (r1, r2, ..., cri, ..., rn) = cdet (r1, r2, ..., ri, ..., rn) If row i is the sum of the two row vectors x and y det (r1, r2, ..., x + y, ..., rn) =det (r1, r2, ..., x, ..., rn) + ... det (r1, r2, ..., y, ..., rn) Meaning The determinant is a linear function of each row. 29 / 42
64. 64. Further Interchanging any two rows of the matrix changes the sign of the determinant det (..., ri, ..., rj, ..., ...) = det (..., rj, ..., ri, ..., ...) Finally The determinant of any identity matrix is 1 30 / 42
65. 65. Further Interchanging any two rows of the matrix changes the sign of the determinant det (..., ri, ..., rj, ..., ...) = det (..., rj, ..., ri, ..., ...) Finally The determinant of any identity matrix is 1 30 / 42
66. 66. Outline 1 Square Matrices Introduction The Inverse Solution to Ax = y Algorithm for the Inverse of a Matrix 2 Determinants Introduction Complexity Increases Reducing the Complexity Some Consequences of the deﬁnition Special Determinants 31 / 42
67. 67. Some Properties Property 1 If A has a row of zeros, then det(A) = 0. Proof 1 if A = (..., 0, ...), also A = (..., c0, ...) 2 det (A) = c × det (A) for any c 3 Thus det (A) = 0 32 / 42
68. 68. Some Properties Property 1 If A has a row of zeros, then det(A) = 0. Proof 1 if A = (..., 0, ...), also A = (..., c0, ...) 2 det (A) = c × det (A) for any c 3 Thus det (A) = 0 32 / 42
69. 69. Some Properties Property 1 If A has a row of zeros, then det(A) = 0. Proof 1 if A = (..., 0, ...), also A = (..., c0, ...) 2 det (A) = c × det (A) for any c 3 Thus det (A) = 0 32 / 42
70. 70. Some Properties Property 1 If A has a row of zeros, then det(A) = 0. Proof 1 if A = (..., 0, ...), also A = (..., c0, ...) 2 det (A) = c × det (A) for any c 3 Thus det (A) = 0 32 / 42
71. 71. Next Property 2 If ri = rj , i = j, then det(A) = 0. Proof Quite easy (Hint sing being reversed). 33 / 42
72. 72. Next Property 2 If ri = rj , i = j, then det(A) = 0. Proof Quite easy (Hint sing being reversed). 33 / 42
73. 73. Finally Proposition 3 If B is obtained from A by replacing ri with ri + crj , then det(B) = det(A) Proof det (B) = det (..., ri + crj, ..., rj, ...) = det (..., ri, ..., rj, ...) + det (..., crj, ..., rj, ...) = det (A) + cdet (..., rj, ..., rj, ...) = det (A) + 0 34 / 42
74. 74. Finally Proposition 3 If B is obtained from A by replacing ri with ri + crj , then det(B) = det(A) Proof det (B) = det (..., ri + crj, ..., rj, ...) = det (..., ri, ..., rj, ...) + det (..., crj, ..., rj, ...) = det (A) + cdet (..., rj, ..., rj, ...) = det (A) + 0 34 / 42
75. 75. Finally Proposition 3 If B is obtained from A by replacing ri with ri + crj , then det(B) = det(A) Proof det (B) = det (..., ri + crj, ..., rj, ...) = det (..., ri, ..., rj, ...) + det (..., crj, ..., rj, ...) = det (A) + cdet (..., rj, ..., rj, ...) = det (A) + 0 34 / 42
76. 76. Finally Proposition 3 If B is obtained from A by replacing ri with ri + crj , then det(B) = det(A) Proof det (B) = det (..., ri + crj, ..., rj, ...) = det (..., ri, ..., rj, ...) + det (..., crj, ..., rj, ...) = det (A) + cdet (..., rj, ..., rj, ...) = det (A) + 0 34 / 42
77. 77. Finally Proposition 3 If B is obtained from A by replacing ri with ri + crj , then det(B) = det(A) Proof det (B) = det (..., ri + crj, ..., rj, ...) = det (..., ri, ..., rj, ...) + det (..., crj, ..., rj, ...) = det (A) + cdet (..., rj, ..., rj, ...) = det (A) + 0 34 / 42
78. 78. Outline 1 Square Matrices Introduction The Inverse Solution to Ax = y Algorithm for the Inverse of a Matrix 2 Determinants Introduction Complexity Increases Reducing the Complexity Some Consequences of the deﬁnition Special Determinants 35 / 42
79. 79. Therefore Theorem The determinant of an upper or lower triangular matrix is equal to the product of the entries on the main diagonal. Proof Suppose A is upper triangular and that none of the entries on the main diagonal is 0. This means all the entries beneath the main diagonal are zero. Using Proposition 3, we can convert it into a diagonal matrix. Then, by property 1 det (Adiag) = [ n i aii] det (I) = n i aii 36 / 42
80. 80. Therefore Theorem The determinant of an upper or lower triangular matrix is equal to the product of the entries on the main diagonal. Proof Suppose A is upper triangular and that none of the entries on the main diagonal is 0. This means all the entries beneath the main diagonal are zero. Using Proposition 3, we can convert it into a diagonal matrix. Then, by property 1 det (Adiag) = [ n i aii] det (I) = n i aii 36 / 42
81. 81. Therefore Theorem The determinant of an upper or lower triangular matrix is equal to the product of the entries on the main diagonal. Proof Suppose A is upper triangular and that none of the entries on the main diagonal is 0. This means all the entries beneath the main diagonal are zero. Using Proposition 3, we can convert it into a diagonal matrix. Then, by property 1 det (Adiag) = [ n i aii] det (I) = n i aii 36 / 42
82. 82. Therefore Theorem The determinant of an upper or lower triangular matrix is equal to the product of the entries on the main diagonal. Proof Suppose A is upper triangular and that none of the entries on the main diagonal is 0. This means all the entries beneath the main diagonal are zero. Using Proposition 3, we can convert it into a diagonal matrix. Then, by property 1 det (Adiag) = [ n i aii] det (I) = n i aii 36 / 42
83. 83. Therefore Theorem The determinant of an upper or lower triangular matrix is equal to the product of the entries on the main diagonal. Proof Suppose A is upper triangular and that none of the entries on the main diagonal is 0. This means all the entries beneath the main diagonal are zero. Using Proposition 3, we can convert it into a diagonal matrix. Then, by property 1 det (Adiag) = [ n i aii] det (I) = n i aii 36 / 42
84. 84. Remark Question This is the property we use to compute determinants!!! How? 37 / 42
85. 85. Example We have 2 1 3 −4 First, we have r1 = (2, 1) = 2 2, 1 2 Then det (A) = 2det 1 1 2 3 −4 38 / 42
86. 86. Example We have 2 1 3 −4 First, we have r1 = (2, 1) = 2 2, 1 2 Then det (A) = 2det 1 1 2 3 −4 38 / 42
87. 87. Example We have 2 1 3 −4 First, we have r1 = (2, 1) = 2 2, 1 2 Then det (A) = 2det 1 1 2 3 −4 38 / 42
88. 88. Further We have by proposition 3 det (A) = 2det 1 1 2 0 −11 2 Using Property 1 det (A) = 2 − 11 2 det 1 1 2 0 1 Therefore det (A) = −11 39 / 42
89. 89. Further We have by proposition 3 det (A) = 2det 1 1 2 0 −11 2 Using Property 1 det (A) = 2 − 11 2 det 1 1 2 0 1 Therefore det (A) = −11 39 / 42
90. 90. Further We have by proposition 3 det (A) = 2det 1 1 2 0 −11 2 Using Property 1 det (A) = 2 − 11 2 det 1 1 2 0 1 Therefore det (A) = −11 39 / 42
91. 91. Further Properties Property 4 The determinant of A is the same as that of its transpose AT . Proof Hint: we do an elementary row operation on A. Then, (EA)T = AT ET 40 / 42
92. 92. Further Properties Property 4 The determinant of A is the same as that of its transpose AT . Proof Hint: we do an elementary row operation on A. Then, (EA)T = AT ET 40 / 42
93. 93. Finally Property 5 If A and B are square matrices of the same size, then det (AB) = det (A) det (B) Therefore If A is invertible: det AA−1 = det (A) det A−1 = det (I) = 1 Thus det A−1 = 1 det (A) 41 / 42
94. 94. Finally Property 5 If A and B are square matrices of the same size, then det (AB) = det (A) det (B) Therefore If A is invertible: det AA−1 = det (A) det A−1 = det (I) = 1 Thus det A−1 = 1 det (A) 41 / 42
95. 95. Finally Property 5 If A and B are square matrices of the same size, then det (AB) = det (A) det (B) Therefore If A is invertible: det AA−1 = det (A) det A−1 = det (I) = 1 Thus det A−1 = 1 det (A) 41 / 42
96. 96. Finally Deﬁnition If the (square) matrix A is invertible, then A is said to be non-singular. Otherwise, A is singular. 42 / 42