T Theorem: AT = A; (AB)T = BT AT : Matrix Transpose Transpose of matrix A is denoted AT , and formed by setting each column in AT from corresponding row in A. 1 4 2 2 1 Let A = ; B = : 2 3 −1 −1 0 2 2 −1 3 1 2 Then AT = ; BT = 2 −1 : 4 3 4 5 1 0 1 / 50 Matrix Transpose Transpose of matrix A is denoted AT , and formed by setting each column in AT from corresponding row in A. 1 4 2 2 1 Let A = ; B = : 2 3 −1 −1 0 2 2 −1 3 1 2 Then AT = ; BT = 2 −1 : 4 3 4 5 1 0 T Theorem: AT = A; (AB)T = BT AT : 1 / 50 I d × (`1) − b × (`2) =) (a d − c b) x1 = d β1 − b β2: I a × (`2) − c × (`1) =) (a d − c b) x2 = a β2 − c β1: Assume a d − c b 6= 0. 1 d β − b β 1 d −b β x = 1 2 = 1 a d − c b a β2 − c β1 a d − c b −c a β2 1 d −b = b def= A−1 b: a d − c b −c a Determinant: det (A) = a d − c b. So A−1 exists () det (A) 6= 0. x2.2 Inverse of Matrix (I) a b x β 2 × 2 system of equations A x = b: 1 = 1 : c d x2 β2 In scalar form: a x1 + b x2 = β1; (`1) c x1 + d x2 = β2: (`2) 2 / 50 Assume a d − c b 6= 0. 1 d β − b β 1 d −b β x = 1 2 = 1 a d − c b a β2 − c β1 a d − c b −c a β2 1 d −b = b def= A−1 b: a d − c b −c a Determinant: det (A) = a d − c b. So A−1 exists () det (A) 6= 0. x2.2 Inverse of Matrix (I) a b x β 2 × 2 system of equations A x = b: 1 = 1 : c d x2 β2 In scalar form: a x1 + b x2 = β1; (`1) c x1 + d x2 = β2: (`2) I d × (`1) − b × (`2) =) (a d − c b) x1 = d β1 − b β2: I a × (`2) − c × (`1) =) (a d − c b) x2 = a β2 − c β1: 2 / 50 1 d −b = b def= A−1 b: a d − c b −c a Determinant: det (A) = a d − c b. So A−1 exists () det (A) 6= 0. x2.2 Inverse of Matrix (I) a b x β 2 × 2 system of equations A x = b: 1 = 1 : c d x2 β2 In scalar form: a x1 + b x2 = β1; (`1) c x1 + d x2 = β2: (`2) I d × (`1) − b × (`2) =) (a d − c b) x1 = d β1 − b β2: I a × (`2) − c × (`1) =) (a d − c b) x2 = a β2 − c β1: Assume a d − c b 6= 0. 1 d β − b β 1 d −b β x = 1 2 = 1 a d − c b a β2 − c β1 a d − c b −c a β2 2 / 50 Determinant: det (A) = a d − c b. So A−1 exists () det (A) 6= 0. x2.2 Inverse of Matrix (I) a b x β 2 × 2 system of equations A x = b: 1 = 1 : c d x2 β2 In scalar form: a x1 + b x2 = β1; (`1) c x1 + d x2 = β2: (`2) I d × (`1) − b × (`2) =) (a d − c b) x1 = d β1 − b β2: I a × (`2) − c × (`1) =) (a d − c b) x2 = a β2 − c β1: Assume a d − c b 6= 0. 1 d β − b β 1 d −b β x = 1 2 = 1 a d − c b a β2 − c β1 a d − c b −c a β2 1 d −b = b def= A−1 b: a d − c b −c a 2 / 50 x2.2 Inverse of Matrix (I) a b x β 2 × 2 system of equations A x = b: 1 = 1 : c d x2 β2 In scalar form: a x1 + b x2 = β1; (`1) c x1 + d x2 = β2: (`2) I d × (`1) − b × (`2) =) (a d − c b) x1 = d β1 − b β2: I a × (`2) − c × (`1) =) (a d − c b) x2 = a β2 − c β1: Assume a d − c b 6= 0. 1 d β − b β 1 d −b β x = 1 2 = 1 a d − c b a β2 − c β1 a d − c b −c a β2 1 d −b = b def= A−1 b: a d − c b −c a Determinant: det (A) = a d − c b. So A−1 exists () det (A) 6= 0. 2 / 50 Definition: Matrix A 2 Rn×n is invertible if there exists matrix 2 1 0 ··· 0 3 6 0 1 ··· 0 7 C 2 Rn×n so that CA = I = AC; with I = 6 7 identity. 6 . .. 7 4 . 5 0 0 ··· 1 C is called inverse of A, denoted as A−1. Inverse of Matrix (I) a b 1 d −b 2 × 2 matrix A = ; A−1 = : c d a d − c b −c a 1 0 Let I = be the identity matrix. Then 0 1 AI = IA = A; I x = x for all A and x. 1 d −b a b A−1 A = = I = AA−1: a d − c b −c a c d 3 / 50 Inverse of Matrix (I) a b 1 d −b 2 × 2 matrix A = ; A−1 = : c d a d − c b −c a 1 0 Let I = be the identity matrix. Then 0 1 AI = IA = A; I x = x for all A and x. 1 d −b a b A−1 A = = I = AA−1: a d − c b −c a c d Definition: Matrix A 2 Rn×n is invertible if there exists matrix 2 1 0 ··· 0 3 6 0 1 ··· 0 7 C 2 Rn×n so that CA = I = AC; with I = 6 7 identity. 6 . .. 7 4 . 5 0 0 ··· 1 C is called inverse of A, denoted as A−1. 3 / 50 2 0 1 2 3 2 −9 14 −3 3 1 EX: A = 1 0 3 ; then (later show) A−1 = −4 8 −2 : 4 5 2 4 5 4 −3 8 3 −4 1 2 1 3 2 1 3 2 −13 3 −1 A x = 4 −1 5 has solution x = A 4 −1 5 = 4 −7 5 : 1 1 4 Inverse of Matrix (II) 3 2 1 4 −2 EX: Let A = ; then A−1 = : 1 4 10 −1 3 3 2 1 1 1 8 x = has solution x = A−1 = : 1 4 −2 −2 10 −7 4 / 50 Inverse of Matrix (II) 3 2 1 4 −2 EX: Let A = ; then A−1 = : 1 4 10 −1 3 3 2 1 1 1 8 x = has solution x = A−1 = : 1 4 −2 −2 10 −7 2 0 1 2 3 2 −9 14 −3 3 1 EX: A = 1 0 3 ; then (later show) A−1 = −4 8 −2 : 4 5 2 4 5 4 −3 8 3 −4 1 2 1 3 2 1 3 2 −13 3 −1 A x = 4 −1 5 has solution x = A 4 −1 5 = 4 −7 5 : 1 1 4 4 / 50 T −1 −1T I A = A −1 −1 −1 I (AB) = B A Proof: B−1 A−1 (AB) = B−1 A−1 A B = B−1 B = I Similarly (AB) B−1 A−1 = I : Therefore (AB)−1 = B−1 A−1 . QED Inverse Matrix (III) Theorem: Let A; B 2 Rn×n be invertible −1−1 I A = A 5 / 50 −1 −1 −1 I (AB) = B A Proof: B−1 A−1 (AB) = B−1 A−1 A B = B−1 B = I Similarly (AB) B−1 A−1 = I : Therefore (AB)−1 = B−1 A−1 . QED Inverse Matrix (III) Theorem: Let A; B 2 Rn×n be invertible −1−1 I A = A T −1 −1T I A = A 5 / 50 Inverse Matrix (III) Theorem: Let A; B 2 Rn×n be invertible −1−1 I A = A T −1 −1T I A = A −1 −1 −1 I (AB) = B A Proof: B−1 A−1 (AB) = B−1 A−1 A B = B−1 B = I Similarly (AB) B−1 A−1 = I : Therefore (AB)−1 = B−1 A−1 . QED 5 / 50 interchange I E1;3 obtained by row1 () row3 on I . Elementary Operation =) Elementary Matrix (E1;3) 2 3 a11 a12 a13 Let A = 4 a21 a22 a23 5 a31 a32 a33 interchange I row1 () row3 2 3 2 3 2 3 a31 a32 a33 0 0 1 a11 a12 a13 A =) 4 a21 a22 a23 5 = 4 0 1 0 5 4 a21 a22 a23 5 a11 a12 a13 1 0 0 a31 a32 a33 def = E1;3 A 6 / 50 Elementary Operation =) Elementary Matrix (E1;3) 2 3 a11 a12 a13 Let A = 4 a21 a22 a23 5 a31 a32 a33 interchange I row1 () row3 2 3 2 3 2 3 a31 a32 a33 0 0 1 a11 a12 a13 A =) 4 a21 a22 a23 5 = 4 0 1 0 5 4 a21 a22 a23 5 a11 a12 a13 1 0 0 a31 a32 a33 def = E1;3 A interchange I E1;3 obtained by row1 () row3 on I . 6 / 50 I Eb2 obtained by row3 − 2 row1 =) row3 on I . Elementary Operation =) Elementary Matrix (Eb2) 2 3 a11 a12 a13 Let A = 4 a21 a22 a23 5 a31 a32 a33 I row3 − 2 row1 =) row3 2 3 a11 a12 a13 A =) 4 a21 a22 a23 5 a31 − 2 a11 a32 − 2 a12 a33 − 2 a13 2 3 2 3 1 0 0 a11 a12 a13 def = 4 0 1 0 5 4 a21 a22 a23 5 = Eb2 A −2 0 1 a31 a32 a33 7 / 50 Elementary Operation =) Elementary Matrix (Eb2) 2 3 a11 a12 a13 Let A = 4 a21 a22 a23 5 a31 a32 a33 I row3 − 2 row1 =) row3 2 3 a11 a12 a13 A =) 4 a21 a22 a23 5 a31 − 2 a11 a32 − 2 a12 a33 − 2 a13 2 3 2 3 1 0 0 a11 a12 a13 def = 4 0 1 0 5 4 a21 a22 a23 5 = Eb2 A −2 0 1 a31 a32 a33 I Eb2 obtained by row3 − 2 row1 =) row3 on I .