3.2, 3.3 Inverting Matrices P. Danziger Properties of Transpose Transpose has higher precedence than multiplica- tion and addition, so T T T T AB = A B and A + B = A + B As opposed to the bracketed expressions (AB)T and (A + B)T Example 1 1 2 1 1 0 1 Let A = ! and B = !. 2 5 2 1 1 0 Find ABT , and (AB)T . T 1 1 1 2 1 1 0 1 1 2 1 0 1 ABT = ! ! = ! 0 1 2 5 2 1 1 0 2 5 2 B C @ 1 0 A 2 3 = ! 4 7 Whereas (AB)T is undefined. 1 3.2, 3.3 Inverting Matrices P. Danziger Theorem 2 (Properties of Transpose) Given ma- trices A and B so that the operations can be pre- formed 1. (AT )T = A 2. (A + B)T = AT + BT and (A B)T = AT BT − − 3. (kA)T = kAT 4. (AB)T = BT AT 2 3.2, 3.3 Inverting Matrices P. Danziger Matrix Algebra Theorem 3 (Algebraic Properties of Matrix Multiplication) 1. (k + `)A = kA + `A (Distributivity of scalar multiplication I) 2. k(A + B) = kA + kB (Distributivity of scalar multiplication II) 3. A(B + C) = AB + AC (Distributivity of matrix multiplication) 4. A(BC) = (AB)C (Associativity of matrix mul- tiplication) 5. A + B = B + A (Commutativity of matrix ad- dition) 6. (A + B) + C = A + (B + C) (Associativity of matrix addition) 7. k(AB) = A(kB) (Commutativity of Scalar Mul- tiplication) 3 3.2, 3.3 Inverting Matrices P. Danziger The matrix 0 is the identity of matrix addition. That is, given a matrix A, A + 0 = 0 + A = A: Further 0A = A0 = 0, where 0 is the appropriately sized 0 matrix. Note that it is possible to have two non-zero ma- trices which multiply to 0. Example 4 1 1 1 1 1 1 1 1 0 0 ! ! = ! = ! 1− 1 1 1 1− + 1 1− + 1 0 0 − − − The matrix I is the identity of matrix multiplica- tion. That is, given an m n matrix A, × AIn = ImA = A Theorem 5 If R is in reduced row echelon form then either R = I, or R has a row of zeros. 4 3.2, 3.3 Inverting Matrices P. Danziger Theorem 6 (Power Laws) For any square ma- trix A, ArAs = Ar+s and (Ar)s = Ars Example 7 2 0 0 1 4 0 0 1 2 1. 0 1 00 1 1 1 0 1 = 1 0 1 B C B 2 2 0 C BB 2 2 0 C C @ A @@ A A 2. Find A6, where 1 0 A = ! 1 1 6 2 4 2 2 2 A = A A = A A . 1 0 Now A2 = !, so 2 1 2 2 1 0 1 0 1 0 1 0 A2 A2 = ! ! = ! ! 2 1 2 1 2 1 3 1 1 0 = ! 5 1 5 3.2, 3.3 Inverting Matrices P. Danziger Inverse of a matrix Given a square matrix A, the inverse of A, denoted 1 A− , is defined to be the matrix such that 1 1 AA− = A− A = I Note that inverses are only defined for square ma- trices Note Not all matrices have inverses. If A has an inverse, it is called invertible. If A is not invertible it is called singular. 6 3.2, 3.3 Inverting Matrices P. Danziger Example 8 1 2 5 2 1:A = ! A 1 = ! 2 5 − 2− 1 − 1 2 5 2 1 0 Check: ! ! = ! 2 5 2− 1 0 1 − 1 2 2:A = ! Has no inverse 2 4 1 1 1 3 1 1 0 1 1 0 − − 1 3:A = 1 2 1 A− = 1 1 0 B C B − C @ 1 1 2 A @ 1 0 1 A − 1 1 1 3 1 1 1 0 0 0 1 0 1 0 1 Check: 1 2 1 1− 1− 0 = 0 1 0 B C B − C B C @ 1 1 2 A @ 1 0 1 A @ 0 0 1 A − 1 2 1 0 1 4:A = 2 1 3 Has no inverse B C @ 3 3 4 A 7 3.2, 3.3 Inverting Matrices P. Danziger Inverses of 2 2 Matrices × Given a 2 2 matrix × a b A = ! c d A is invertible if and only if ad bc = 0 and − 6 1 1 d b A = − ! − ad bc c a − − The quantity ad bc is called the determinant of − the matrix and is written det(A), or A . j j Example 9 2 1 2 1 3 2 1 A = ! A 1 = ! = − 3 ! 3 3 − 3 3− 1 1 1 − − −3 1 3 2 1 2 1 3 0 Check: − ! ! = − ! = I 3 3 1 3 3 −3 0 3 − − − 8 3.2, 3.3 Inverting Matrices P. Danziger Algebra of Invertibility Theorem 10 Given an invertible matrix A: 1 1 1. (A− )− = A, n 1 1 n n 2. (A )− = (A− ) = A− , 1 1 1 3. (kA)− = kA− , T 1 1 T 4. (A )− = (A− ) , 9 3.2, 3.3 Inverting Matrices P. Danziger Theorem 11 Given two invertible matrices A and B 1 1 1 (AB)− = B− A− : Proof: Let A and B be invertible matricies and 1 1 let C = AB, so C− = (AB)− . Consider C = AB. 1 Multiply both sides on the left by A− : 1 1 A− C = A− AB = B: 1 Multiply both sides on the left by B− . 1 1 1 B− A− C = B− B = I: 1 1 So, B− A− is the matrix you need to multiply C by to get the identity. Thus, by the definition of inverse 1 1 1 1 B− A− = C− = (AB)− : 10 3.2, 3.3 Inverting Matrices P. Danziger A Method for Inverses Given a square matrix A and a vector b Rn, 2 consider the equation Ax = b This represents a system of equations with coeffi- cient matrix A. 1 Multiply both sides by A− on the left, to get 1 1 A− Ax = A− b: 1 But A− A = In and Ix = x, so we have 1 x = A− b: Note that we have a unique solution. The as- sumption that A is invertible is equaivalent to the assumption that Ax = b has unique solution. 11 3.2, 3.3 Inverting Matrices P. Danziger During the course of Gauss-Jordan elimination on the augmented matrix (A b) we reduce A I and j ! b A 1b, so (A b) I A 1b . ! − j ! j − If we instead augment A with I, row reducing will 1 produce (hopefully) I on the left and A− on the right, so (A I) I A 1 . j ! j − The Method: 1. Augment A with I 2. Use Gauss-Jordan to obtain (I A 1). j − 3. If I does not appear on the left, A is not in- vertable. 1 Otherwise, A− is given on the right. 12 3.2, 3.3 Inverting Matrices P. Danziger Example 12 1 1. Find A− , where 1 2 3 0 1 A = 2 5 5 B C @ 3 5 8 A 13 3.2, 3.3 Inverting Matrices P. Danziger Augment with I and row reduce: 1 2 3 1 0 0 0 1 R2 R2 2R1 2 5 5 0 1 0 ! − B C R3 R3 3R1 @ 3 5 8 0 0 1 A ! − 1 2 3 1 0 0 0 1 0 1 1 2 1 0 R3 R3 + R2 B − − C ! @ 0 1 1 3 0 1 A − − − 1 2 3 1 0 0 0 1 1 0 1 1 2 1 0 R3 2R3 B − − C ! − @ 0 0 2 5 1 1 A − − 1 2 3 1 0 0 0 1 R1 R1 3R3 0 1 1 2 1 0 ! − B − − C R2 R2 + R3 @ 0 0 1 5=2 1=2 1=2 A ! − − 1 2 0 13=2 3=2 3=2 0 − 1 0 1 0 1=2 1=2 1=2 R1 R1 2R2 B − C ! − @ 0 0 1 5=2 1=2 1=2 A − − 1 0 0 15=2 1=2 5=2 0 1 0 1 0 −1=2 1=2 1=2 B − C @ 0 0 1 5=2 1=2 1=2 A − − 14 3.2, 3.3 Inverting Matrices P. Danziger So 15 1 5 1 1 0 − 1 A− = 1 1 1 2 B − C @ 5 1 1 A − − To check inverse multiply together: 1 2 3 15 1 5 1 0 1 1 0 − 1 AA− = 2 5 5 2 1 1 1 B C B − C @ 3 5 8 A @ 5 1 1 A 2 0 0 − − 1 0 1 = 2 0 2 0 = I B C @ 0 0 2 A 2. Solve Ax = b in the case where b = (2; 2; 4)T . 15 1 5 2 1 1 0 − 1 0 1 x = A− b = 2 1 1 1 2 B − C B C @ 5 1 1 A @ 4 A 18 9− − 1 0 − 1 0 − 1 = 2 0 = 0 B C B C @ 4 A @ 2 A 15 3.2, 3.3 Inverting Matrices P. Danziger 3. Solve Ax = b in the case where b = (2; 0; 2)T . 15 1 5 2 1 1 0 − 1 0 1 x = A− b = 2 1 1 1 0 B − C B C @ 5 1 1 A @ 2 A 20 9− − 1 0 − 1 0 − 1 = 2 0 = 0 B C B C @ 8 A @ 4 A 4.