Math 191 Applied Linear Algebra Lecture 8: Inverse of a Matrix Stephen Billups University of Colorado at Denver Math 191Applied Linear Algebra p.1/0
Announcements We will not make it to section. tonight, so... You do not need to turn in problems from sec.. for Tuesday. The quiz on Tuesday will not cover problems from section. Reminder Exam 1 will be Sept.. Math 191Applied Linear Algebra p./0
Outline Discuss Quiz and HWK Finish Sec..1 Matrix Operations. Sec.., Inverse of a Matrix. Math 191Applied Linear Algebra p./0
Section.1: Review: Matrix Notation. A ij or a ij = entry in i-th row, j-th column of A. [a ij ] = matrix, whose ij-th entry is a ij. Matrix Operations. Matrix addition and scalar multiplication just like vector addition and scalar multiplication. A + B = [A ij + B ij ], ca = [ca ij ]. Matrix multiplication defined so that (AB)x = A(Bx). Two ways to calculate AB: i) by definition, ii) Row-column rule. Compatible Dimensions. Math 191Applied Linear Algebra p./0
Tonight Practice multiplying matrices. Properties of matrix multiplication. Powers of a matrix Transpose of a matrix. Math 191Applied Linear Algebra p.5/0
Practice Let A = 1. (AB) 11 = 1 0 0 1 h 5 and B = i 1 1 0 5. 5 =.. (AB) 1 =.. AB = 5.. BA =. Math 191Applied Linear Algebra p./0
Properties of Matrix Multiplication THEOREM Let A be m n and let B and C have sizes for which the indicated sums and products are defined. a. A (BC) = (AB)C (associative law of multiplication) b. A (B + C) = AB + AC (left - distributive law) c. (B + C) A = BA + CA (right-distributive law) d. r(ab) = (ra)b = A(rB) for any scalar r e. I m A = A = AI n (identity for matrix multiplication) Math 191Applied Linear Algebra p./0
WARNINGS Properties above are analogous to properties of real numbers. But NOT ALL real number properties correspond to matrix properties. 1. It is not the case that AB always equal BA. (see Example, page 11). Even if AB = AC, then B may not equal C. (see Exercise 10, page 11). It is possible for AB = 0 even if A 0 and B 0. (see Exercise 1, page 11) Math 191Applied Linear Algebra p.8/0
Powers of A A k = A A }{{} k EXAMPLE: [ ] [ ] [ 1 = [ ] [ ] 1 0 = 1 0 [ = ] [ 1 0 ] 1 0 1 8 ] Math 191Applied Linear Algebra p.9/0
Transpose If A is m n, the transpose of A is the n m matrix, denoted by A T, whose columns are formed from the corresponding rows of A. EXAMPLE: A = 1 5 8 9 8 5 = A T = 1 8 5 9 5 8 Math 191Applied Linear Algebra p.10/0
EXAMPLE: Let A = [ 1 0 0 1 A T B T and B T A T. ], B = 1 0 1. Compute AB, (AB) T, Solution: [ AB = 1 0 0 1 ] 1 0 1 (AB) T = = [ ] [ ] Math 191Applied Linear Algebra p.11/0
Solution, (cont.) A T B T = B T A T = [ 1 0 0 1 [ 1 0 1 1 0 1 ] 1 0 0 1 ] = = 10 0 1 [ ] Math 191Applied Linear Algebra p.1/0
THEOREM Let A and B denote matrices whose sizes are appropriate for the following sums and products. a. ( A T ) T = A (I.e., the transpose of A T is A) b. (A + B) T = A T + B T c. For any scalar r, (ra) T = ra T d. (AB) T = B T A T (I.e. the transpose of a product of matrices equals the product of their transposes in reverse order. ) Math 191Applied Linear Algebra p.1/0
EXAMPLE: Prove that (ABC) T =. Solution: By Theorem d, (ABC) T = ((AB) C) T = C T ( ) T = C T ( ) =. Math 191Applied Linear Algebra p.1/0
Sec.: The Inverse of a Matrix The inverse of a real number a is denoted by a 1. For example, 1 = 1/ and 1 = 1 = 1 An n n matrix A is said to be invertible if there is an n n matrix C satisfying CA = AC = I n where I n is the n n identity matrix. We call C the inverse of A. FACT If A is invertible, then the inverse is unique. Proof: Assume B and C are both inverses of A. Then B = BI = B ( ) = ( ) = I = C. So the inverse is unique since any two inverses coincide. Math 191Applied Linear Algebra p.15/0
Notation and Terminology The inverse of A is usually denoted by A 1. We have AA 1 = A 1 A = I n Not all n n matrices are invertible. A matrix that is not invertible is sometimes called a singular matrix. An invertible matrix is called a nonsingular matrix. Memory Trick: There is only a single real number that is not invertible (namely zero), so non-invertible numbers are singular. Interesting Fact: Almost all matrices are invertible. Math 191Applied Linear Algebra p.1/0
Theorem Let A = a b c d 5. If ad bc 0, then A is invertible and A 1 = 1 ad bc d b c a 5. If ad bc = 0, then A is not invertible. Assume A is any invertible matrix and we wish to solve Ax = b. Then Ax = b and so Ix = or x =. Suppose w is also a solution to Ax = b. Then Aw = b and Aw = b which means w =A 1 b. So, w =A 1 b, which is in fact the same solution. Math 191Applied Linear Algebra p.1/0
Theorem 5 We have proved the following result: Theorem 5 If A is an invertible n n matrix, then for each b in R n, the equation Ax = b has the unique solution x = A 1 b. EXAMPLE: A = Solution: A 1 = [ 5 1 1 15 x =A 1 b = [ [ ] 5 5, b = [ 1 ] ] [ ] ] =. [ = 5 [ ] ]. Math 191Applied Linear Algebra p.18/0
Theorem Suppose A and B are invertible. Then the following results hold: a. A 1 is invertible and `A 1 1 = A (i.e. A is the inverse of A 1 ). b. AB is invertible and (AB) 1 = B 1 A 1 c. A T is invertible and `A T 1 `A 1 T = Partial proof of part b: (AB) `B 1 A 1 = A ( ) A 1 = A ( ) A 1 = =. Similarly, one can show that `B 1 A 1 (AB) = I. Math 191Applied Linear Algebra p.19/0
More Matrices... Theorem, part b can be generalized to three or more invertible matrices: (ABC) 1 =. Proof: (ABC) 1 = ((AB) C) 1 = C 1 ( ) 1 = C 1 ( ) =. Math 191Applied Linear Algebra p.0/0
Elementary Matrices To calculate inverses of larger than matrices, we first need to look at elementary matrices. Definiton: An elementary matrix is one that is obtained by performing a single elementary row operation on an identity matrix. EX: Let E 1 = 0 0 5, E = 0 0 1 0 0 1 and E are elementary matrices. Why? 5, E = 0 1 5. E 1, E, Math 191Applied Linear Algebra p.1/0
Multiplying by an Elementary Matrix Let A = E 1 A = a b c d e f g h i 0 0 0 0 1. a b c d e f g h i = Describe the effect multiplying by E 1 had on A. Math 191Applied Linear Algebra p./0
Effect of Multiplying by Elementary Matrix If an elementary row operation is performed on an m n matrix A, the resulting matrix can be written as EA, where the m m matrix E is created by performing the same row operations on I m. Examples: E A = 0 0 1 5 a b c d e f g h i 5 = a b c g h i d e f 5 E A = 0 1 5 a b c d e f g h i 5 = a b c d e f a + g b + h c + i 5 Math 191Applied Linear Algebra p./0
Inverses of Elementary Matrices Elementary matrices are invertible because row operations are reversible. To determine the inverse of an elementary matrix E, determine the elementary row operation needed to transform E back into I and apply this operation to I to find the inverse. For example, E = E = 0 1 0 0 1 E 1 = E 1 = Math 191Applied Linear Algebra p./0
Inverting A with Elementary Matrices Example: Let A = E 1 A = E (E 1 A) = 0 0 0 0 1 E (E E 1 A) = 1 0 5 0 0 1 0 1 1 0 5 5 5. Then 5 = 0 1 0 1 5 = 0 1 5 = 5 0 1 0 0 1 5 5 Math 191Applied Linear Algebra p.5/0
So Then multiplying on the right by A 1, we get E E E 1 A = I. So E E E 1 A = I. E E E 1 I = A 1 E E E 1 I = 0 1 5 0 0 1 5 0 0 0 0 1 5 I = 0 1 5 0 1 5 = 0 1 5 0 0 1 0 0 5 = 0 0 1 0 5 Math 191Applied Linear Algebra p./0
The elementary row operations that row reduce A to I n are the same elementary row operations that transform I n into A 1. Theorem An n n matrix A is invertible if and only if A is row equivalent to I n, and in this case, any sequence of elementary row operations that reduces A to I n will also transform I n to A 1. Math 191Applied Linear Algebra p./0
Algorithm for finding A 1 Place A and I side-by-side to form an augmented matrix [A I]. Then perform row operations on this matrix (which will produce identical operations on A and I). So by Theorem :[A I] will row reduce to ˆI A 1 (or A is not invertible). EXAMPLE: Find the inverse of A = 0 0 0 1 5, if it exists. Solution:[A I] = 0 0 1 0 0 0 1 5 1 0 0 0 0 1 0 0 1 1 0 5 So A 1 = 1 0 0 0 0 1 1 0 5 Math 191Applied Linear Algebra p.8/0
Bad Example Find the inverse of A = Solution: [A I] = 1 1 5 1 8 5. 1 1 5 1 8 0 0 1 5 1 0 1 1 1 0 0 1 0 1 5 1 0 1 1 1 0 0 0 0 1 1 5 What went wrong? A is. Math 191Applied Linear Algebra p.9/0
Order of multiplication is important! EXAMPLE Suppose A,B,C, and D are invertible n n matrices and A = B(D I n )C. Solve for D in terms of A, B, C and D. Solution: A = B(D I n )C D I n = B 1 AC 1 D I n + = B 1 AC 1 + D = Math 191Applied Linear Algebra p.0/0