Lecture 6. Inverse of Matrix


 Bethanie McDonald
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1 Lecture 6 Inverse of Matrix Recall that any linear system can be written as a matrix equation In one dimension case, ie, A is 1 1, then can be easily solved as A x b Ax b x b A 1 A b A 1 b provided that A 6 0 In this lecture, we intend to extend this simple method to matrix equations De nition 71 A square matrix A n n is said to be invertible if there exists a unique matrix C n n of the same size such that AC CA I n The matrix C is called the inverse of A, and is denoted by C A 1 Suppose now A n n is invertible and C A 1 is its inverse matrix Then the matrix equation A x b can be easily solved as follows Leftmultipling the matrix equation by the inverse matrix C A 1, we have CA x C b By de nition, CA A 1 A I n It leads to which is the same as I n x C b, x A 1 b (1) We have just solve the matrix equation and obtained a unique solution (1) discussion is summarized as Theorem 71 Let A be an invertible matrix Then matrix equation A x b The above has a unique solution x A 1 b 1
2 Example 71 (a) Show A is invertible and A 1 C, where A, C 7 (b) Solve (c) Show that the matrix x 1 + 5x 1 x 1 7x 4 B is NOT invertible Solution: (a) Direct calculations lead to AC I CA I By de nition, C A 1 (b) The system has the coe cient matrix A, ie, the matrix equation is 1 A x 4 Therefore, According to Theorem 1, the solution is 1 x A (c) By calculation, we nd that B Now, suppose that B is invertible Then there exists a matrix D such that BD I
3 Multiplying the above equation by B from the left, we nd which implies B (BC) BI, B C B, 0 B, since B 0 This is obviously a contradiction Therefore, the assumption at the beginning "suppose that B is invertible" is false, and consequently, B is NOT invertible Theorem 7 A matrix a b A c d is invertible i When ad bc 6 0, the inverse is Example 7 (a) Find A 1 if (b) Solve det (A) def ad bc 6 0 A 1 1 ad bc A d b c a 5 6 x + 5y 1 x 6y Solution: (a) a, b 5, c, d 6 ad bc 1 ( 15) So A 1 1 d b ad bc c a / 1 / We may verify the above solution as follows: 5 5/ 6 1 / (b) To solve the system, we write its matrix equation: A x b, where 5 1 A, 6 b The solution is x A 1 5/ 1 b 1 / 16 7
4 ² Properties if Invertible Matrix: Theorem 7 Suppose that A is an invertible square matrix Then 1 A 1 is also invertible and (A 1 ) 1 A A T is also invertible and A T 1 (A 1 ) T If B is another invertible matrix of the same size, then AB is also invertible, and (AB) 1 B 1 A 1 4 The reduced Echelon of A is the identity matrix I of the same size, ie, A! I Proof (1) By de nition, A 1 if we can nd a matrix C such that A 1 C C A 1 I The above is indeed true if C A () Take transposes of all three sides of (A 1 ) A A (A 1 ) I, A 1 A T A A 1 T I T ) A T A 1 T A 1 T A T I () Let C B 1 A 1 Since ) A T C CA T I, where C A 1 T is the inverse of A T C (AB) (CA) B B 1 A 1 A B B 1 A 1 A B B 1 I B B 1 B I (AB) C A (BC) A B B 1 A 1 A BB 1 A 1 A IA 1 AA 1 I it follows from the de nition that this C is the inverse of AB, ie, (4) Since A x 0 has the only solution (AB) 1 C B 1 A 1 x A 1 0 0, there is no nontrivial solution Consequently, A has no free variable All columns are pivot columns Since A is a square matrix, this means that r (A) number of columns number of rows Therefore, the reduced Echelon form of A has a nonzero entry in each row and thus has to be the identity matrix We next develop an algorithm to nd inverse matrices De nition 7 A matrix is called an elementary matrix if it is obtained by performing one single elementary row operation on an identity matrix 4
5 Example 7 Let us look at elementary matrices for corresponding row operations A type (1) elementary matrix E 1 is obtained by performing one type (1) row operation For instance, R + λr 1! R λ 1 05 E! We call E 1 is the elementary matrix associated with the row operation R + λr 1! R For any matrix A, performing the above row operation is the same as left multiplying by E 1 For instance, we see that R + ( ) R 1! R ! On the other hand, left multiplication by E 1 with λ yields E 1 A E A type () elementary matrix E is obtained by performing one type () row operation For instance, R! R E! For any matrix A, performing the above row operation is the same as left multiplying by E For instance, we see that R! R ! On the other hand, left multiplication by E yield E A E
6 A type () elementary matrix E is obtained by performing one type () row operation For instance, λr! R E! λ Similarly, performing the above row operation is the same as left multiplying by E For instance, we see that R! R ! 1 0 1/4 On the other hand, left multiplication by E with λ 1/4 leads to 1 1 E A E / /4 Conclusion Performing an elementary row operation on an identity matrix produces an elementary matrix corresponding to that elementary row operation Any elementary row operation is equivalent to left multiplying by the corresponding elementary matrix ² Justi cation of LU Decomposition Algorithm Recall in Lecture, we introduced LU Decomposition as follows: Any matrix A may be decomposed as the product A m n L m m U m n of a lower triangle matrix L and a upper triangle matrix U by the following algorithm: 1 Reduce A to an echelon form from U by a sequence of type one row operations (row replacement row operation) Place entries in L such that the same sequence of row operations reduces L to the identity matrix We can now justify the algorithm We know that reducing A to U by a sequence of row operations is equivalent to multiplying A from the left by a sequence of elementary matrices, ie, E p E E 1 A U It follows that A (E p E E 1 ) 1 U 6
7 This means that or equivalently L (E p E E 1 ) 1 E p E E 1 L I This shows that the same sequence of row operations reduces L to I ² Inverse matrix Algorithm: Suppose now A is invertible Then, its reduced Echelon form is the identity matrix In other words, by a series of successive row operations, the matrix A is reduced to I Since each row operation is equivalent to left multiplication by an elementary matrix, this means that there exist elementary matrices E 1, E,, E k, such that By de nition, this implies (E k E E 1 ) A I A 1 E k E E 1 E k E E 1 (I) The very last equation says that the exact same row operations that reduce A to the identity I, in the same time, also transforms the identity matrix I to A 1 We create a n (n) matrix by adding the identity to the left of A : [A I] 6 4 a 11 a 1 a 1n a 1 a a n a m1 a mn We then perform row operations till the rst n columns form the identity matrix When the rst n columns form the identity matrix, the remaining columns form the inverse A 1 : Example 74 Find A 1 if [A I] elementary row operations! A I A 1 7
8 Solution: [A, I] R R 1! R R + R 1! R! R 1 R! R R 6R! R! R 1 + 7R! R R R! R! A We may verify our answer directly: ² Extended Inverse Matrix Algorithm Using the same idea, the inverse matrix algorithm can be generalized to nd A 1 B Let A n n be a n n square invertible matrix and B be a n m matrix We combine the columns of A and columns of B to create a n (n + m) matrix [A B], and perform row operations till the rst n columns of the augmented matrix [A B] form the identity matrix of dimension n When the rst n columns of the augmented matrix [A B] form the identity matrix, the last m columns forms A 1 B In short, ² Characterization of Invertible Matrices [A B]! I A 1 B We summarize this entire lecture by the following theorem Theorem 74 (Invertible matrix theorem) Let A n n be a square matrix The following statements are equivalent: 1 The matrix A is invertible A T is invertible 8
9 AB AC ) B C (ie, cancellation law holds for A 4 A is leftinvertible, ie, there exists C such that CA I 5 A is rightinvertible, ie, there exists D such that AD I 6 The reduced Echelon form of A is identity matrix I 7 A is rowequivalent to identity matrix I 8 A has n pivot positions 9 A x 0 has only trivial solution x 0 10 Column vectors of A are linearly independent 11 Column vectors of A span R n 1 Row vectors of A span R n 1 A x b is consistent for all b in R n 14 The columns of A form a basis for Col (A) 15 Col (A) R n 16 Rank(A) n 17 Rank(Null (A)) 0 Proof (Very brie y) (1) () () : because A T 1 (A 1 ) T (1) ) () : AB AC ) A (B C) 0 ) A 1 A (B C) 0 ) (B C) 0 (1) ) (4) and (1) ) (5) : obviously; (1) ) (5) : E p E E 1 A I ) E p E E 1 A 1 (6) () (7) : A» I (7) () (8) : I has n pivots (8) ) (9) : no free variable (9) ) (10) : any linear relation x is a solution of A x 0 (8) ) (11) : A x b is always consistent since by (10), the last column in the augmented matrix is nonpivot () ) (1) : columns of A T are exactly the rows of A (11) () (1) : restatements (11) () (15) : restatements (16) () (17) : dimension theorem 9
10 ² Homework #6 1 Let A 1, 5 1 b 1 1, b 1, 5 b, 5 b 4 1 (a) Find A 1, and use it to solve A x 1 b 1, A x b, A x b, A x 4 b 4 (b) Consider the augmented matrix ha i b 1 b b b Perform row operations to reduced the the augmented matrix to a matrix whose rst two columns are identity When the rst two columns form identity matrix, nd the last four columns (c) Are the answers from part (a) and (b) above the same? Can you explain why? Use Inverse Matrix Algorithm to nd the inverse of (a) A (b) B Given A , B Find (AB) 1 and (BA) 1 4 Find A 1 B if A , B For each statement below, determine whether it is true or false If it is false, provide a counterexample If true, explain why (a) Suppose A is a n n matrix, and B and C are n m matrices If AB AC, then B C 10
11 (b) let A and B be n n matrices Suppose that AB is invertible Then both A and B are invertible (c) (AB) 1 A 1 B 1 (d) If a square matrix A has an echelon form whose diagonal entries are all pivot, then A is invertible (e) If A is a square matrix and if A x b has a unique solution for some b, then A is invertible (f) Let A be a n n matrix, and B a n m matrix If AB 0, then B 0 11
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