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Lecture 6. Inverse of Matrix

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Lecture 6. Inverse of Matrix Lecture 6. Inverse of Matrix Recall that any linear system can be written as a matrix equation A~x = ~b. In one dimension case, i.e., A is 1 1, then £ Ax = b can be easily solved as b 1 1 x = = b = A¡ b provided that A = 0. A A 6 In this lecture, we intend to extend this simple method to matrix equations. De…nition 7.1. A square matrix An n is said to be invertible if there exists a unique £ matrix Cn n of the same size such that £ AC = CA = In. The matrix C is called the inverse of A, and is denoted by 1 C = A¡ 1 Suppose now An n is invertible and C = A¡ is its inverse matrix. Then the matrix equation £ A~x = ~b can be easily solved as follows. Left-multipling the matrix equation by the inverse matrix 1 C = A¡ , we have CA~x = C~b. 1 By de…nition, CA = A¡ A = In. It leads to ~ In~x = Cb, which is the same as 1 ~x = A¡ ~b. (1) We have just solve the matrix equation and obtained a unique solution (1). The above discussion is summarized as Theorem 7.1 Let A be an invertible matrix. Then matrix equation A~x = ~b has a unique solution 1 ~x = A¡ ~b. 1 1 Example 7.1 (a) Show A is invertible and A¡ = C, where 2 5 7 5 A = , C = . 3 7 ¡3 ¡2 ·¡ ¡ ¸ · ¸ (b) Solve 2x1 + 5x2 = 1 3x 7x = 4. ¡ 1 ¡ 2 (c) Show that the matrix 0 2 B = 0 0 · ¸ is NOT invertible. Solution: (a) Direct calculations lead to 2 5 7 5 1 0 AC = = = I 3 7 ¡3 ¡2 0 1 2 ·¡ ¡ ¸ · ¸ · ¸ 7 5 2 5 1 0 CA = = = I . ¡3 ¡2 3 7 0 1 2 · ¸ ·¡ ¡ ¸ · ¸ By de…nition, 1 C = A¡ . (b) The system has the coe¢cient matrix A, .i.e., the matrix equation is 1 A~x = . 4 · ¸ Therefore, According to Theorem 1, the solution is 1 ~x = A 1 ¡ 4 · ¸ 7 5 1 = ¡3 ¡2 4 · ¸ · ¸ 27 = . ¡11 · ¸ (c) By calculation, we …nd that 0 2 0 2 0 0 B2 = = . 0 0 0 0 0 0 · ¸ · ¸ · ¸ Now, suppose that B is invertible. Then there exists a matrix D such that BD = I2. 2 Multiplying the above equation by B from the left, we …nd B (BC) = BI2, B2C = B, which implies 0 = B, since B2 = 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 7.2 A 2 2 matrix £ a b A = c d · ¸ is invertible i¤ def det (A) = ad bc = 0. ¡ 6 When ad bc = 0, the inverse is ¡ 6 1 1 d b A¡ = ¡ . ad bc c a ¡ ·¡ ¸ 1 Example 7.2. (a) Find A¡ if 2 5 A = . 3 6 ·¡ ¡ ¸ (b) Solve 2x + 5y = 1 3x 6y = 2 ¡ ¡ Solution: (a) a = 2, b = 5, c = 3, d = 6. ad bc = 12 ( 15) = 3. So ¡ ¡ ¡ ¡ ¡ ¡ 1 1 d b A¡ = ¡ ad bc c a ¡ ·¡ ¸ 1 6 5 2 5/3 = ¡ ¡ = ¡ ¡ . 3 3 2 1 2/3 · ¸ · ¸ We may verify the above solution as follows: 2 5 2 5/3 1 0 = . 3 6 ¡1 ¡2/3 0 1 ·¡ ¡ ¸ · ¸ · ¸ (b) To solve the system, we write its matrix equation: A~x = ~b, where 2 5 1 A = , ~b = . 3 6 2 ·¡ ¡ ¸ · ¸ The solution is 2 5/3 1 16 ~x = A 1~b = = 3 . ¡ ¡1 ¡2/3 2 ¡7 · ¸ · ¸ · 3 ¸ 3 Properties if Invertible Matrix: ² Theorem 7.3 Suppose that A is an invertible square matrix. Then 1 1 1 1. A¡ is also invertible and (A¡ )¡ = A. T T 1 1 T 2. A is also invertible and A ¡ = (A¡ ) . 3. If B is another invertible¡mat¢rix of the same size, then AB is also invertible, and 1 1 1 (AB)¡ = B¡ A¡ . 4. The reduced Echelon of A is the identity matrix I of the same size, i.e., A I. ¡! 1 Proof. (1) By de…nition, A¡ if we can …nd a matrix C such that 1 1 A¡ C = C A¡ = I. The above is indeed true if C = A¡. ¢ ¡ ¢ 1 1 (2) Take transposes of all three sides of (A¡ ) A = A (A¡ ) = I, 1 T 1 T T T 1 T 1 T T A¡ A = A A¡ = I = A A¡ = A¡ A = I ) T T 1 T T ¡¡ ¢ ¢ =¡ ¡A C¢=¢ CA = I, wh¡ere C¢ = A¡¡ ¢ is the inverse of A . ) 1 1 (3) Let C = B¡ A¡ . Since ¡ ¢ 1 1 1 1 1 1 C (AB) = (CA) B = B¡ A¡ A B = B¡ A¡ A B = B¡ I B = B¡ B = I 1 1 1 1 1 1 (AB) C = A (BC) = ¡A¡ B B¡ ¢A¡¢ = ¡A B¡B¡ A¢¢¡ = ¡A IA¡¢ = A¡ A¡ ¢= I. it follows from the de…niti¡on t¡hat this ¢C¢ is the¡¡inverse¢of A¢B, i.e.¡, ¢ 1 1 1 (AB)¡ = C = B¡ A¡ . (4) Since A~x = ~0 has the only solution 1 ~x = A¡ ~0 = ~0, there is no non-trivial 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 non-zero entry in each row and thus has to be the identity matrix. We next develop an algorithm to …nd inverse matrices. De…nition 7.2 A matrix is called an elementary matrix if it is obtained by performing one single elementary row operation on an identity matrix. 4 Example 7.3 Let us look at 3 3 elementary matrices for corresponding row operations. £ A type (1) elementary matrix E1 is obtained by performing one type (1) row operation. For instance, 1 0 0 1 0 0 0 1 0 R + ¸R R ¸ 1 0 = E . 2 3 2 1 ! 2 2 3 1 0 0 1 ¡¡¡¡¡¡¡¡¡¡¡! 0 0 1 We call E1 is the elemen4tary mat5rix associated wit4h the row5operation R2 + ¸R1 R2. ! For any matrix A, performing the above row operation is the same as left multiplying by E1. For instance, we see that 1 3 1 1 3 1 ¡ ¡ 2 1 0 R2 + ( 2) R1 R2 0 5 2 . 2 3 ¡ ! 2 ¡ 3 4 0 1 ¡¡¡¡¡¡¡¡¡¡¡¡¡¡! 4 0 1 4 5 4 5 On the other hand, left multiplication by E1 with ¸= 2 yields ¡ 1 3 1 ¡ E1A = E1 2 1 0 24 0 1 3 14 0 0 51 3 1 1 3 1 = 2 1 0 2 1 ¡0 = 0 5 ¡2 . 2¡0 0 13 24 0 1 3 24 ¡0 1 3 4 5 4 5 4 5 A type (2) elementary matrix E2 is obtained by performing one type (2) row operation. For instance, 1 0 0 0 1 0 0 1 0 R2 R1 1 0 0 = E2. 2 3 ! 2 3 0 0 1 ¡¡¡¡¡¡! 0 0 1 For any matrix A, performi4ng the ab5ove row op4eration is5the same as left multiplying by E2. For instance, we see that 1 3 1 2 1 0 ¡ 2 1 0 R2 R1 1 3 1 2 3 ! 2 ¡ 3 4 0 1 ¡¡¡¡¡¡! 4 0 1 4 5 4 5 On the other hand, left multiplication by E2 yield 1 3 1 ¡ E2A = E2 2 1 0 24 0 1 3 0 41 0 153 1 2 1 0 = 1 0 0 2 1 ¡0 = 1 3 1 . 20 0 13 24 0 1 3 24 0 ¡1 3 4 5 4 5 4 5 5 A type (3) elementary matrix E3 is obtained by performing one type (3) row operation. For instance, 1 0 0 1 0 0 0 1 0 ¸R R 0 1 0 = E . 2 3 3 ! 3 2 3 3 0 0 1 ¡¡¡¡¡¡¡! 0 0 ¸ 4 5 4 5 Similarly, performing the above row operation is the same as left multiplying by E2. For instance, we see that 1 3 1 1 3 1 ¡ 1 ¡ 2 1 0 R3 R3 2 1 0 . 24 0 1 3 4 ! 21 0 1/43 ¡¡¡¡¡¡¡! 4 5 4 5 On the other hand, left multiplication by E3 with ¸= 1/4 leads to 1 3 1 ¡ E3A = E3 2 1 0 24 0 1 3 1 40 0 51 3 1 1 3 1 = 0 1 0 2 1 ¡0 = 2 1 ¡0 . 20 0 1/43 24 0 1 3 21 0 1/43 4 5 4 5 4 5 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. Justi…cation of LU Decomposition Algorithm ² Recall in Lecture 2, we introduced LU Decomposition as follows: Any matrix A may be decomposed as the product Am n = Lm mUm 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) 2. 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, i.e., E E E A = U.
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