Matrices: 2.3 The Inverse of Matrices


 Anna Powell
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1 September 4
2 Goals Define inverse of a matrix. Point out that not every matrix A has an inverse. Discuss uniqueness of inverse of a matrix A. Discuss methods of computing inverses, particularly by row operations. Discuss properties of inverses. Apply them to solve systems of linear equations.
3 s : Nonexistance Definition: Let A be a square matrix A (of size n n). A is said to be invertible (or nonsingular) if there exists a matrix B such that AB = BA = I n where I n is the identity matrix of order n. Subsequently, we will see that such a B is unique (if exists), which will be (is) called the inverse of A. Note we assumed that A is a square matrix. We will see, not all square matrices have an inverse.
4 s : Nonexistance Theorem. Suppose A is an invetible matrix. Then, its inverse is unique. This unique inverse is denoted by A 1. Proof. Since A is invertible, it has at least one inverse. Suppose it has two inverses, B and C. By definition AB = BA = I n = AC = CA. So, B = BI n = B(AC) = (BA)C = I n C = C. So, B = C. The proof is complete. Reading Assignment: 1, 2 from 2.3.
5 Inverse of a given Matrix s : Nonexistance Let A = [ Then AB = BA = I 2. So, A 1 = B. ] [ 1 1, B = ]
6 Inverse by solving Preview s : Nonexistance Let A = and let A 1 = a x u b y v c z w Then AA 1 = a x u b y v c z w =
7 So, Preview a+b x +y u +v a+c x +z u +w b +c y +z v +w s : Nonexistance = This gives three systems of linear equations a+b = 1 a+c = 0 b +c = 0 We solve them as before: a =.5 b =.5 c =.5 x +y = 0 x +z = 1 y +z = 0 x =.5 y =.5 z = u +v = 0 u +w = 0 v +w = 1 u =.5 v =.5 w =.5
8 s : Nonexistance So, A 1 = a x u b y v c z w = It is obvious AA 1 = I n. We should also check A 1 A = I n, which we skip.
9 An Algorithm to find Inverse s : Nonexistance In the above example, we solved three systems of linear equations to find the inverse. An algothrithm to do the same by GaussJordan Elimination is as follows: let A be a matrix of size n n. Let I be the identity matrix of order n Form the n 2n matrix [A I] by adjoining I to A. By row operations,try to reduce [A I] to the form [I B]. If is possible then A 1 = B. If not, then A is not invertible. Check (ideally) that AB = BA = I. (Subsequently, we will see that this step is not necessary.) Reading Assignment: Exercise 3, 4 from 2.3
10 s : Nonexistance Computing Inverse using GaussJordan We will use the above algorithm to compute inverse of A = We adjoin the identity matrix I 3 to A, and get Now we apply row operations.
11 s : Nonexistance subtract 3 times first row from second and add first row to third:
12 s : Nonexistance Subtract 2 times the second row from the first and add 2 times the second row to the last:
13 s : Nonexistance Add 4 times the last row to the first and subtract 3 times the last row from the second: This is in the form [I 3 B]. So, A 1 = B =
14 An example that has no inverse s : Nonexistance Let A = We use GaussJordan Elimination: adjoin the identity matrix I 3 to A:
15 Switch the last and first row: s : Nonexistance
16 Divide the first row by 2: s : Nonexistance
17 s : Nonexistance Subtract 4 times the first row from second and subtract 3 times the first from third: Divide second row by 8:
18 s : Nonexistance Add second row to first and subtract 5 times second row to the last: The first half of this matrix does not reduce to the identity I 3. So, A does not have an inverse.
19 s : Nonexistance Determinant and Inverse of 2 2 Matrices Let A = [ a b c d ] Define determinant of A as det(a) = ad bc. Check directly that [ ] A 1 1 d b = if ad bc 0. ad bc c a In next chapter is devoted to determinant of matrices of higher order. It also generalizes this formula for inverse.
20 s : Nonexistance Let A,B be two invertible matrices (of size n n), c 0 is a scalar and k is a positive integer. Then, (A 1 ) 1 = A. ( ) A k 1 = (A 1 ) k. (ca) 1 = 1 c A 1 (A T ) 1 = (A 1 ) T (AB) 1 = B 1 A 1
21 Proof. Preview s : Nonexistance In each case, we need to verify the definition of inverse. I will only prove the last one and leave the rest as exercises. We have (AB) ( B 1 A 1) = A(BB 1 )A 1 = A(I)A 1 = AA 1 = I and similarly (B 1 A 1 )AB = I. So, the last statement is proved.
22 s : Nonexistance Cancellation Recall, in general, for matrices, AC = BC does not necessarily imples A = B. (Please review the example in my notes on 2.2.) But invertible matrices has the cancellation property: Let C be an invertible matrix. Then, AC = BC = A = B. CA = CB = A = B.
23 Proof. Preview s : Nonexistance Suppose AC = BC. On the right side each of this equation, multiply by C 1. Then we have, (AC)C 1 = (BC)C 1 = A(CC 1 ) = B(CC 1 ) A(I) = B(I) = A = B. So, the first statement is established. Similiarly, prove the second statement.
24 Theorem. Suppose Preview s : Nonexistance Ax = b Systems of Linear Equations Proof. If A is invertible, then x = A 1 b. Ax = b A 1 Ax = A 1 b I n x = A 1 b x = A 1 b The proof is complete. Reading Assignment: 8 from 2.3 of the textbook.
25 Exercise 16 Exercise 16. Compute inverse of A = Solution: Augment I 2 to A. We have [A I 3 ] =
26 Continued Subtract 3 times the third row from first: Add 5 times first row to second; then subtract 3 times first row from third:
27 Continued Interchange Second and third row: Add third row to second row:
28 Continued Add second row to the first and then add 4 times the second row to the third: Multiply third row by 1:
29 Continued Add third row to the first: This has the form [B I] So, A 1 =
30 Exercise 44. Exercise 44. A 1 = , B 1 = (a) Compute (AB) 1 (b) Compute (A T ) 1 (c) Compute (2A) 1
31 Solution Solution: (a) (AB) 1 = B 1 A 1 = =
32 Solution Solution: (b) (A T ) 1 = (A 1 ) T = T =
33 Solution Solution: (c) (2A) 1 = 1 2 A 1 = =
34 Exercise 47b Use inverse of matrices to solve x 1 +2x 2 +x 3 = 1 x 1 +2x 2 x 3 = 3 x 1 2x 2 +x 3 = 3 Solution: In matrix form, the equation is x x 2 = 3 notationslly it is : Ax = b x 3 3
35 Solution: In matrix notation solution is x = A 1 b, if A 1 exists. First, we find the inverse co the coefficient matrix A = Augment I 3 to A : [A I 3 ] =
36 Solution: Subtract first row from second; then first row from third: Interchange second and third rows:
37 Solution: Divide second row by 4, then divide third row by 2; Subtract 2 times the second row from first:
38 Solution: Subtract third row from first: This has the form [I B] So, A 1 =
39 Solution: So, the solution of the system is: So, x 1 = 0,x 2 = 1,x 3 = 1. x 1 x 2 x = x = A 1 b = = 0 1 1
40 See the homework site
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