ERG2013_Lecture_9

# Is a common factor we can factor it out and because

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Unformatted text preview: have established the multiplicative property, the forward part of (*) is cntl-C-cntl-V of the 2x2 case. Suppose that A is an invertible 3x3 matrix. Let the inverse be A-1. Apply the determinant function to both sides of 1 = 1 1 we obtain det( ∙ ) = 1. Using the multiplicative property, det( )det( ) = 1. If det(A) is zero, the left hand side is 0. This yields a contradiction to the above equation. Therefore, det(A) is non-zero. Another formula for 3x3 determinant 17. Definitions on permutations. In general, for a positive integer n, a permutation of 1,2,…,n is represented by a bijective function, say , from {1,2,…,n} to itself. 1 ↓ (1) 2 … ↓ … (2) … n ↓ ( n) There are n! possible bijective functions from {1,2,…,n} to itself. The collection of all n! of them is denoted by Sn (called the symmetric group in mathematics). We use the short-hand notation 8 1 (1) to represent the permutation 2 … (2) … n (1) → ( ). For example The permutation 1 → 2, 2 → 3 and 3 → 1 is represented by 12 23 3 . 1 A transposition is a permutation which exchanges two items, while keeping the other fixed. For example, the swapping of 1 and 2 is represented by 12 21 3 . 3 The sign (or signature) of the permutation represented by is defined as sgn( ) = (−1) where N is the number of transpositions required to transform (1,2,…,n) to ((1), (2), …, (n)). This is equal to +1 if N is even and -1 if N is odd. Example for n=3. For the permutation specified by (1)=1, (2)=2, (3)=3, (the identity permutation), we have N = 0, and hence sgn() = 1. If is a permutation specified by (1)=2, (2)=1, (3)=3, it is the exchange of 1 and 2, and sgn() = -1. If (1)=3, (2)=1, (3)=2, we have N = 2, and sgn() = 1. 18. Using the notation introduced above, we have another formula for 3x3 determinant: For 3x3 matrix A= [aij], det( ) = sgn( ) () () () ∈ There are six terms in the above summation, each term corresponds to a bijective function in S3. nxn determinant in general 19. Given an nxn matrix M = [mij,], we can either define the determinant of M recursively by Laplace expansion: Expansion on the ith row (i=1,2,…,n). (−1) + (−1) 9 + ⋯ + (−1) Expansion on the jth column (j=1,2,…,n). (−1) + (−1) + ⋯ + (− 1) or by generalizing the formula in (18), sgn( ) ( )… () () ∈ with the summation running over all n! permutations in Sn. The first approach is adopted in Kreyszig ’s book, but the second is also common (see http://en.wikipedia.org/wiki/Determinant). The definition via expansion has the drawback that we need to check that all expansions yield the same number. (This is done in Appendix 4 in Kreysizig ’s book.) The definition in terms of permutations does not have the problem of well-definedness, but is less friendly to numerical calculations. The main result (*) holds for nxn matrix in general, the proof goes along the same pathway as in the 3x3 case. The properties in (13) are also satisfied by nxn matrices in general. 20. Practical method for evaluating determinant. Computing an nxn determinant directly from the two definitions in (19) require n! steps, which increases exponentially as n increases. A more practical approach is to apply elementary row operations to transform the matrix to an upper (or lower) matrix, and calculate the product of the diagonal elements. This method is based on the properties Subtracting a scalar multiple of a row from another row does not change the determinant. The determinant of an upper (or lower) triangular matrix is equal to the product of the diagonal entries. 21. (Caution) When we multiply a matrix by a constant, we multiply all entries in the matrix. When we multiply a determinant by a constant, we multiply only the entries in a row or in a column. For constant c, 1 3 1 3 2 = 4 3 2 2 1 = = 4 34 3 10 2 , 4 2 = 4 3 2 1 = 4 3 2 . 4...
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## This document was uploaded on 09/16/2013.

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