Write a report on a daring bank robbery.
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Determinant of a Square Matrix
A determinant could be thought of as a function from Fn� n to F: Let A = (aij) be an n� n matrix. We define its determinant, written as
,
by
.
where Sn is the group of all n! permutations on the symbols{1,2,3,4,...,n} and sgn (s ) for a permutation s Î Sn is defined as follows: Let s written as a function be
.
Let Ni (1 � i < n) denote the number of indices j > i, for which s (j) < s (i), and let N(s ) = S 1� i<n Ni. Ni is called the number of inversions in the permutation s corresponding to the index i, and N is called the total number of inversions in the permutation s . Finally, we define sgn (s ) by the relation sgn (s) � (-1)N(s ) . Thus, sgn (s ) is +1, or -1 according as N(s ) is even, or odd, and accordingly, we call the permutation itself to be even, or odd permutation.
The above notion of determinant remains useful in many more situations, e.g., when the entries of A are from a commutative ring G with identity (e.g, the ring F [x] of polynomials over a field F). In the sequel, however, unless otherwise specified we restrict to the case of matrices over a field.
Theorem. If A is a square block upper triangular matrix over a ring G, in which the diagonal blocks are square, then the determinant of A equals the product of the determinants of its diagonal blocks.
Proof: Suppose the n� n A is partitioned into m2 blocks Apq, where Apq is square if p = q and Apq = 0 if p > q, 1 � p, q � m:
Let Sp denote the set of row or column numbers pertaining to the block App and let c(Sp) denote the number of elements in Sp. Consider the right hand side of
.
If s Î Sn is such that for some i, the (i,s (i))-th entry of A lies in the (p,q)-th block with q > p, then for some j Î Sk with k � p, s -1(j) Î Sr for an r > p, so that A being block upper triangular the contribution of the permutation s to |A| is zero. This is because if {j : s (j) Î È t� p St} Ì È t� p St, then i Î Sp and s (i) Î Sq is contradicted. Hence the only surviving terms in |A| correspond to those s Î Sn which may be written as a product s = s 1s 2�s m, where s p(i) = i if i Ï Sp, i.e., s p permutes withing Sp only, 1 � p � m. Using sgn s = sgn s 1 sgn s 2 ... sgn s m, we thus have
.
PROBLEMS
1. What is the determinant of a 1� 1 matrix?
2. If A and B are 2� 2 matrices, verify that |AB|= |A||B|.
3. If A is 2� 3 and B is 3� 2, prove that |BA| = 0. Construct two pairs of A and B of sizes 2� 3 and 3� 2, respectively, such that in one case |AB| = 1 and in the other |AB| = 0.
4. Use the definition of the determinant to derive the formulae
(i) ,
(ii) )
(iii) If A = (aij ) is n� n and is triangular (i.e., either upper, or lower triangular), then det A = a11 a22 a33�ann.
Properties of the Determinant Function
Let A be a square matrix with entries from F . We describe below some properties of the determinant function, mainly related with certain row operations on A:
(1) If B is obtained from A by multiplying some row of it by a scalar c, then det (B) = c det (A).
(2) If B is obtained by interchanging some two rows of A, then det (B) = - det (A).
(3) If B is obtained by adding c-times a row of A to a different row, the det (B) = det (A).
(4) The determinant of the identity matrix I is 1.
(5) If a row of A is zero, det (A) = 0.
(6) If two rows of A are identical, det (A) = 0.
(7) |A| = |A� |.
Of these properties: (1) follows from the definition, since in each product P 1� i� n ais (i) precisely one element from each row participates, and it does so linearly. For (2), if i-th and (i+1)-th rows are interchanged, the respective permutations s and s � , say, look like
,
.
where the omitted entries in the two permutations are the same. For these two permutations the inversions Nj for j < i and for j > i+1 are the same, while Ni + Ni+1 for the two differ by one. Hence sgn (s � ) = - sgn s and (2) follows when two consecutive rows are interchanged. Since, however, an interchange of any two rows, i-th and j-th, say, can be effected by 2|i-j| - 1, an odd number of consecutive row interchanges, the result follows in general. Next we prove (6): In view of (2), via row interchanges, w.l.g. (without loss of generality) let these rows be the 1-st and the 2-nd ones. Then
.
Explanation:
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