Characteristic #1:1.3.5 The Determinant Of A Square Matrix In section 1.3.4 we have seen that the condition of existence and uniqueness for solutions to A x = b involves whether KA = 0, i.e. only w = 0 has the property that Aw = 0. To use this result, we need a method by which we can examine the elements of A to determine if KA = 0. For N = 1, this is simple. For the single equation Ax = b (1.3.5-1) If a0, we have a single (unique) solution x = ≠ab. If a = 0, then if b = 0, there exists an infinite number of solutions. If b≠0, there is no solution. For N > 1, we want a similar rule. Given an N x N real matrix A, we want a rule to calculate a scalar called the determinant, det(A), such that =≠==solution unique a has bxA then 0,cc,soltuion unique no has bxA then 0,det(A) (1.3.5-2) Since this determinant is to be used to determine whether a system Ax = b will have a unique solution, we can identify some characteristics that a suitable functional form of det(A) must possess.Characteristic #1: If we multiply any equation in our system, say the jth aj1x1 + aj2x2 + … + ajNxN = bj (1.3.5-3) by a scalar c 0, we obtain an equation ≠ caj1x1 + caj2x2 + … + cajNxN = cbj (1.3.5-4) As this new equation is completely equivalent to the first one, the determinants of the following 2 matrices should either both be zero or both be non-zero. NNN2N1jNj2jN1N1211NNN2N1jNj2jN1N1211a ... a a: : : ca ... ca ca: : : a ... a a and a ... a a: : : a ... a a: : : a ... a a (1.3.5-5) Moreover, if c = 0, then even if det(A) ≠0, the determinant of the 2nd matrix in (1.3.5-5) should be zero. We note that we can satisfy these requirements if our determinant function has the property that the determinant of the 2nd matrix is c * det(A). Characteristic #2. The existence of a solution to Ax = b does not depend upon the order in which we write the equations. Therefore, we must be able to exchange any 2 rows in a matrix without affecting whether the determinant is zero or non-zero. One way to satisfy this is if A’ is the matrix obtained from A by interchanging any 2 rows, then our determinant should satisfy det(A’) = ±det(A). (1.3.5-6)Characteristic #3: We can write the following 3 equations x + y + z = 4 2x + y + 3z = 7 3x + y + 6z = 2 (1.3.5-7) in matrix form with the labels x1 = x x2 = y x3 = z (1.3.5-8) to yield the matrix A = 6 1 33 1 21 1 1 We could just as well label the unknowns by x1 = x x2 = z x3 = y (1.3.5-9) In which case we obtain a matrix A’ = (1.3.5-10) 1 6 31 3 21 1 1 Obviously, such an interchange of columns does nothing to affect the existence and uniqueness of solutions. Therefore either det(A) and det(A’) are both zero, or det(A) and det(A’) are both non-zero. One way to satisfy this is to make det(A’) = ±det(A). (1.3.5-11)Characteristic #4: We can select any 2 equations, say # i and #j, ai1x1 + ai2x2 + … + aiNxN = bi aj1x1 + aj2x2+ … + ajNxN = bj (1.3.5-11) and replace them by the following 2, with c≠0 ai1x1 + ai2x2 + … + aiNxN = bi (cai1 + aj1)x1 + (cai2 + aj2)x2 + … + (caiN + ajN)xN = (cbi + bj) (1.3.5-12) If A is the original matrix of the system, and A’ is the new matrix obtained after making this replacement, than either det(A) and det(A’) are both zero or they are both non-zero. Characteristic #5: If C = AB, the viewing Cx = b as RN RN RN B A x Bx b C We see that det(C) 0 if and only if both det(A) ≠≠0 (so a unique Bx exists) and if det(B) 0. ≠ One way to ensure this is if det(C) = det(A) x det(B) (1.3.5-13) Characteristic #6: If any 2 rows of A are identical, the equations that they represent are dependent. We therefore do not have a unique solution, and must have det(A) = 0. Similarly if all elements of a given row are zero, we have the equation 0 = bj, which is inconsistent if bj≠ 0. Therefore, we must have det(A) = 0.Characteristic #7: If any 2 rows of A are equal, say columns #i and #j, then for all M∈[1,N] aMi = aMj. We can therefore write each equation as aM1x1 + aM2x2 + … + aMixi + … + aMjxj + … + aMNxN = aM1x1 + aM2x2 + … + aMi(xi + xj) + … + aMNxN (1.3.5-14) Since xi and xj only appear together in this system of equations as the sum xi + xj, we could make the following change for any c that would not affect Ax, xi Å xi + c xj Å xj – c (1.3.5-15) Therefore, we must have det(A) = 0. Similarly, if any column of A contains all zeros, det(A) = 0. We now have identified a number of properties that any functional form for det(A) must have to be a proper measure of existence and uniqueness for Ax = b. We now propose a functional form for the determinant, and show that it does satisfy these characteristics.We define the determinant of the N x N matrix A as det(A) = ∑∑ (1.3.5-16) ∑== =N1iN1iN1iN,i,2i,1i...iii12 NN21N21...aaaE.... Where −+npermutatioparity oddan is )i...,i,(i if 1,npermutatioparity even an is )i...,i,(i if 1,equal are }i...,i,{i of any two if 0,EN21N21N21...iiiN21 (1.3.5-17) By “even parity permutation” we mean the following. Since E = 0 if any two of the set {i }, we know that the ordered set ( i ), if is to be non-zero, must be related to the ordered set (1, 2, 3, …, N) by some shuffling of the order. N21...iiiN21...iiiEN21i...,i,N21i...,i, For example, consider i1 = 3, i2 = 2, i3 = 4, i = 1, so (i1, i2, i3, i4) = (3, 2, 4, 1) (1.3.5-18) We want to perform a sequence of interchanges to put it in the order (1, 2, 3, 4). Interchange #1, (3, 2, 4, 1) Æ (3, 2, 1, 4) (1.3.5-19) Interchange #2, (3, 2, 1, 4) Æ (3, 1, 2, 4) Interchange #3, (3, 1, 2, 4) Æ (1, 3, 2, 4) Interchange #4, (1, 3, 2, 4) Æ (1, 2, 3, 4) (1.3.4-19) So we have put (3, 2, 4, 1) into order (1, 2, 3, 4) with four interchanges.Note that we could do the same thing with only 2 interchanges: (3, 2, 4, 1) Æ (1, 2, 4, 3) Æ (1, 2, 3, 4) (1.3.5-20) or less efficiently, with six (3, 2, 4, 1) Æ (4, 2, 3, 1) Æ (2, 4, 3, 1) Æ ( 1, 4, 3, 2) Æ (1, 4, 2, 3) Æ (1, 4, 3, 2) Æ (1, …