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 b If a 0 then if b 0 there exists an a infinite number of solutions If b 0 there is no solution If a 0 we have a single unique solution x 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 0 then A x b has no unique soltuion det A c c 0 then A x b has a unique solution 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 a 11 a 12 a 1N a jN a j2 a jN a N1 a N2 a NN and a 11 a 12 a 1N ca jN ca j2 ca jN a N1 a N2 a NN 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 1 1 A 2 1 3 1 1 3 6 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 1 1 1 A 2 3 1 3 6 1 1 3 5 10 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 B RN 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 xi c xj 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 N N N E i1i2 i N a i1 1a i2 2 a i N N i1 1 i 2 1 1 3 5 16 i N 1 Where 0 if any two of i1 i 2 i N are equal E i1i 2 i N 1 if i1 i 2 i N is an even parity permutation 1 if i i i is an odd parity permutation 1 2 N 1 3 5 17 By even parity permutation we mean the following Since E i1i 2 i N 0 if any two of the set i1 i 2 i N we know that the ordered set i1 i 2 i N if E i1i 2 i N is to be non zero must be related to the ordered set 1 2 3 N by some shuffling of the order For example consider i1 3 i2 2 i3 4 i 1 so 1 3 5 18 i1 i2 i3 i4 3 2 4 1 We want to perform a sequence of interchanges to put it in the order 1 2 3 4 Interchange 1 3 2 4 1 3 2 1 4 Interchange 2 3 2 1 4 3 1 2 4 Interchange 3 3 1 2 4 1 3 2 4 Interchange 4 1 3 2 4 1 2 3 4 1 3 5 19 1 3 4 19 So we have put 3 2 4 …
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