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1 DETERMINANTS 1 Systems of two equations in two unknowns A system of two equations in two unknowns has the form a 11 x 1 + a 12 x 2 = b 1 a 21 x 1 + a 22 x 2 = b 2 This can be written more concisely in matrix notation [ ] [ ] [ ] a11 a 12 x1 b1 = a 21 a 22 or simply Ax = b Let s try to work out a general formula for the solution to such a system To solve for x 1, you need to eliminate x 2 Multiplying the first equation by a 22 and the second by a 12, and subtracting gives x 2 b 2 (a 11 a 22 a 12 a 21 )x 1 = a 22 b 1 a 12 b 2 If the coefficient of x 1 on the left is non-zero, this gives x 1 = a 22b 1 a 12 b 2 a 11 a 22 a 12 a 21 A similar method can be used to solve for x 2 To eliminate x 1, multiply the first equation by a 21 and the second by a 11 and subtract, obtaining (a 11 a 22 a 12 a 21 )x 2 = a 11 b 2 a 21 b 1 If the coefficient of x 2 on the left is non-zero, you get x 2 = a 11b 2 a 21 b 1 a 11 a 22 a 12 a 21 Notice that the number we had to divide by to solve for either x 1 or x 2, namely a 11 a 22 a 12 a 21, was the same This number is called the erminant of the matrix [ a11 a A = 12 a 21 a 22 It is denoted by either A or A Roughly speaking, it is the ] number you must divide by to solve the system Ax = b Notice that there is a theorem implicit in the above calculations As long as the erminant A is non-zero, the above calculations can be carried out, leading to a unique solution Theorem 1 The 2 2 system Ax = b has a unique solution if A 0, and therefore the coefficient matrix A is invertible The proof of the following complementary result is left as an exercise Theorem 2 If A = 0 then A is not invertible, so the system Ax = b either has no solutions or it has infinitely many solutions Our goal in the following sections is to develop an analogous theory of erminants for square matrices of arbitrary size 1

2 DETERMINANTS 2 2 Permutations 21 What is a permutation? A permutation of the numbers {1, 2,, n} is an arrangement of those numbers in a particular order There are two permutations of the numbers {1, 2}, namely, (1, 2) and (2, 1) There are six permutations of the numbers {1, 2, 3}, namely, (1, 2, 3), (1, 3, 2), (2, 1, 3), (2, 3, 1), (3, 1, 2), and (3, 2, 1) There are 24 permutations of the numbers {1, 2, 3, 4}, which you may list if you care to In general, the number of permutations of the numbers {1, 2,, n} is n! 22 Parity A inversion in a permutation is an occurrence of a larger number before a smaller one Thus the permutation (3, 1, 2) contains two inversions, since 3 precedes the smaller numbers 1 and 2 The permutation (3, 2, 1) contains three inversions, since 3 precedes 1 and 2, and 2 precedes 1 A permutation is called even if it contains an even number of inversions, and odd if it contains an odd number of inversions Thus the permutations (1, 2, 3), (2, 3, 1), and (3, 1, 2) are even, while the permutations (1, 3, 2), (3, 1, 2), and (2, 1, 3) are odd The sign of a permutation J = (j 1, j 2,, j n ), denoted sgn(j), is 1 if J is even and 1 if J is odd Thus sgn(1, 2, 3) = sgn(2, 3, 1) = sgn(3, 1, 2) = 1 and sgn(1, 3, 2) = sgn(3, 2, 1) = sgn(2, 1, 3) = 1 What happens to the parity of a permutation if you interchange two entries? For example, the permutation (2, 3, 1) is even If you switch the first and third entry, you get (1, 3, 2), which is odd, so the parity reverses This always happens Theorem 3 If the permutation J is obtained from J by interchanging two entries, then J and J have opposite parities: sgn(j ) = sgn(j) Proof: Consider first the case when the two entries that are exchanged are adjacent, say j k and j k+1 There are two cases to consider If j k < j k+1, then interchanging them will increase the number of of inversions by 1, thus reversing the parity On the other hand, if j k > j k+1, then interchanging them will decrease the number of inversions by 1, again reversing the parity Now consider the case of interchanging two entries that are separated by l other entries This can be accomplished by repeatedly interchanging adjacent entries Let s count the number of interchanges of adjacent entries that are required First, take one of the two entries to be exchanged, and move it past each of the l inbetween entries, so that it becomes adjacent to the other entry that is to be moved This requires exactly l interchanges of adjacent entries Now move the other entry to be moved to its final destination, by exchanging it past the one you ve already moved, and then past the l intermediate entries This requires exactly l + 1 interchanges of adjacent entries Thus the total number of interchanges of adjacent entries is l + (l + 1) = 2l + 1 Since this number is odd, the number of parity changes is odd, and so the parity is reversed

3 DETERMINANTS 3 3 Determinants of n n matrices 31 The 3 3 case The erminant of a 3 3 matrix A = a 11 a 12 a 13 a 21 a 22 a 23 a 31 a 32 a 33 is A = a 11 a 22 a 33 + a 12 a 23 a 31 + a 13 a 21 a 32 a 13 a 22 a 31 a 11 a 23 a 32 a 12 a 21 a 33 Notice that each term is of the form ±a 1j1 a 2j2 a 3j3, where (j 1, j 2, j 3 ) is a permutation of {1, 2, 3}, and the sign is the sign of the permutation Thus we can write A = sgn(j 1, j 2, j 3 )a 1j1 a 2j2 a 3j3 where the sum runs over the six permutations (j 1, j 2, j 3 ) of {1, 2, 3} 32 The general case The definition of the erminant of an n n matrix follows the same pattern as a 3 3 matrix: (1) A = sgn(j)a 1j1 a 2j2 a njn where the sum runs over all permutations J = (j 1, j 2, j n ) of the numbers {1, 2,, n} 1 Before proceeding, you should note that this definition does not provide a practical method for calculating erminants of large matrices The calculation of each term in the sum requires n 1 multiplications, and since there are n! terms, the number of multiplications required is n!(n 1) This number grows very rapidly as n increases If you could perform one million multiplications each second, it would take about 15 million years to calculate the erminant of a matrix directly from the definition 4 Properties of erminants 41 Primitive properties There are three primitive properties of erminants that are deduced directly from the definition Other properties are derived from these three primitive properties Property (i):the erminant function depends linearly on each individual row when the other rows are held fixed To be explicit, write the n n matrix A as a column of row vectors A = and suppose that the ith row is a linear combination of two other row vectors:, A i = t A i + t A i 1 You should check that this is consistent with the definition given previously for 2 2 matrices

4 DETERMINANTS 4 Then A = t A i + t A i To verify Property (i), write = t A i + t A i A i = [a i1 a i2 a in] and A i = [a i1 a i2 a in] so that the ith row of A is A i = t A i +t A i From the definition of the erminant, A = sgn(j)a 1j1 (t a ij i + t a ij i ) a nj n = ( t sgn(j)a 1j1 a ij i a njn + t sgn(j)a 1j1 a ij i a njn ) = t sgn(j)a 1j1 a ij i a njn + t sgn(j)a 1j1 a ij i a njn = t A i + t A i Property (ii): A square matrix with two equal rows has erminant 0 To verify property (ii), let A be a square matrix with two equal rows, say rows k and l For each permutation J, let J be the permutation obtained by switching the entries in positions k and l Since rows k and l are equal, we have a kjk = a ljk = a lj l, while a iji = a ij i when i is not equal to k or l Therefore, the terms in (1) corresponding to the permutations J and J are the same, except for the factors sgn(j) and sgn(j ), which have opposite sign by Theorem 3 Therefore, these two terms sum to 0 Thus the terms of (1) can be paired into terms that sum to 0 Property (iii):the erminant of the n n identity matrix is 1 To verify Property (iii), let δ ij be the entries of I n, so that δ ij = 0 if i j, and δ ii = 1 From the definition of the erminant, I = sgn(j)δ 1j1 δ njn However, if j i i for even one value of i, then the corresponding term in the above sum has a factor of 0, and so the product is 0 Therefore, the only non-zero term comes from the permutation (1, 2,, n), so I = sgn(1, 2,, n)δ 11 δ nn = 1

5 DETERMINANTS 5 42 Derived properties Let s now deduce some additional properties of erminants from the three primitive properties above Property (iv)multiplying a row of A by a constant c changes the erminant by a factor of c This is really just a special case of Property (i) Property (v):if A has a row consisting entirely of zeroes, then A = 0 To see this, notice that multiplying the row of zeroes by 0 has no effect on the matrix, and therefor no effect on its erminant On the other hand, Property (iv) says that the erminant changes by a factor of 0, so A = 0 Property (vi):if B is obtained from A by interchanging two rows, then B = A This is a consequence of Property (ii) Suppose B is obtained from A by interchanging rows k and l with k < l Write By Property (ii), A = A l + A l + A l = 0 Applying Property (i) twice, once in row k and again in row l, we obtain + A l + A l + A l A l = 0 By Property (ii), the first and last terms in this sum are 0, while the second and third are A and B, respectively We have therefore shown that A + B = 0, so B = A

6 DETERMINANTS 6 Property (vii):adding a multiple of one row of A to another row of A will not change the erminant To verify this, let,, be the row vectors of A, and let B be the matrix obtained by adding a multiple of row k of A to row i, with i k Thus row i of B can be written B i = A i + t, while the other rows of B are the same as the corresponding rows of A By Property (i), we have B = A i + t = A i + t Notice that the first term on the right is just A, while the second term is t times the erminant of a matrix with two equal rows, which is zero by Property (ii) Thus the right hand side reduces to A 5 The effect of elementary row operations Properties (iv), (vi) and (vii) tell you how elementary row operations affect the erminant Theorem 4 Let A be an n n matrix (1) If B is obtained from A by multiplying a single row of A by a non-zero constant λ, then B = λ A (2) If B is obtained from A by interchanging two rows, then B = A (3) If B is obtained from A by adding a multiple of one row to another row, then B = A Notice in every case, an elementary row operation changes the erminant by a non-zero multiple It follows that a sequence of several elementary row operations changes the erminant by a non-zero multiple, so we obtain Corollary 5 If B is row equivalent to A, then B = c A for some non-zero scalar c In particular, the above corollary applies when B is the reduced row echelon form of A Let s examine two cases If A is invertible, then the reduced row echelon form of A is an identity matrix, which has erminant 1 by Property (iii) Thus, in this case, the corollary gives A = c I = c 0 On the other hand, if A is singular, the reduced row echelon form E of A contains a row of zeroes, and Property (v) gives A = c E = c 0 = 0 Therefore you get Theorem 6 A is invertible if and only if A 0 51 Calculating erminants using EROs In principle, Theorems 4 and 6 give you a new way to calculate erminants using elementary row operations You can reduce to reduced row echelon form, keeping track of the effect on the erminant using Theorem 4, and then plug in the value of the erminant of the reduced matrix using either Property (iii) (if the reduced matrix is an identity matrix) or Property (v) (if the reduced matrix has a row of zeroes) In practice, you can avoid the necessity of completely reducing the matrix by first working out the erminant of an upper triangular matrix The result is

7 DETERMINANTS 7 Theorem 7 The erminant of an upper triangular matrix is the product of its diagonal entries Proof Let s first consider the case when one of the diagonal entries is zero In this case, the assertion is that A = 0, so we want to show that any upper triangular matrix A with a zero on the diagonal must have erminant zero By Theorem 6, it s enough to show that A is singular For this, consider the homogeneous system Ax = 0 Let a ii be the first zero on the diagonal Then the variable x i does not occur from the ith equation on Set x i = 1 and x k = 0 for k > i, and ermine x 1,, x i 1 by back substitution in the first i 1 equations Then x is a non-trivial solution to the homogeneous system Ax = 0, and since the system has a non-trivial solution, its coefficient matrix A is singular It remains to consider the case when all diagonal entries are non-zero In this case, by applying Property (i) to each row in succession, we obtain (2) A = a 11 a 22 a nn B where the entries of B are b ij = a ij /a ii In particular, B has a zero in every position where A has one, so B is also triangular Moreover, B has all ones along the diagonal It follows that B can be row reduced to an identity matrix I using only the operation of adding multiples of rows to other rows Since this type of elementary row operation does not affect the erminant, it follows that B = I = 1, so (2) gives A = a 11 a 22 a nn, as required The procedure for calculating erminants using elementary row operations is now the following (1) Perform elementary row operations to reduce A to an upper triangular matrix, using Theorem 4 to keep track of the effect on the erminant (2) Calculate the erminant of the resulting upper triangular matrix by multiplying its diagonal entries This method us usually much more efficient than calculating from the definition of erminants We refer to your textbook for examples illustrating the technique The goal in this section is to establish 6 Determinants of products Theorem 8 Let A and B be n n matrices Then (AB) = ( A)( B) We begin with some facts about elementary matrices Recall that an elementary matrix is one obtained from an identity matrix by a single elementary row operation Determinants of elementary matrices can be read off from Theorem 4 and the fact that the erminant of an identity matrix is 1 (Property (iii)) Lemma 9 (1) If E is obtained by multiplying a row of an identity matrix by the scalar λ, then E = λ (2) If E is obtained by exchanging two rows of an identity matrix, then E = 1 (3) If E is obtained by adding a multiple of a row of an identity matrix to another row, then E = 1 In view of Lemma 9, we can reformulate Theorem 4 as follows: Lemma 10 If B is an n n matrix, and E is an elementary n n matrix, then (EB) = ( E)( B)

8 DETERMINANTS 8 Iterating this result gives Lemma 11 If B is an n n matrix and E 1,, E K are elementary n n matrices, then (E 1 E K B) = ( E 1 ) ( E K )( B) In particular, when B = I n, we obtain Corollary 12 If E 1,, E K are elementary n n matrices, then (E 1 E K ) = ( E 1 ) ( E K ) Proof of Theorem 8 Consider first the case when the first factor A is singular In this case, the product AB is also singular, so by Theorem 6, it follows that (AB) = 0 and A = 0 Therefore (AB) = 0 = 0 B = ( A)( B), so the theorem holds in this case To complete the proof, we must establish the theorem when A is invertible In this case, we can write A as a product of elementary matrices: A = E 1 E K By Lemma 11 and Corollary 12, we obtain (AB) = (E 1 E K B) = ( E 1 ) ( E K )( B) = ((E 1 E K ))( B) = ( A)( B) and the theorem is proved As a special case of Theorem 8, suppose A is invertible, and take B = A 1 We obtain 1 = I = (AA 1 ) = ( A)( A 1 ) Dividing by A gives Corollary 13 If A is invertible, then (A 1 ) = 1 A In this section, we establish 7 Transposition Theorem 14 For any n n matrix A, we have (A t ) = A We begin with some special cases First, if A is singular, then A t is also singular, so both A and A t have erminant 0: Lemma 15 If A is singular, then A t = A = 0 For elementary matrices, we have Lemma 16 If E is an elementary matrix, then (E t ) = E Proof There are three types of elementary matrices We ll take them one at a time If E is obtained from an identity matrix by adding a multiple of row k to row l, then E = 1, by Theorem 4 But E t is obtained by adding a multiple of row

9 DETERMINANTS 9 l of an identity matrix to row k, so (E t ) = 1, again by Theorem 4 Therefore (E t ) = 1 = E If E is obtained by multiplying a row of the identity matrix by a constant, then E is diagonal, and hence symmetric, so E t = E, and so (E t ) = (E) Finally, if E is obtained by switching rows k and l of an identity matrix, then the only non-zero off diagonal entries of E are ones in positions (k, l) and (l, k), and therefore E is symmetric, so again (E t ) = E Proof of Theorem 14 In view of Lemma 15, we may assume that A is invertible Therefore, A can be expressed as a product of elementary matrices: Theorem 8 and Lemma 16, you get A = E 1 E 2 E K A t = ((E 1 E 2 E K ) t ) = (E t K E t 1) = (E t K) (E t 1) = (E K ) (E 1 ) = (E 1 ) (E K ) = (E 1 E K ) = A 71 Consequences of the Transposition Theorem Theorem 14 allows us to immediately deduce transposed forms of results we have already discussed For example, transposing Theorem 7 gives Theorem 17 The erminant of a lower triangular matrix is the product of the diagonal entries In addition, the properties of erminants that we have formulated for rows have analogues for columns In particular, Theorem 4 has an exact analogue for elementary column operations, so you can use both row and column operations in erminant calculations 8 Cofactor expansions Let A be an n n matrix The minor M ij is the (n 1) (n 1) matrix obtained by deleting the ith row and jth column of A The cofactors of A are the scalars For example, the 3 3 matrix cof ij (A) = ( 1) i+j M ij A = has nine minors and nine cofactors A typical minor is [ ] 4 6 M 12 = 7 9 The corresponding cofactor is, cof 12 (A) = ( 1) 1+2 M 12 = (36 42) = 6 Theorem 18 Let A be an n n matrix (1) (Cofactor expansion along the ith row) For any fixed i we have A = a ij cof ij (A) = ( 1) i+j a ij M ij j=1 j=1

11 DETERMINANTS 11 Theorem 19 (Adjoint Formula) For any n n matrix A we have A adj(a) = ( A)I n If A 0, dividing through by A gives a formula for the inverse of A Corollary 20 If A 0 then A 1 = 1 A adj(a) We will finish our discussion of erminants by applying the above corollary to the solution of an n n system of equations Consider the system Ax = b, where the coefficient matrix A is invertible Then the unique solution is x = A 1 B = adj(a)b A In particular, this gives n j=1 x i = cof ji(a)b j A But the sum in the numerator on the right may be viewed as a cofactor expansion along the ith column of the matrix A i obtained by replacing the ith column of A by b Therefore, we have established Theorem 21 (Cramer s Rule) Consider the n n system Ax = b, with invertible coefficient matrix A The unique solution is given by x i = A i A, where A i is the matrix obtained by replacing the ith column of A with b Let s write out Cramer s Rule explicitly for the 2 2 system a 11 x 1 + a 12 x 2 = b 1 a 21 x 1 + a 22 x 2 = b 2 We have [ ] [ ] b1 a 12 a11 b 1 b 2 a 22 a 21 b 2 x 1 = [ ], x 2 = [ ] a11 a 12 a11 a 12 a 21 a 22 a 21 a 22 You should check that this is consistent with the formulas obtained in Section 1 for solutions to 2 2 systems You should view Cramer s Rule as a generalization of those formulas to n n systems Finally, it should be noted that, although Cramer s Rule is of theoretical interest, it is rarely an efficient method for solving linear systems Except for very small systems, Gaussian elimination is usually much more computationally efficient

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