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If we now let Zp = {0, 1, . . . , p − 1}, then it can be proved that Zp forms a field under the operations of modular addition and multiplication mod p. For example, the additive inverse of 3 in Z7 is 4, so we write −3 = 4 when calculating in Z7 . Also the multiplicative inverse of 3 in Z7 is 5 , so we write 3−1 = 5 when calculating in Z7 . In practice, we write a ⊕ b and a ⊗ b as a + b and ab or a × b when dealing with linear equations over Zp . The simplest field is Z2 , which consists of two elements 0, 1 with addition satisfying 1 + 1 = 0. So in Z2 , −1 = 1 and the arithmetic involved in solving equations over Z2 is very simple. EXAMPLE 1.1.4 Solve the following system over Z2 : x+y+z = 0 x + z = 1. Solution. We add the first equation to the second to get y = 1. Then x = 1 − z = 1 + z , with z arbitrary. Hence the solutions are (x, y , z ) = (1, 1, 0) and (0, 1, 1). We use Q and R to denote the fields of rational and real numbers, respectively. Unless otherwise stated, the field used will be Q....
4x + y + (a2 − 14)z = a + 2. [Answer: a = −4, no solution; a = 4, infinitely many solutions; a = ±4, exactly one solution.] 12. Solve the following system of homogeneous equations over Z2 : x1 + x 3 + x 5 = 0 x2 + x 4 + x 5 = 0 x1 + x 2 + x 3 + x 4 = 0 x3 + x4 = 0. [Answer: x1 = x2 = x4 + x5 , x3 = x4 , with x4 and x5 arbitrary elements of Z2 .] 13. Solve the following systems of linear equations over Z5 : (a) 2x + y + 3z = 4 4x + y + 4z = 1 3x + y + 2z = 0 (b) 2x + y + 3z = 4 4x + y + 4z = 1 x + y = 3....
. − 101 Show that if B is a 3 × 2 such that AB = I2 , 011 a b B = −a − 1 1 − b a+1 b...
and consequently AX = 0 has a non–trivial solution x = −1, y = −1, z = 1. REMARK 2.5.5 More generally, if A is row–equivalent to a matrix containing a zero row, then A is singular. For then the homogeneous system AX = 0 has a non–trivial solution. An important class of non–singular matrices is that of the elementary row matrices. DEFINITION 2.5.2 (Elementary row matrices) There are three types, Eij , Ei (t), Eij (t), corresponding to the three kinds of elementary row operation: 1. Eij , (i = j ) is obtained from the identity matrix In by interchanging rows i and j . 2. Ei (t), (t = 0) is obtained by multiplying the i–th row of In by t. 3. Eij (t), (i = j ) is obtained from In by adding t times the j –th row of In to the i–th row. EXAMPLE 2.5.6 (n = 3.) 100 00 1 0 10 E23 = 0 0 1 , E2 (−1) = 0 −1 0 , E23 (−1) = 0 1 −1 . 010 0 01 00 1...
We also observe that E12 (−2)E2 (−1)E21 (−1)A = I2 . Hence A−1 = E12 (−2)E2 (−1)E21 (−1) A = E21 (1)E2 (−1)E12 (2)....
A = E21 (−3)E2 (13)E12 (4) is one such decomposition.] 2. A square matrix D = [dij ] is called diagonal if dij = 0 for i = j . (That is the off–diagonal elements are zero.) Prove that pre–multiplication of a matrix A by a diagonal matrix D results in matrix DA whose rows are the rows of A multiplied by the respective diagonal elements of D. State and prove a similar result for post–multiplication by a diagonal matrix. Let diag (a1 , . . . , an ) denote the diagonal matrix whose diagonal elements dii are a1 , . . . , an , respectively. Show that diag (a1 , . . . , an )diag (b1 , . . . , bn ) = diag (a1 b1 , . . . , an bn ) and deduce that if a1 . . . an = 0, then diag (a1 , . . . , an ) is non–singular and − (diag (a1 , . . . , an ))−1 = diag (a1 1 , . . . , a−1 ). n Also prove that diag (a1 , . . . , an ) is singular if ai = 0 for some i. 0 3. Let A = 1 3 express A as a 02 2 6 . Prove that A is non–singular, find A−1 and 79 product of elementary row matrices. −12
9 2 1 2...
THEOREM 3.2.2 If A is row equivalent to B , then R(A) = R(B ). Proof. Suppose that B is obtained from A by a sequence of elementary row operations. Then it is easy to see that each row of B is a linear combination of the rows of A. But A can be obtained from B by a sequence of elementary operations, so each row of A is a linear combination of the rows of B . Hence by Theorem 3.2.1, R(A) = R(B )....
An important special case is when A is a diagonal matrix. If A =diag (a1 , . . . , an ) then det A = a1 . . . an . In particular, for a scalar matrix tIn , we have det (tIn ) = tn . Pro of. Use induction on the size n of the matrix. The result is true for n = 2. Now let n > 2 and assume the result true for matrices of size n − 1. If A is n × n, then expanding det A along row 1 gives a = 0 ... 0 22 a32 a33 . . . 0 det A = a11 . . . an1 an2 . . . ann a11 (a22 . . . ann ) by the induction hypothesis. If A is upper triangular, equation 4.1 remains true and the proof is again an exercise in induction, with the slight difference that the column version of theorem 4.0.1 is needed. REMARK 4.0.1 It can be shown that the expanded form of the determinant of an n × n matrix A consists of n! signed products ±a1i1 a2i2 . . . anin , where (i1 , i2 , . . . , in ) is a permutation of (1, 2, . . . , n), the sign being 1 or −1, according as the number of inversions of (i1 , i2 , . . . , in ) is even or odd. An inversion occurs when ir > is but r < s. (The proof is not easy and is omitted.) The definition of the determinant of an n × n matrix was given in terms of the first–row expansion. The next theorem says that we can expand the determinant along any row or column. (The proof is not easy and is omitted.)...
81 1 2 1 0 1 1 −1 = −2 0 −1 −13 −5 0 0 1 3 = 1 1 2 1 01 1 −1 −2 0 0 −12 −6 1 3 00 1 = 1 2 1 01 1 −1 2 00 1 3 0 0 −12 −6 = 1 12 1 0 1 1 −1 2 60. 001 3 0 0 0 30 EXAMPLE 4.0.6 (Vandermonde determinant) Prove that = 1 11 abc (b − a)(c − a)(c − b). a2 b2 c2 Solution. Subtracting column 1 from columns 2 and 3 , then expanding along row 1, gives = 1 1 0 0 11 a b−a c−a abc a2 b2 − a2 c2 − a2 a2 b2 c2 b = −a c−a 2 − a 2 c2 − a 2 b = 1 1 (b − a)(c − a)(c − b). (b − a)(c − a) b+a c+a REMARK 4.0.4 From theorems 4.0.6, 4.0.10 and corollary 4.0.2, we deduce (a) det (Eij A) = −det A, (b) det (Ei (t)A) = t det A, if t = 0, = 1...
and use theorem 2.5.8 and induction, to prove that det (B A) = det B det A, if B is non–singular. Also prove that the formula holds when B is singular....
It is customary to blur the distinction between the real complex number {x} and the real number x and write {x} as x. Thus we write the complex number {x} + i{y } simply as x + iy . More generally, the sum of two complex numbers is a complex number: (x1 + iy1 ) + (x2 + iy2 ) = (x1 + x2 ) + i(y1 + y2 ); (5.1)...
The geometrical representation of complex numbers can be very useful when complex number methods are used to investigate properties of triangles and circles. It is very important in the branch of calculus known as Complex Function theory, where geometric methods play an important role. We mention that the line through two distinct points P1 = (x1 , y1 ) and P2 = (x2 , y2 ) has the form z = (1 − t)z1 + tz2 , t ∈ R, where z = x + iy is any point on the line and zi = xi + iyi , i = 1, 2. For the line has parametric equations x = (1 − t)x1 + tx2 , y = (1 − t)y1 + ty2...
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