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1.1. INTRODUCTION TO LINEAR EQUATIONS Then addition and multiplication mod p are defined by a ⊕ b = (a + b) (mod p)...
Hence if t = 2 the system is inconsistent. If t = 2 the system is consistent and 112 101 B = 0 1 1 → 0 1 1 . 000 000 We read off the solution x = 1, y = 1. EXAMPLE 1.4.6 For which rationals a and b does the following system have (i) no solution, (ii) a unique solution, (iii) infinitely many solutions? x − 2y + 3z = 4 ...
. − 101 Show that if B is a 3 × 2 such that AB = I2 , 011 a b B = −a − 1 1 − b a+1 b...
3. (Eij (t))−1 = Eij (−t). Pro of. Taking A = In in the above theorem, we deduce the following equations: Eij Eij Ei (t)Ei (t
−1...
6. X t X = x2 + . . . + x2 if X = [x1 , . . . , xn ]t is a column vector. n 1 We prove only the fourth property. First check that both (AB )t and B t At have the same size (p × m). Moreover, corresponding elements of both matrices are equal. For if A = [aij ] and B = [bj k ], we have ( k AB )t i = (AB )ik jn = aij bj k
=1...
(ii) Express A4 in terms of A2 , A and I3 and hence calculate A4 explicitly. (iii) Use (i) to prove that A is non–singular and find A−1 explicitly. −11 −8 −4 9 4 ; [Answers: (ii) A4 = 6A2 − 8A + 3I3 = 12 20 16 5 (iii) A−1 −1 −3 1 4 −1 .] = A2 − 3A + 3I3 = 2 0 1 0 ...
3.3. LINEAR DEPENDENCE REMARK 3.3.1 If X1 , . . . , Xm are linearly independent and x 1 X1 + · · · + x m Xm = y 1 X1 + · · · + y m Xm , then x1 = y1 , . . . , xm = ym . For the equation can be rewritten as (x1 − y1 )X1 + · · · + (xm − ym )Xm = 0 and so x1 − y1 = 0, . . . , xm − ym = 0....
= δik det A = ((det A)In )ik . Hence A(adj A) = (det A)In . The other equation is proved similarly. COROLLARY 4.0.1 (Formula for the inverse) If det A = 0, then A is non–singular and 1 A−1 = adj A. det A EXAMPLE 4.0.3 The matrix 123 A= 4 5 6 889 4 − 6 6 2 89 89 −3 + 24 − 24 5 + 4 5 88 ...
However the i–th component of the last vector is the column i. Hence ∆1 /∆ ∆1 x1 ∆2 ∆2 /∆ x2 1 . = . = . . . ∆ . . . . xn ∆n ∆n /∆...
Calculating with complex numb ers
We can now do all the standard linear algebra calculations over the field of complex numbers – find the reduced row–echelon form of an matrix whose elements are complex numbers, solve systems of linear equations, find inverses and calculate determinants. For example, = 1 +i 2−i (1 + i)(8 − 2i) − 7(2 − i) 7 8 − 2i = (8 − 2i) + i(8 − 2i) − 14 + 7i = −4 + 13i = 0...
In numerical examples, we can write (i), for example, as Arg (z1 z2 ) ≡ Arg z1 + Arg z2 . EXAMPLE 5.6.3 Find the modulus and principal argument of √ 17 3+i z= 1+i and hence express z in modulus–argument form. √ 217 | 3 + i|17 =√ = 217/2 . Solution. |z | = 17 |1 + i|17 ( 2) = √ 3+i Arg z ≡ 17Arg 1+i √ 17(Arg ( 3 + i) − Arg (1 + i)) π π = −17π = 17 − . 6 4 12 −17 + 2k π , where k is an integer. We see that k = 1 and Hence Arg z = 12 π . c 7π hence Arg z = 12 . Consequently z = 217/2 os 7π + i sin 7π 12 12 DEFINITION 5.6.2 If θ is a real number, then we define eiθ by eiθ = cos θ + i sin θ. More generally, if z = x + iy , then we define ez by ez = ex eiy ...
Let a and b be distinct complex numbers and 0 < α < π . (i) Prove that each of the following sets in the complex plane represents a circular arc and sketch the circular arcs on the same diagram:
114 Arg
CHAPTER 5. COMPLEX NUMBERS z−a = α, −α, π − α, α − π . z−b z−a = π represents the line segment joining Also show that Arg z−b z−a a and b, while Arg = 0 represents the remaining portion of z−b the line through a and b. (ii) Use (i) to prove that four distinct points z1 , z2 , z3 , z4 are concyclic or collinear, if and only if the cross–ratio z4 − z 1 z3 − z 1 / z4 − z 2 z3 − z 2 is real. (iii) Use (ii) to derive Ptolemy’s Theorem: Four distinct points A, B , C, D are concyclic or collinear, if and only if one of the following holds: AB · CD + B C · AD = AC · B D
B D · AC + AD · B C = AB · CD
B D · AC + AB · CD = AD · B C...
This equation has real roots ( ( a+b± a + b)2 − 4(ab − h2 ) a − b)2 + 4h2 a+b± = (6.2) λ= 2 2 (The roots are distinct if a = b or h = 0. The case a = b and h = 0 needs no investigation, as it gives an equation of a circle.) The equation λ2 − (a + b)λ + ab − h2 = 0 is called the eigenvalue equation of the matrix A....
Multiplying equation 6.3 by λ1 and subtracting from equation 6.4 gives (λ2 − λ1 )y X2 = 0. Hence y = 0, as (λ2 − λ1 ) = 0 and X2 = 0. Then from equation 6.3, xX1 = 0 and hence x = 0. Then the equations AX1 = λ1 X1 and AX2 = λ2 X2 give AP = A[X1 |X2 ] = [AX1 |AX2 ] = [λ1 X1 |λ2 X2 ] λ = λ , 0 0 1 1 = [X1 |X2 ] P 0 λ2 0 λ2 so λ . 0 1 −1 P AP = 0 λ2 2 b 1 EXAMPLE 6.2.2 Let A = e the matrix of example 6.2.1. Then 12 a 1a − 1 re eigenvectors corresponding to eigenvalues nd X2 = X1 = 1 1 − , 11 1 and 3, respectively. Hence if P = we have 11 . 1 0 −1 P AP = 03 There are two immediate applications of theorem 6.2.1. The first is to the calculation of An : If P −1 AP = diag (λ1 , λ2 ), then A = P diag (λ1 , λ2 )P −1 and n P n λn P λ Pλ 0 0 0 1 1 −1 −1 −1 n 1 . P =P =P A= 0 λn 0 λ2 0 λ2 2 The second application is to solving a system of linear differential equations dx dt dy dt a = ax + by = cx + dy ,...
EXAMPLE 7.1.1 Let A be the symmetric matrix 1 . 2 −6 A= −6 7 Find a proper orthogonal matrix P such that P t AP is diagonal. Solution. The characteristic equation of A is λ2 − 19λ + 48 = 0, or (λ − 16)(λ − 3) = 0. Hence A has distinct eigenvalues λ1 = 16 and λ2 = 3. We find corresponding eigenvectors a 2. − 3 nd X2 = X1 = 3 2 √ Now ||X1 || = ||X2 || = 13. So we take 1 X1 = √ 13 − 3 2 a 1 nd X2 = √ 13 3 2 ....
Then under the rotation X = P Y , our original quadratic equation becomes 60 38 2 16x2 + 3y1 + √ (3x1 + 2y1 ) − √ (−2x1 + 3y1 ) + 31 = 0, 1 13 13 6 256 2 2 16x1 + 3y1 + √ x1 + √ y1 + 31 = 0. 13 13 Now complete the square in x1 and y1 : x + y + 16 2 2 16 2 + √ x1 3 1 + √ y1 31 = 0, 1 13 13 16 x 8 1+√ 13 2 +3 y 1 1+√ 13 2 = 16 = 48. 8 2 √ +3 13 1 2 √ − 31 13 (7.9) or...
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