SOLUTIONS TO PROBLEMS ELEMENTARY LINEAR ALGEBRA

Transcript

1 SOLUTIONS TO PROBLEMS ELEMENTARY LINEAR ALGEBRA K R MATTHEWS DEPARTMENT OF MATHEMATICS UNIVERSITY OF QUEENSLAND First Printing, 99

2 CONTENTS PROBLEMS 6 PROBLEMS 4 PROBLEMS 7 8 PROBLEMS 36 3 PROBLEMS 4 45 PROBLEMS PROBLEMS PROBLEMS PROBLEMS 88 9 i

3 [ (i) 4 0 [ 0 3 (ii) 4 (iii) R R + R 3 R 3 R 3 R R 3 (iv) (a) SECTION 6 [ 4 0 R R [ 4 R R 0 3 R R R R 3 R 3 R R R R R 3 R 3 + R R R 4R 3 R R + 3R 3 R 3 R 3 + R R R R R [ ; [ 0 R R R R R + R 3 R 3 R R R R R 3 R 3 R R ; R The augmented matrix has been converted to reduced row echelon form and we read off the unique solution x 3, y 9 4, z 4 0 (b) R 0 R 3R R R 3 + 5R R 3 R 3 + R From the last matrix we see that the original system is inconsistent ;

4 (c) R R R R 3 R 3 3R R 4 R 4 6R R R R R + R R 4 R 4 R 3 R 3 R 3 R The augmented matrix has been converted to reduced row echelon form and we read off the complete solution x 3z, y 3 z, with z arbitrary 3 a 3 a b R R R 8 b a 5 5 c 5 5 c 8 b a R R 3 a R 8 b a R R b + 3a R 5 5 c 3 R 3 + 5R b 5a + c 8 b a R 3 R 3 + R R 5 R 9 b 3a b a + c R R R 0 (b+a) b 3a b a + c From the last matrix we see that the original system is inconsistent if 3b a + c 0 If 3b a + c 0, the system is consistent and the solution is x where z is arbitrary 5 t t + t 3 R 3 R 3 R Case (b + a) 5 + (b 3a) z, y z, R R tr R 3 R 3 ( + t)r 0 t t t No solution B 0 t 0 0 t t

5 Case t B We read off the unique solution x, y 0 6 Method R R R 4 R R R 4 R 3 R 3 R 4 R 4 R 4 R 3 R R Hence the given homogeneous system has complete solution with x 4 arbitrary Method Write the system as x x 4, x x 4, x 3 x 4, x + x + x 3 + x 4 4x x + x + x 3 + x 4 4x x + x + x 3 + x 4 4x 3 x + x + x 3 + x 4 4x 4 Then it is immediate that any solution must satisfy x x x 3 x 4 Conversely, if x, x, x 3, x 4 satisfy x x x 3 x 4, we get a solution 7 [ λ 3 λ 3 R R [ λ 3 λ 3 R R (λ 3)R [ λ 3 0 λ + 6λ 8 B Case : λ + 6λ 8 0 [ That is (λ )(λ 4) 0 or λ, 4 Here B is 0 row equivalent to : 0 [ [ λ 3 0 R λ +6λ 8 R R 0 R (λ 3)R 0 Hence we get the trivial solution x 0, y 0 3

6 Case : λ Then B arbitrary Case 3: λ 4 Then B arbitrary [ 0 0 [ 0 0 and the solution is x y, with y and the solution is x y, with y 8 [ 3 5 R [ 3 R [ R R 5R R 3 [ 8 R R R 3 R 4 [ Hence the solution of the associated homogeneous system is 3 x 4 x 3, x 4 x 3 x 4, with x 3 and x 4 arbitrary 9 A n n n R R R n R R R n R n R n R n n 0 n 0 n n n R 0 n R n R n R n The last matrix is in reduced row echelon form Consequently the homogeneous system with coefficient matrix A has the solution x x n, x x n,,x n x n, 4

7 with x n arbitrary Alternatively, writing the system in the form x + + x n nx x + + x n nx x + + x n nx n shows that any solution must satisfy nx nx nx n, so x x x n Conversely if x x n,,x n x n, we see that x,,x n is a solution [ a b 0 Let A and assume that ad bc 0 c d Case : a 0 [ a b c d R [ R b a R a c d [ a b ad bc R a 0 R R cr [ b a 0 ad bc a [ R R b 0 a R 0 Case : a 0 Then bc 0 and hence c 0 [ [ [ [ 0 b c d d A R c d R c 0 0 b 0 0 So in both cases, A has reduced row echelon form equal to [ 0 0 We simplify the augmented matrix of the system using row operations: R 3 4 R 3R a R 4 a + 3 R 3 4R 0 7 a a 4 R 3 R 3 R 3 4 R 7 R R R R R R R 0 0 a 6 a 4 Denote the last matrix by B a 6 a 4 5

8 Case : a 6 0 ie a ±4 Then R 3 a 6 R 3 R R R 3 R R + R 3 and we get the unique solution x Case : a 4 Then B Case 3: a 4 Then B 0 0 8a+5 7(a+4) 0 0 0a+54 7(a+4) 0 0 a+4 8a + 5 0a + 54, y 7(a + 4) 7(a + 4), z a , so our system is inconsistent We read off that the system is consistent, with complete solution x 8 7 z, y z, where z is arbitrary We reduce the augmented array of the system to reduced row echelon form: R 3 R 3 + R R 3 R 3 + R R R + R 4 R 3 R The last matrix is in reduced row echelon form and we read off the solution of the corresponding homogeneous system: x x 4 x 5 x 4 + x 5 x x 4 x 5 x 4 + x 5 x 3 x 4 x 4, 6

9 where x 4 and x 5 are arbitrary elements of Z Hence there are four solutions: x x x 3 x 4 x (a) We reduce the augmented matrix to reduced row echelon form: R 3R R R + R R 3 R 3 + R R R + R R 3 R 3 + 3R R 4R R R + R 3 R R + 3R Consequently the system has the unique solution x, y, z 0 (b) Again we reduce the augmented matrix to reduced row echelon form: R R R R + R R 3 R 3 + 3R R R + 4R R 3 R 3 + R We read off the complete solution R 3R x 3z + z y z + 3z, where z is an arbitrary element of Z

10 4 Suppose that (α,,α n ) and (β,,β n ) are solutions of the system of linear equations n a ij x j b i, i m Then j n a ij α j b i j and n a ij β j b i for i m Let γ i ( t)α i + tβ i for i m Then (γ,,γ n ) is a solution of the given system For n a ij γ j j j n a ij {( t)α j + tβ j } j n a ij ( t)α j + j ( t)b i + tb i b i n a ij tβ j 5 Suppose that (α,,α n ) is a solution of the system of linear equations j n a ij x j b i, i m () j Then the system can be rewritten as or equivalently So we have n a ij x j j n a ij α j, i m, j n a ij (x j α j ) 0, i m j n a ij y j 0, i m j where x j α j y j Hence x j α j + y j, j n, where (y,,y n ) is a solution of the associated homogeneous system Conversely if (y,,y n ) 8

11 is a solution of the associated homogeneous system and x j α j + y j, j n, then reversing the argument shows that (x,,x n ) is a solution of the system 6 We simplify the augmented matrix using row operations, working towards row echelon form: a b R R ar 0 a + a a b a R 3 0 a + a 3 R 3 3R 0 3 a 3 a R R 3 R R R 3 R 3 + (a )R a a 0 a + a a b a a a a ( a)(a ) a + a + b B Case : a Then 4 a 0 and B a a a 0 0 a +a+b 4 a Hence we can solve for x, y and z in terms of the arbitrary variable w Case : a Then B b Hence there is no solution if b However if b, then 0 0 B and we get the solution x z, y 3z w, where w is arbitrary 7(a) We first prove that Observe that the elements + 0, +, + a, + b 9

12 are distinct elements of F by virtue of the cancellation law for addition For this law states that +x +y x y and hence x y +x +y Hence the above four elements are just the elements 0,, a, b in some order Consequently ( + 0) + ( + ) + ( + a) + ( + b) a + b ( ) + (0 + + a + b) 0 + (0 + + a + b), so after cancellation Now ( + )( + ), so we have x 0, where x + Hence x 0 Then a + a a( + ) a 0 0 Next a + b For a + b must be one of 0,, a, b Clearly we can t have a + b a or b; also if a + b 0, then a + b a + a and hence b a; hence a + b Then a + a + (a + b) (a + a) + b 0 + b b Similarly b + a Consequently the addition table for F is + 0 a b 0 0 a b 0 b a a a b 0 b b a 0 We now find the multiplication table First, ab must be one of, a, b; however we can t have ab a or b, so this leaves ab Next a b For a must be one of, a, b; however a a a 0 or a ; also a a 0 (a )(a + ) 0 (a ) 0 a ; hence a b Similarly b a Consequently the multiplication table for F is 0 a b a b a 0 a b b 0 b a (b) We use the addition and multiplication tables for F: a b a A a b b R a b a R + ar 0 0 a a R a 3 R 3 + R 0 b a 0 0

13 R R 3 R R + ar a b a 0 b a a a 0 a a 0 b R ar R 3 br 3 R R + ar 3 R R + br 3 The last matrix is in reduced row echelon form a b a 0 b b 0 0

14 Section 4 a b Suppose B c d and that AB I Then e f [ a b [ [ 0 c d 0 a + e b + f 0 0 c + e d + f e f Hence Next, a + e c + e 0, b + f 0 d + f ; e a + c e (a + ), f b d f b ; a b B a b a + b (BA) B (BA)(BA)B B(AB)(AB) BI I BI B 4 Let p n denote the statement A n (3n ) A + (3 3n ) I Then p asserts that A (3 ) A + (3 3) I, which is true So let n and assume p n Then from (), A n+ { } A A n A (3 n ) A + (3 3n ) I (3n ) A + (3 3n ) A (3n ) (4A 3I ) + (3 3n ) A (3n )4+(3 3 n ) A + (3n )( 3) I (4 3n 3 n ) A + (3 3n+ ) I (3n+ ) A + (3 3n+ ) I Hence p n+ is true and the induction proceeds 5 The equation x n+ ax n + bx n is seen to be equivalent to [ [ [ xn+ a b xn 0 x n x n

15 or where X n [ xn+ x n X n AX n, [ a b and A Then 0 X n A n X 0 if n Hence by Question 3, [ { xn+ (3 n ) A + (3 } [ 3n ) x I x n x 0 { (3 n [ [ ) n (3n ) + 3 3n (3 n )( 3) 3 n 3 3 n n [ x x 0 } [ x x 0 Hence, equating the (, ) elements gives x n (3n ) x + (3 3n ) x 0 if n 7 Note: λ + λ a + d and λ λ ad bc Then (λ + λ )k n λ λ k n (λ + λ )(λ n + λ n λ + + λ λ n + λ n ) λ λ (λ n + λ n 3 λ + + λ λ n 3 + λ n ) If λ λ, we see (λ n + λ n λ + + λ λ n ) +(λ n λ + + λ λ n + λ n ) (λ n λ + + λ λ n ) λ n + λ n λ + + λ λ n + λ n k n+ k n λ n + λ n λ + + λ λ n + λ n λ n + λ n λ + + λ λ n + λ n nλ n 3

16 If λ λ, we see that (λ λ )k n (λ λ )(λ n + λ n λ + + λ λ n + λ n ) λ n + λ n λ + + λ λ n (λ n λ + + λ λ n + λ n ) λ n λ n Hence k n λn λn λ λ We have to prove n: A n k n A λ λ k n I A A; also k A λ λ k 0 I k A λ λ 0I A Let n and assume equation holds Then A n+ A n A (k n A λ λ k n I )A k n A λ λ k n A Now A (a + d)a (ad bc)i (λ + λ )A λ λ I Hence A n+ k n (λ + λ )A λ λ I λ λ k n A {k n (λ + λ ) λ λ k n }A λ λ k n I k n+ A λ λ k n I, and the induction goes through 8 Here λ, λ are the roots of the polynomial x x 3 (x 3)(x + ) So we can take λ 3, λ Then k n 3n ( ) n 3 ( ) 3n + ( ) n+ 4 Hence A n { 3 n + ( ) n+ 4 3n + ( ) n+ 4 } A ( 3) [ + 3 { 3 n + ( ) n 4 { 3 n + ( ) n 4 } I } [ 0 0, 4

17 which is equivalent to the stated result 9 In terms of matrices, we have [ [ [ Fn+ Fn 0 [ Fn+ F n F n [ 0 n [ F F 0 F n for n [ 0 n [ 0 Now λ, λ are the roots of the polynomial x x here Hence λ + 5 and λ 5 and k n ( + 5 ( + 5 ) n ( 5 ) + ( 5 5 ) n ( 5 ) n ) n 5 Hence A n k n A λ λ k n I k n A + k n I So [ Fn+ F n [ (k n A + k n I ) 0 [ [ [ kn + k k n + k n n 0 k n Hence F n k n ( + 5 ) n ( 5 ) n 5 0 From Question 5, we know that [ [ xn r y n n [ a b 5

18 [ r Now by Question 7, with A, A n k n A λ λ k n I k n A ( r)k n I, where λ + r and λ r are the roots of the polynomial x x + ( r) and Hence [ xn y n k n λn λn r [ a (k n A ( r)k n I ) b ([ [ ) [ kn k n r ( r)kn 0 a k n k n 0 ( r)k n b [ [ kn ( r)k n k n r a k n k n ( r)k n b [ a(kn ( r)k n ) + bk n r ak n + b(k n ( r)k n ) Hence, in view of the fact that we have k n λn λn k n λ n λ n [ xn y n λn ( {λ λ } n ) λ n ( { λ λ } n ) λ, as n, a(k n ( r)k n ) + bk n r ak n + b(k n ( r)k n ) kn a( k n ( r)) + b kn k n r a kn k n + b( kn k n ( r)) a(λ ( r)) + bλ r aλ + b(λ ( r)) a( r + r) + b( + r)r a( + r) + b( r + r) r{a( + r) + b( + r) r} a( + r) + b( r + r) r 6

19 [ [A I R 3 R so Hence 4 3 [ Section 7 0 3/3 /3 Hence A is non singular and A Moreover [ 4 R R + 3R [ 0 R R 4R 0 [ /3 4/3 3/3 /3 E ( 4)E (/3)E (3)A I, A E ( 4)E (/3)E (3) /3 4/3 3/3 /3 A {E (3)} {E (/3)} {E ( 4)} E ( 3)E (3)E (4) Let D [d ij be an m m diagonal matrix and let A [a jk be an m n matrix Then n (DA) ik d ij a jk d ii a ik, j as d ij 0 if i j It follows that the ith row of DA is obtained by multiplying the ith row of A by d ii Similarly, post multiplication of a matrix by a diagonal matrix D results in a matrix whose columns are those of A, multiplied by the respective diagonal elements of D In particular, diag (a,,a n )diag (b,,b n ) diag (a b,,a n b n ), as the left hand side can be regarded as pre multiplication of the matrix diag (b,,b n ) by the diagonal matrix diag (a,,a n ) Finally, suppose that each of a,,a n is non zero Then a,,a n all exist and we have diag (a,,a n )diag (a,,a n ) diag (a a,,a na n ) diag (,,) I n Hence diag (a,,a n ) is non singular and its inverse is diag (a,,a n ) 7

20 Next suppose that a i 0 Then diag (a,,a n ) is row equivalent to a matix containing a zero row and is hence singular [A I R R R 3 R 3 3R R R R 3 R R R R / / R R 4R / 3 R R + 9R / 0 0 Hence A is non singular and A Also Hence 7 9/ 3 / 0 0 E 3 (9)E 3 ( 4)E ( )E 3 (/)E 3 E 3 ( 3)E A I 3 A E 3 (9)E 3 ( 4)E ( )E 3 (/)E 3 E 3 ( 3)E, so A E E 3 (3)E 3 E 3 ()E ()E 3 (4)E 3 ( 9) 4 A k k 0 7 3k k k 0 7 3k k B Hence if 6 k 0, ie if k 3, we see that B can be reduced to I 3 and hence A is non singular 3 If k 3, then B B and consequently A is singular, as it is row equivalent to a matrix containing a zero row 8

21 [ 5 E () 4 [ is singular 4 [ 0 0 Hence, as in the previous question, 6 Starting from the equation A A + 3I 0, we deduce A(A I ) 3I (A I )A Hence AB BA I, where B 3 (A I ) Consequently A is non singular and A B 7 We assume the equation A 3 3A 3A + I 3 (ii) A 4 A 3 A (3A 3A + I 3 )A 3A 3 3A + A 3(3A 3A + I 3 ) 3A + A 6A 8A + 3I 3 (iii) A 3 3A + 3A I 3 Hence Hence A is non singular and A(A 3A + 3I 3 ) I 3 (A 3A + 3I 3 )A A A 3A + 3I (i) If B 3 0 then (I n B)(I n + B + B ) I n (I n + B + B ) B(I n + B + B ) (I n + B + B ) (B + B + B 3 ) I n B 3 I n 0 I n Similarly (I n + B + B )(I n B) I n Hence A I n B is non singular and A I n + B + B It follows that the system AX b has the unique solution X A b (I n + B + B )b b + Bb + B b 9

22 (ii) Let B 0 r s 0 0 t from the preceding question Then B 0 0 rt and B 3 0 Hence (I 3 B) I 3 + B + B r s t r s + rt 0 t rt (i) Suppose that A 0 Then if A exists, we deduce that A (AA) A 0, which gives A 0 and this is a contradiction, as the zero matrix is singular We conclude that A does not have an inverse (ii) Suppose that A A and that A exists Then A (AA) A A, which gives A I n Equivalently, if A A and A I n, then A does not have an inverse 0 The system of linear equations x + y z a z b x + y + z c is equivalent to the matrix equation AX B, where x A 0 0, X y, B z By Question 7, A exists and hence the system has the unique solution 3 a a 3b + c X 4 b a + 4b c 0 0 c b Hence x a 3b + c, y a + 4b c, z b a b c 0

23 A E 3 ()E 4 E 4 (3) E 3 ()E 4 E 3 () Also A (E 3 ()E 4 E 4 (3)) (E 4 (3)) E4 (E 3()) E 4 ( 3)E 4 E 3 (/) E 4 ( 3)E 4 E 4 ( 3) / / / (All matrices in this question are over Z ) (a)

24 Hence A is non singular and (b) A 4 (a) R 3 R 3 R A R 4 R 4 + R R 3 R 3 R R R 3 R R + R 3 R R / 0 /, so A is singular / 0 / 0 / Hence A exists and A 0 0 / 0 / (b) R 3 R 3 R R 3 R R R R R R R R /

25 R R R / 0 0 / Hence A exists and (c) A R 7 R R 3 5 R 3 / 0 0 / Hence A is singular by virtue of the zero row R (d) R 0 0 R R 0 0 R 3 7 R 3 R 3 R 3 R / / /7 Hence A exists and A diag (/, /5, /7) (Of course this was also immediate from Question ) (e) R R R R R R R R 3R R R + R R 3 R 3 R R 4 R / Hence A exists and A /

26 (f) R R 4R R R 3 5R R 3 R 3 R Hence A is singular by virtue of the zero row 5 Suppose that A is non singular Then Taking transposes throughout gives AA I n A A (AA ) t I t n (A A) t (A ) t A t I n A t (A ) t, so A t is non singular and (A t ) (A ) t [ a b 6 Let A, where ad bc 0 Then the equation c d A (a + d)a + (ad bc)i 0 reduces to A (a + d)a 0 and hence A (a + d)a From the last equation, if A exists, we deduce that A (a + d)i, or [ [ a b a + d 0 c d 0 a + d Hence a a + d, b 0, c 0, d a + d and a b c d 0, which contradicts the assumption that A is non singular 7 a b A a c b c R R + ar R 3 R 3 + br R +a R R 3 R 3 (ab c)r 4 a b 0 + a c + ab 0 ab c + b a b c+ab 0 +a 0 ab c + b a b 0 c+ab +a b + (c ab)(c+ab) +a B

27 Now + b + (c ab)(c + ab) + a + b + c (ab) + a + a + b + c + a 0 Hence B can be reduced to I 3 using four more row operations and consequently A is non singular 8 The proposition is clearly true when n So let n and assume (P AP) n P A n P Then (P AP) n+ (P AP) n (P AP) (P A n P)(P AP) P A n (PP )AP P A n IAP P (A n A)P P A n+ P Then from the previous ques- and the induction goes through [ [ /3 /4 9 Let A and P /3 3/4 [ 5/ 0 We then verify that P AP 0 tion, P A n P (P AP) n [ 5/ n [ Then P [ (5/) n 0 0 n Hence [ [ [ A n (5/) n 0 P P 3 (5/) n [ (5/) n [ (5/) n 4 [ 4(5/) n + 3 ( 3)(5/) n (5/) n + 4 3(5/) n + 4 [ [ (5/)n 4 3 [ (5/) n 0 0 [

28 [ Notice that A n as n This problem is a special case of 4 4 a more general result about Markov matrices [ a b 0 Let A be a matrix whose elements are non negative real c d numbers satisfying a 0, b 0, c 0, d 0, a + c b + d [ b Also let P and suppose that A I c (i) det P b c (b + c) Now b + c 0 Also if b + c 0, then we would have b c 0 and hence d a, resulting in A I Hence detp < 0 and P is non singular Next, P AP [ [ [ a b b b + c c b c d c [ [ a c b d b b + c ac + bc cb + bd c Now Also Hence b + c b + c [ [ b ac + bc cb + bd c [ b c 0 ( ac + bc)b + ( cb + bd)c ac + bc + cb bd acb + b c c b + bdc cb(a + c) + bc(b + d) cb + bc 0 (a + d )(b + c) ab ac db dc + b + c P AP b + c ac + b( a) + c( d) bd ac + bc + cb bd [ (b + c) 0 0 (a + d )(b + c) [ 0 0 a + d 6

29 [ 0 (ii) We next prove that if we impose the extra restriction that A 0 then a + d < This will then have the following consequence: [ 0 A P P 0 a + d [ n A n 0 P P 0 a + d [ 0 P 0 (a + d ) n P [ 0 P P 0 0 [ [ [ b 0 c 0 0 b + c c b [ [ b 0 b + c c 0 c b b + c b + c [ b b c c [ b b, c c where we have used the fact that (a + d ) n 0 as n We first prove the inequality a + d : a + d + d d a + d Next, if a + d, we have a + d ; so a d and hence c 0 b, contradicting our assumption that A I Also if a + d [, then 0 a + d 0; so a 0 d and hence c b and hence A 0 The system is inconsistent: We work towards reducing the augmented matrix: 4 5 R R R R 3 R 3 3R R 3 R 3 R ,

30 The last row reveals inconsistency The system in matrix form is AX B, where [ A x, X, B y 3 5 The normal equations are given by the matrix equation Now A t A A t B [ 3 5 [ 3 5 Hence the normal equations are A t AX A t B x + 8y 45 8x + 30y [ [ These may be solved, for example, by Cramer s rule: x y Substituting the coordinates of the five points into the parabola equation gives the following equations: a 0 a + b + c 0 a + b + 4c a + 3b + 9c 4 a + 4b + 6c 8 8

31 The associated normal equations are given by a b c 60 which have the solution a /5, b, c 4 Suppose that A is symmetric, ie A t A and that AB is defined Then so B t AB is also symmetric (B t AB) t B t A t (B t ) t B t AB, 5 Let A be m n and B be n m, where m > n Then the homogeneous system BX 0 has a non trivial solution X 0, as the number of unknowns is greater than the number of equations Then (AB)X 0 A(BX 0 ) A0 0 and the m m matrix AB is therefore singular, as X (i) Let B be a singular n n matrix Then BX 0 for some non zero column vector X Then (AB)X A(BX) A0 0 and hence AB is also singular (ii) Suppose A is a singular n n matrix Then A t is also singular and hence by (i) so is B t A t (AB) t Consequently AB is also singular, 9

32 Section 36 (a) Let S be the set of vectors [x, y satisfying x y Then S is a vector subspace of R For (i) [0, 0 S as x y holds with x 0 and y 0 (ii) S is closed under addition For let [x, y and [x, y belong to S Then x y and x y Hence x + x y + y (y + y ) and hence belongs to S [x + x, y + y [x, y + [x, y (iii) S is closed under scalar multiplication For let [x, y S and t R Then x y and hence tx (ty) Consequently [tx, ty t[x, y S (b) Let S be the set of vectors [x, y satisfying x y and x y Then S is a subspace of R This can be proved in the same way as (a), or alternatively we see that x y and x y imply x 4x and hence x 0 y Hence S {[0, 0}, the set consisting of the zero vector This is always a subspace (c) Let S be the set of vectors [x, y satisfying x y + Then S doesn t contain the zero vector and consequently fails to be a vector subspace (d) Let S be the set of vectors [x, y satisfying xy 0 Then S is not closed under addition of vectors For example [, 0 S and [0, S, but [, 0 + [0, [, S (e) Let S be the set of vectors [x, y satisfying x 0 and y 0 Then S is not closed under scalar multiplication For example [, 0 S and R, but ( )[, 0 [, 0 S Let X, Y, Z be vectors in R n Then by Lemma 3 X + Y, X + Z, Y + Z X, Y, Z, as each of X + Y, X + Z, Y + Z is a linear combination of X, Y, Z 30

33 Also X (X + Y ) + (X + Z) (Y + Z), so Hence 3 Let X Y (X + Y ) (X + Z) + (Y + Z), Z (X + Y ) + (X + Z) + (Y + Z), 0 X, Y, Z X + Y, X + Z, Y + Z X, Y, Z X + Y, X + Z, Y + Z, X 0 and X 3 3 We have to decide if X, X, X 3 are linearly independent, that is if the equation xx + yx + zx 3 0 has only the trivial solution This equation is equivalent to the folowing homogeneous system x + 0y + z 0 0x + y + z 0 x + y + z 0 x + y + 3z 0 We reduce the coefficient matrix to reduced row echelon form: and consequently the system has only the trivial solution x 0, y 0, z 0 Hence the given vectors are linearly independent 4 The vectors X λ, X λ, X 3 λ 3

34 are linearly dependent for precisely those values of λ for which the equation xx +yx +zx 3 0 has a non trivial solution This equation is equivalent to the system of homogeneous equations λx y z 0 x + λy z 0 x y + λz 0 Now the coefficient determinant of this system is λ λ λ (λ + ) (λ ) So the values of λ which make X, X, X 3 linearly independent are those λ satisfying λ and λ 5 Let A be the following matrix of rationals: 0 A Then A has reduced row echelon form B From B we read off the following: (a) The rows of B form a basis for R(A) (Consequently the rows of A also form a basis for R(A)) (b) The first four columns of A form a basis for C(A) (c) To find a basis for N(A), we solve AX 0 and equivalently BX 0 From B we see that the solution is x x 5 x 0 x 3 x 5 x 4 3x 5, 3

35 with x 5 arbitrary Then X x 5 0 x 5 3x 5 x 5 x 5 so [, 0,, 3, t is a basis for N(A) 0 3, 6 In Section 6, problem, we found that the matrix 0 0 A has reduced row echelon form B From B we read off the following: (a) The three non zero rows of B form a basis for R(A) (b) The first three columns of A form a basis for C(A) (c) To find a basis for N(A), we solve AX 0 and equivalently BX 0 From B we see that the solution is x x 4 x 5 x 4 + x 5 x x 4 x 5 x 4 + x 5 x 3 x 4 x 4, with x 4 and x 5 arbitrary elements of Z Hence x 4 + x 5 x 4 + x 5 X x 4 x 4 x 4 + x 5 x Hence [,,,, 0 t and [,, 0, 0, t form a basis for N(A) 33

36 7 Let A be the following matrix over Z 5 : A We find that A has reduced row echelon form B: B From B we read off the following: (a) The four rows of B form a basis for R(A) (Consequently the rows of A also form a basis for R(A) (b) The first four columns of A form a basis for C(A) (c) To find a basis for N(A), we solve AX 0 and equivalently BX 0 From B we see that the solution is x x 5 4x 6 3x 5 + x 6 x 4x 5 4x 6 x 5 + x 6 x 3 0 x 4 3x 5 x 5, where x 5 and x 6 are arbitrary elements of Z 5 Hence 3 X x x 6 0 0, 0 0 so [3,, 0,,, 0 t and [,, 0, 0, 0, t form a basis for N(A) 8 Let F {0,, a, b} be a field and let A be the following matrix over F: a b a A a b b a 34

37 In Section 6, problem 7, we found that A had reduced row echelon form B 0 0 b 0 0 From B we read off the following: (a) The rows of B form a basis for R(A) (Consequently the rows of A also form a basis for R(A) (b) The first three columns of A form a basis for C(A) (c) To find a basis for N(A), we solve AX 0 and equivalently BX 0 From B we see that the solution is x 0 x bx 4 bx 4 x 3 x 4 x 4, where x 4 is an arbitrary element of F Hence 0 X x 4 b, so [0, b,, t is a basis for N(A) 9 Suppose that X,,X m form a basis for a subspace S We have to prove that X, X + X,,X + + X m also form a basis for S First we prove the independence of the family: Suppose Then x X + x (X + X ) + + x m (X + + X m ) 0 (x + x + + x m )X + + x m X m 0 Then the linear independence of X,,X m gives x + x + + x m 0,,x m 0, 35

38 form which we deduce that x 0,,x m 0 Secondly we have to prove that every vector of S is expressible as a linear combination of X, X + X,,X + + X m Suppose X S Then We have to find x,,x m such that Then X a X + + a m X m X x X + x (X + X ) + + x m (X + + X m ) (x + x + + x m )X + + x m X m a X + + a m X m (x + x + + x m )X + + x m X m So if we can solve the system x + x + + x m a,,x m a m, we are finished Clearly these equations have the unique solution x a a,,x m a m a m, x m a m [ a b c 0 Let A If [a, b, c is a multiple of [,,, (that is, a b c), then ranka For if then [a, b, c t[,,, R(A) [a, b, c, [,, t[,,, [,, [,,, so [,, is a basis for R(A) However if [a, b, c is not a multiple of [,,, (that is at least two of a, b, c are distinct), then the left to right test shows that [a, b, c and [,, are linearly independent and hence form a basis for R(A) Consequently ranka in this case Let S be a subspace of F n with dims m Also suppose that X,,X m are vectors in S such that S X,,X m We have to prove that X,,X m form a basis for S; in other words, we must prove that X,,X m are linearly independent 36

39 However if X,,X m were linearly dependent, then one of these vectors would be a linear combination of the remaining vectors Consequently S would be spanned by m vectors But there exist a family of m linearly independent vectors in S Then by Theorem 33, we would have the contradiction m m Let [x, y, z t S Then x + y + 3z 0 Hence x y 3z and x y z y 3z y z y 0 + z Hence [,, 0 t and [ 3, 0, t form a basis for S Next ( ) + ( ) + 3() 0, so [,, t S To find a basis for S which includes [,, t, we note that [,, 0 t is not a multiple of [,, t Hence we have found a linearly independent family of two vectors in S, a subspace of dimension equal to Consequently these two vectors form a basis for S 3 Without loss of generality, suppose that X X Then we have the non trivial dependency relation: X + ( )X + 0X X m (a) Suppose that X m+ is a linear combination of X,,X m Then X,,X m, X m+ X,,X m and hence dim X,,X m, X m+ dim X,,X m (b) Suppose that X m+ is not a linear combination of X,,X m If not all of X,,X m are zero, there will be a subfamily X c,,x cr which is a basis for X,,X m Then as X m+ is not a linear combination of X c,,x cr, it follows that X c,,x cr, X m+ are linearly independent Also Consequently X,,X m, X m+ X c,,x cr, X m+ dim X,,X m, X m+ r + dim X,,X m + 37

40 Our result can be rephrased in a form suitable for the second part of the problem: dim X,,X m, X m+ dim X,,X m if and only if X m+ is a linear combination of X,,X m If X [x,,x n t, then AX B is equivalent to B x A + + x n A n So AX B is soluble for X if and only if B is a linear combination of the columns of A, that is B C(A) However by the first part of this question, B C(A) if and only if dimc([a B) dimc(a), that is, rank [A B rank A 5 Let a,,a n be elements of F, not all zero Let S denote the set of vectors [x,,x n t, where x,,x n satisfy a x + + a n x n 0 Then S N(A), where A is the row matrix [a,,a n Now ranka as A 0 So by the rank + nullity theorem, noting that the number of columns of A equals n, we have dim N(A) nullity (A) n ranka n 6 (a) (Proof of Lemma 3) Suppose that each of X,,X r is a linear combination of Y,,Y s Then s X i a ij Y j, j ( i r) Now let X r i x ix i be a linear combination of X,,X r Then X x (a Y + + a s Y s ) + + x r (a r Y + + a rs Y s ) y Y + + y s Y s, where y j a j x + +a rj x r Hence X is a linear combination of Y,,Y s Another way of stating Lemma 3 is X,,X r Y,,Y s, () 38

41 if each of X,,X r is a linear combination of Y,,Y s (b) (Proof of Theorem 3) Suppose that each of X,,X r is a linear combination of Y,,Y s and that each of Y,,Y s is a linear combination of X,,X r Then by (a) equation () above X,,X r Y,,Y s and Hence Y,,Y s X,,X r X,,X r Y,,Y s (c) (Proof of Corollary 3) Suppose that each of Z,,Z t is a linear combination of X,,X r Then each of X,,X r, Z,,Z t is a linear combination of X,,X r Also each of X,,X r is a linear combination of X,,X r, Z,,Z t, so by Theorem 3 X,,X r, Z,,Z t X,,X r (d) (Proof of Theorem 33) Let Y,,Y s be vectors in X,,X r and assume that s > r We have to prove that Y,,Y s are linearly dependent So we consider the equation x Y + + x s Y s 0 Now Y i r j a ijx j, for i s Hence x Y + + x s Y s x (a X + + a r X r ) + + x r (a s X + + a sr X r ) y X + + y r X r, () where y j a j x + + a sj x s However the homogeneous system y 0,, y r 0 has a non trivial solution x,,x s, as s > r and from (), this results in a non trivial solution of the equation x Y + + x s Y s 0 39

42 Hence Y,,Y s are linearly dependent 7 Let R and S be subspaces of F n, with R S We first prove dimr dims Let X,,X r be a basis for R Now by Theorem 35, because X,,X r form a linearly independent family lying in S, this family can be extended to a basis X,,X r,,x s for S Then dim S s r dimr Next suppose that dimr dims Let X,,X r be a basis for R Then because X,,X r form a linearly independent family in S and S is a subspace whose dimension is r, it follows from Theorem 343 that X,,X r form a basis for S Then S X,,X r R 8 Suppose that R and S are subspaces of F n with the property that R S is also a subspace of F n We have to prove that R S or S R We argue by contradiction: Suppose that R S and S R Then there exist vectors u and v such that u R and u S, v S and v R Consider the vector u+v As we are assuming R S is a subspace, R S is closed under addition Hence u + v R S and so u + v R or u + v S However if u + v R, then v (u + v) u R, which is a contradiction; similarly if u + v S Hence we have derived a contradiction on the asumption that R S and S R Consequently at least one of these must be false In other words R S or S R 9 Let X,,X r be a basis for S (i) First let Y a X + + a r X r () Y r a r X + + a rr X r, 40

43 where A [a ij is non singular Then the above system of equations can be solved for X,,X r in terms of Y,,Y r Consequently by Theorem 3 Y,,Y r X,,X r S It follows from problem that Y,,Y r is a basis for S (ii) We show that all bases for S are given by equations So suppose that Y,,Y r forms a basis for S Then because X,,X r form a basis for S, we can express Y,,Y r in terms of X,,X r as in, for some matrix A [a ij We show A is non singular by demonstrating that the linear independence of Y,,Y r implies that the rows of A are linearly independent So assume x [a,,a r + + x r [a r,,a rr [0,,0 Then on equating components, we have Hence a x + + a r x r 0 a r x + + a rr x r 0 x Y + + x r Y r x (a X + + a r X r ) + + x r (a r X + + a rr X r ) (a x + + a r x r )X + + (a r x + + a rr x r )X r 0X + + 0X r 0 Then the linear independence of Y,,Y r implies x 0,,x r 0 (We mention that the last argument is reversible and provides an alternative proof of part (i)) 4

44 P P 3 P O Section 4 We first prove that the area of a triangle P P P 3, where the points are in anti clockwise orientation, is given by the formula { } x x y y + x x 3 y y 3 + x 3 x y 3 y Referring to the above diagram, we have Area P P P 3 AreaOP P + AreaOP P 3 Area OP P 3 x x y y + x x 3 y y 3 x x 3 y y 3, which gives the desired formula We now turn to the area of a quadrilateral One possible configuration occurs when the quadrilateral is convex as in figure (a) below The interior diagonal breaks the quadrilateral into two triangles P P P 3 and P P 3 P 4 Then Area P P P 3 P 4 AreaP P P 3 + AreaP P 3 P 4 { x x y y + x x 3 y y 3 + x 3 x y 3 y } 4

45 P 4 P 3 (a) P P P 3 (b) P 4 P P + { } x x 3 y y 3 + x 3 x 4 y 3 y 4 + x 4 x y 4 y { } x x y y + x x 3 y y 3 + x 3 x 4 y 3 y 4 + x 4 x y 4 y, after cancellation Another possible configuration for the quadrilateral occurs when it is not convex, as in figure (b) The interior diagonal P P 4 then gives two triangles P P P 4 and P P 3 P 4 and we can proceed similarly as before a + x b + y c + z x + u y + v z + w u + a v + b w + c Now a b c x + u y + v z + w u + a v + b w + c a b c x + u y + v z + w u + a v + b w + c a b c x y z u + a v + b w + c + + x y z x + u y + v z + w u + a v + b w + c a b c u v w u + a v + b w + c a b c x y z u v w + a b c x y z a b c + a b c u v w u v w + a b c u v w a b c a b c x y z u v w Similarly x y z x + u y + v z + w u + a v + b w + c x y z u v w a b c x y z a b c u v w a b c x y z u v w 43

46 a b c Hence x y z u v w n (n + ) (n + ) 3 (n + ) (n + ) (n + 3) (n + ) (n + 3) (n + 4) C 3 C 3 C R 3 R 3 R R R R 4 (a) ( ) C 3 C 3 C C C C n n + (n + ) n + 3 (n + ) n + 5 n n + n + 0 n n n + n + 3 (n + ) n + 3 n + 5 (n + ) n + 5 n (b) ( ) det A

47 Hence A is non singular and A 3 adja C C C 3 C C C 3 3 C 3 C 3 C (i) a b b c b a a + c a + b a + b b (a+b) b a a + c a + b a + b b (a + b)(a b) (ii) b + c b c c c + a a b a a + b C 3 C 3 C R 3 R 3 R R R + R C C C a + b b C C C a + b b + a b + a b a a + c a + b a + b b (a+b) (a + b)(a b) 0 (b a) a a + c 0 a + b b c b c a c + a a b a a a + b c b 0 a c + a a b a a a a c b 0 a c + a b a a c b 0 a a c + a b c 0 a c b b c a(c + b ) 7 Suppose that the curve y ax + bx + c passes through the points (x, y ), (x, y ), (x 3, y 3 ), where x i x j if i j Then ax + bx + c y ax + bx + c y ax 3 + bx 3 + c y 3 The coefficient determinant is essentially a Vandermonde determinant: x x x x x 3 x 3 x x x 3 x x x 3 x x x 3 x x x (x x )(x 3 x )(x 3 x ) 3 45

48 Hence the coefficient determinant is non zero and by Cramer s rule, there is a unique solution for a, b, c 8 Let det A 3 k k 3 Then C 3 C 3 + C C C C 0 0 k + k 4 k + k 4 4 (k )(k + ) (k k 6) (k + 3)(k ) Hence det A 0 if and only if k 3 or k Consequently if k 3 and k, then deta 0 and the given system x + y z x + 3y + kz 3 x + ky + 3z has a unique solution We consider the cases k 3 and k separately k 3 : AM R R R 0 R R 3 R R 3 R 3 + 4R from which we read off inconsistency k : AM 3 3 R R R R 3 3 R 3 R R 3 R 3 R , We read off the complete solution x 5z, y 4z, where z is arbitrary 46

49 Finally we have to determine the solution for which x +y +z is least x + y + z (5z) + ( 4z) + z 4z 8z + { ( 4(z 4 z + 4 ) 4 z { ( 4 z ) } ) + 4 ( ) } We see that the least value of x +y +z is 4 3 and this occurs when z /, with corresponding values x 0/ and y 4 3/ b 9 Let a 0 be the coefficient determinant of the given system 5 0 Then expanding along column gives a 5 0 b a 0 ( ab) ab 4 (ab ) 88 3 Hence 0 if and only if ab Hence if ab, the given system has a unique solution If ab we must argue with care: AM b 3 a b 3 0 a ab 3a 0 5b 4 b 3 b 3 5b b ab 0 a ab 3a b 3 5b B 6 a Hence if 6 a 0, ie a 3, the system has no solution If a 3 (and hence b 4), then 4 3 B /3 / a 3

50 Consequently the complete solution of the system is x z, y z, where z is arbitrary Hence there are infinitely many solutions t t t 3 3 t t t t t t R 4 R 4 R R 3 R 3 R R R R R R R 3 0 t 0 t t (t ) t (t )(t ) t t t Hence 0 if and only if t or t Consequently the given matrix B is non singular if and only if t and t Let A be a 3 3 matrix with deta 0 Then (i) A adja (det A)I 3 () (det A)det (adja) det (deta I 3 ) (det A) 3 Hence, as det A 0, dividing out by deta in the last equation gives det (adja) (det A) (ii) Also from equation () ( ) deta A adja I 3, so adj A is non singular and Finally ( adja) deta A A adj (A ) (det A )I 3 and multiplying both sides of the last equation by A gives adj (A ) A(det A )I 3 det A A 48

51 Let A be a real 3 3 matrix satisfying A t A I 3 Then (i) A t (A I 3 ) A t A A t I 3 A t Taking determinants of both sides then gives (A t I 3 ) (A t I t 3) (A I 3 ) t deta t det(a I 3 ) det( (A I 3 ) t ) detadet (A I 3 ) ( ) 3 det (A I 3 ) t (ii) Also detaa t deti 3, so det(a I 3 ) () deta t det A (deta) Hence det A ± (iii) Suppose that deta Then equation () gives det(a I 3 ) det (A I 3 ), so ( + )det (A I 3 ) 0 and hence det(a I 3 ) 0 3 Suppose that column is a linear combination of the remaining columns: Then det A A x A + + x n A n x a + + x n a n a a n x a + + x n a n a a n x a n + + x n a nn a n a nn Now deta is unchanged in value if we perform the operation C C x C x n C n : 0 a a n 0 a a n det A 0 0 a n a nn 49

52 Conversely, suppose that deta 0 Then the homogeneous system AX 0 has a non trivial solution X [x,,x n t So x A + + x n A n 0 Suppose for example that x 0 Then ( A x x ) + + ( x ) n A n x and the first column of A is a linear combination of the remaining columns 4 Consider the system x + 3y z x + y z 4 x y + z 3 3 Let Hence the system has a unique solution which can be calculated using Cramer s rule: x, y, z 3, where 3 4, 6, Hence x 4 6 8, y 3, z 4 5 In Remark 404, take A I n Then we deduce (a) dete ij ; (b) dete i (t) t; 50

53 (c) dete ij (t) Now suppose that B is a non singular n n matrix Then we know that B is a product of elementary row matrices: Consequently we have to prove that B E E m dete E m A dete E m deta We prove this by induction on m First the case m We have to prove dete A dete deta if E is an elementary row matrix This follows form Remark 404: (a) dete ij A deta det E ij deta; (b) dete i (t)a t deta det E i (t) deta; (c) dete ij (t)a det A dete ij (t)det A Let m and assume the proposition holds for products of m elementary row matrices Then det E E m E m+ A det (E E m )(E m+ A) det (E E m ) det(e m+ A) det (E E m ) dete m+ det A det ((E E m )E m+ )det A and the induction goes through Hence det BA det B det A if B is non singular If B is singular, problem 6, Chapter 7 tells us that BA is also singlular However singular matrices have zero determinant, so det B 0 det BA 0, so the equation detba detb deta holds trivially in this case 6 a + b + c a + b a a a + b a + b + c a a a a a + b + c a + b a a a + b a + b + c 5

54 C C + C C 3 C 3 + C 7 Let we have R R R c c 0 0 R R R 3 b b + c b c b R 3 R 3 R c c a a a + b a + b + c c b b + c b c b b + c b c b 0 0 c c c 0 c c a a a + b a + b + c a a + b a + b + c b + c b c b c c 0 c 0 a a + b a + b + c c b + c b c a a + b + c c (b + c) a a + b + c c (b + c)(4a + b + c) + u u u u u + u u u u 3 u 3 + u 3 u 3 u 4 u 4 u 4 + u 4 (where t + u + u + u 3 + u 4 ) ( + u + u + u 3 + u 4 ) R R + R + R 3 + R 4 t t t t u + u u u u 3 u 3 + u 3 u 3 u 4 u 4 u 4 + u 4 Then using the operation u + u u u u 3 u 3 + u 3 u 3 u 4 u 4 u 4 + u 4 The last determinant equals C C C C 3 C 3 C C 4 C 4 C u 0 0 u u

55 8 Suppose that A t A, that A M n n (F), where n is odd Then det A t det( A) det A ( ) n deta det A Hence ( + )det A 0 and consequently deta 0 if + 0 in F 9 r r r r r r C 4 C 4 C 3 C 3 C 3 C C C C r r 0 0 r 0 r 0 r 0 0 r ( r) 3 0 a bc a 4 b ca b 4 c ab c 4 R R R R 3 R 3 R a bc a 4 0 b ca a + bc b 4 a 4 0 c ab a + bc c 4 a 4 b ca a + bc b 4 a 4 c ab a + bc c 4 a 4 (b a)(b + a) + c(b a) (b a)(b + a)(b + a ) (c a)(c + a) + b(c a) (c a)(c + a)(c + a ) (b a)(b + a + c) (b a)(b + a)(b + a ) (c a)(c + a + b) (c a)(c + a)(c + a ) (b a)(c a) b + a + c (b + a)(b + a ) c + a + b (c + a)(c + a ) (b a)(c a)(a + b + c) (b + a)(b + a ) (c + a)(c + a ) Finally (b + a)(b + a ) (c + a)(c + a ) (c 3 + ac + ca + a 3 ) (b 3 + ab + ba + a 3 ) (c 3 b 3 ) + a(c b ) + a (c b) (c b)(c + cb + b + a(c + b) + a ) (c b)(c + cb + b + ac + ab + a ) 53

56 Section 58 (i) ( 3 + i)(4 i) ( 3)(4 i) + i(4 i) {( 3)4 ( 3)(i)} + i(4) i(i) ( 4 + 6i) + (4i + ) i (ii) + 3i 4i ( + 3i)( + 4i) ( 4i)( + 4i) (( + 3i) + ( + 3i)(4i) i i (iii) ( + i) i + 4i + (i) i + 4i i i i ( 3 + 4i)( + i) 7 + i 7 + i (i) iz + ( 0i)z 3z + i z(i + 0i 3) i z( 9i) i z i + 9i i( 9i) i 8 9 i 4 (ii) The coefficient determinant is + i i + i 3 + i ( + i)(3 + i) ( i)( + i) + i 0 Hence Cramer s rule applies: there is a unique solution given by 3i i + i 3 + i 3 i z + 5i + i + i + i 3i + i + i 6 + 7i 9 8i w + i + i 5 54

57 3 + ( + i) + + ( + i) 99 ( + i)00 ( + i) Now ( + i) i Hence ( + i)00 i i { ( + i) 00 } ( + i) 00 (i) i ( ) 5 50 Hence i { ( + i) 00 } i( 50 ) ( 50 + )i 4 (i) Let z 8 6i and write zx+iy, where x and y are real Then z x y + xyi 8 6i, so x y 8 and xy 6 Hence y 3/x, x ( ) 3 8, x so x 4 + 8x 9 0 This is a quadratic in x Hence x or 9 and consequently x Hence x, y 3 or x and y 3 Hence z 3i or z + 3i (ii) z (3 + i)z i 0 has the solutions z (3 + i ± d)/, where d is any complex number satisfying d (3 + i) 4(4 + 3i) 8 6i Hence by part (i) we can take d 3i Consequently z 3 + i ± ( 3i) (i) The number lies in the first quadrant of the complex plane 4 + i Also Arg (4 + i) α, where tan α /4 and 0 < α < π/ Hence α tan (/4) i or + i y 4 + i α x 55

58 (ii) The number lies in the third quadrant of the complex plane 3 i 3 i ( 3) + ( ) Also Arg ( 3 i ) π + α, where tan α /3 /3 and 0 < α < π/ Hence α tan (/3) (iii) The number lies in the second quadrant of the complex plane + i ( ) + 5 Also Arg ( +i) π α, where tan α and 0 < α < π/ Hence α tan (iv) The number lies in the second quadrant of the complex plane + i 3 + i 3 ( ) + ( 3) + 3 Also Arg ( + 3 i) π α, where tan α 3 / 3 and 0 < α < π/ Hence α π/3 6 (i) Let z ( + i)( + 3i)( 3 i) Then 3 i α y + i y α + 3 i α y x x x z + i + 3i 3 i + + ( 3) ( 3) + ( ) Argz Arg ( + i) + Arg ( + 3) + Arg ( 3 i) (mod π) 56

59 Hence Arg z 5 π 4 + π 3 π 6 5 and the polar decomposition of z is z 4 ( cos 5π ) 5π + i sin (ii) Let z (+i)5 ( i 3) 5 ( 3+i) 4 Then z ( + i) 5 ( i ( ) 3) ( 3 + i) 4 4 7/ Argz Arg ( + i) 5 + Arg ( 3i) 5 Arg ( 3 + i) 4 (mod π) 5Arg ( + i) + 5Arg ( 3i) 4Arg ( 3 + i) 5 π ( ) π π 3 6 3π π Hence Arg z π and the polar decomposition of z is ( z 7/ cos π ) π + i sin 7 (i) Let z (cos π 4 + i sin π 4 ) and w 3(cos π 6 + i sin π 6 ) (Both of these numbers are already in polar form) (a) zw 6(cos ( π 4 + π 6 ) + i sin(π 4 + π 6 )) 6(cos 5π 5π + i sin ) (b) z w 3 (cos (π 4 π 6 ) + i sin(π 4 π 6 )) 3 (cos π + i sin π ) (c) w z 3 (cos (π 6 π 4 ) + i sin(π 6 π 4 )) 3 π π (cos ( ) + i sin( )) (d) z5 5 (cos ( 5π w 3 4 π 6 ) + i sin(5π 4 π 6 )) 3 9 (cos π (a) ( + i) i, so + i sin π ) ( + i) (i) 6 6 i 6 64(i ) 3 64( )

60 (b) ( i ) i, so ( i ) 6 ( ( i ) ) 3 ( i) 3 i 3 i i i 8 (i) To solve the equation z + 3i, we write + 3i in modulus argument form: + 3i (cos π 3 + i sin π 3 ) Then the solutions are z k ( ( π 3 cos + kπ ) + i sin Now k 0 gives the solution Clearly z z 0 z 0 (cos π 6 + i sin π 6 ) ( π 3 + kπ )), k 0, ( ) 3 + i 3 + i (ii) To solve the equation z 4 i, we write i in modulus argument form: Then the solutions are ( π z k cos + kπ ) + i sin 4 ( π ) Now cos +kπ 4 cos ( π 8 + kπ ), so z k ( π cos i cos π + i sin π 8 + kπ ( cos π + i sin π ( π + kπ 4 ) ( π + sin i k (cos π 8 + i sin π 8 ) ), k 0,,, 3 ) 8 + kπ ) k π (cos 8 + i sin π 8 ) 58

61 Geometrically, the solutions lie equi spaced on the unit circle at arguments π 8, π 8 + π 5π 8, π 8 + π 9π 8, π 8 + 3π 3π 8 Also z z 0 and z 3 z (iii) To solve the equation z 3 8i, we rewrite the equation as Then ( ) z, i ( ) z 3 i + 3i, or 3i Hence z i, 3 + i or 3 + i Geometrically, the solutions lie equi spaced on the circle z, at arguments π 6, π 6 + π 3 5π 6, π 6 + π 3 3π (iv) To solve z 4 i, we write i in modulus argument form: ( i 3/ cos π ) π + i sin 4 4 Hence the solutions are z k 3/8 cos ( π 4 + kπ 4 ) ( π + i sin 4 + kπ 4 We see the solutions can also be written as ( z k 3/8 i k cos π ) π + i sin ( 6 6 3/8 i k cos π 6 i sin π ) 6 ), k 0,,, 3 Geometrically, the solutions lie equi spaced on the circle z 3/8, at arguments π 6, π 6 + π 7π 6, π 6 + π 5π 6, π 6 + 3π 3π 6 Also z z 0 and z 3 z 59

62 9 + i + i + i + i + i + i + i R R ( + i)r R 3 R 3 ir R R R R 3 R 3 R i 0 0 i R R R R ir i The last matrix is in reduced row echelon form 0 (i) Let p l + im and z x + iy Then i + i + i i i pz + pz (l im)(x + iy) + (l + im)(x iy) i (lx + liy imx + my) + (lx liy + imx + my) (lx + my) Hence pz + pz n lx + my n (ii) Let w be the complex number which results from reflecting the complex number z in the line lx + my n Then because p is perpendicular to the given line, we have w z tp, t R (a) Also the midpoint w+z of the segment joining w and z lies on the given line, so ( ) ( ) w + z w + z p + p n, ( ) ( ) w + z w + z p + p n (b) Taking conjugates of equation (a) gives w z tp (c) Then substituting in (b), using (a) and (c), gives ( ) ( ) w tp z + tp p + p n 60

63 and hence pw + pz n (iii) Let p b a and n b a Then z a z b z a z b (z a)(z a) (z b)(z b) (z a)(z a) (z b)(z b) zz az za + aa zz bz zb + bb (b a)z + (b a)z b a pz + pz n Suppose z lies on the circle z a z b and let w be the reflection of z in the line pz + pz n Then by part (ii) pw + pz n Taking conjugates gives pw + pz n and hence z n pw p (a) Substituting for z in the circle equation, using (a) gives n pw p a λ b n pw pa n pw pb n pw p (b) However n pa b a (b a)a bb aa ba + aa b(b a) bp Similarly n pb ap Consequently (b) simplifies to λ bp pw ap pw b w a w w b w a, which gives λ w a w b 6

64 Let a and b be distinct complex numbers and 0 < α < π (i) When z lies on the circular arc shown, it subtends a constant angle α This angle is given by Arg (z a) Arg (z b) However ( ) z a Arg Arg (z a) Arg (z b) + kπ z b α + kπ It follows that k 0, as 0 < α < π and π < Argθ π Hence ( ) z a Arg α z b Similarly if z lies on the circular arc shown, then ( ) z a Arg γ (π α) α π z b Replacing α by π α, we deduce that if z 4 lies on the circular arc shown, then ( ) z4 a Arg π α, z 4 b while if z 3 lies on the circular arc shown, then ( ) z3 a Arg α z 3 b The straight line through a and b has the equation z ( t)a + tb, 6

65 where t is real Then 0 < t < describes the segment ab Also z a z b t t Hence z a z b is real and negative if z is on the segment a, but is real and positive if z is on the remaining part of the line, with corresponding values ( ) z a Arg π, 0, z b respectively (ii) Case (a) Suppose z, z and z 3 are not collinear Then these points determine a circle Now z and z partition this circle into two arcs If z 3 and z 4 lie on the same arc, then ( ) ( ) z3 z z4 z Arg Arg ; z 3 z z 4 z whereas if z 3 and z 4 lie on opposite arcs, then ( ) z3 z Arg α z 3 z and Hence in both cases ( z3 z Arg / z ) 4 z z 3 z z 4 z In other words, the cross ratio ( ) z4 z Arg α π z 4 z ( ) ( ) z3 z z4 z Arg Arg z 3 z z 4 z 0 or π z 3 z z 3 z / z 4 z z 4 z is real (b) If z, z and z 3 are collinear, then again the cross ratio is real The argument is reversible (mod π) (iii) Assume that A, B, C, D are distinct points such that the cross ratio r z 3 z z 3 z / z 4 z z 4 z is real Now r cannot be 0 or Then there are three cases: 63

66 (i) 0 < r < ; (ii) r < 0; (iii) r > Case (i) Here r + r So z 4 z z3 z z 4 z z 3 z + ( ) z4 z z3 z z 4 z z 3 z Multiplying both sides by the denominator z 4 z z 3 z gives after simplification or z 4 z z 3 z + z z z 4 z 3 z 4 z z 3 z, (a) AD BC + AB CD BD AC Case (ii) Here + r r This leads to the equation (b) BD AC + AD BC+ AB CD Case (iii) Here + r r This leads to the equation (c) BD AC + AB CD AD BC Conversely if (a), (b) or (c) hold, then we can reverse the argument to deduce that r is a complex number satisfying one of the equations r + r, + r r, + r r, from which we deduce that r is real 64

67 Section 63 [ 4 3 Let A Then A has characteristic equation λ 0 4λ or (λ 3)(λ ) 0 Hence the eigenvalues of A are λ 3 and λ λ 3 The corresponding eigenvectors satisfy (A λ I )X 0, or [ 3 3 or equivalently x 3y 0 Hence [ [ x 3y y y and we take X [ 3 [ 0 0, [ 3 y Similarly for λ we find the eigenvector X [ 3 Hence if P [X X P AP [, then P is non singular and [ Hence and consequently A n A P [ P [ 3 n 0 P 0 n P [ [ [ 3 3 n 0 0 n 3 [ [ 3 n+ 3 n 3 [ 3 n+ 3 n n 3 n + 3 3n A + 3 3n I 65

68 [ 3/5 4/5 Let A /5 /5 λ /5, with corresponding eigenvectors [ X and X Then if P [X X, P is non singular and [ P 0 AP and A P 0 /5 Then we find that the eigenvalues are λ and [ [ 0 0 /5 P Hence [ A n 0 P 0 ( /5) n P [ 0 P P 0 0 [ [ [ [ [ [ [ /3 /3 3 /3 /3 3 The given system of differential equations is equivalent to Ẋ AX, where [ [ 3 x A and X 5 4 y [ The matrix P is a non-singular matrix of eigenvectors corresponding to eigenvalues λ and λ Then 5 [ P 0 AP 0 The substitution X PY, where Y [x, y t, gives Ẏ [ 0 0 Y, 66

69 or equivalently x x and y y Hence x x (0)e t and y y (0)e t To determine x (0) and y (0), we note that [ [ x (0) P x(0) [ [ [ 3 3 y (0) y(0) Hence x 3e t and y 7e t Consequently x x + y 6e t + 7e t and y 5x + y 5e t + 7e t 4 Introducing the vector X n [ xn y n, the system of recurrence relations x n+ 3x n y n y n+ x n + 3y n, [ 3 becomes X n+ AX n, where A Hence X 3 n A n X 0, where [ X 0 To find A n we can use the eigenvalue method We get A n [ n + 4 n n 4 n n 4 n n + 4 n Hence [ n + 4 n n 4 n [ X n n 4 n n + 4 n [ n + 4 n + ( n 4 n ) n 4 n + ( n + 4 n ) [ 3 n 4 n 3 n + 4 n [ (3 n 4 n )/ (3 n + 4 n )/ Hence x n (3 n 4 n ) and y n (3 n + 4 n ) [ a b 5 Let A be a real or complex matrix with distinct eigenvalues c d λ, λ and corresponding eigenvectors X, X Also let P [X X (a) The system of recurrence relations x n+ y n+ ax n + by n cx n + dy n 67