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PROBLEMS 1.6 ............................................ 1

PROBLEMS 2.4 ............................................ 12

PROBLEMS 2.7 ............................................ 18

PROBLEMS 3.6 ............................................ 32

PROBLEMS 4.1 ............................................ 45

PROBLEMS 5.8 ............................................ 58

PROBLEMS 6.3 ............................................ 69

PROBLEMS 7.3 ............................................ 83

PROBLEMS 8.8 ............................................ 91

SECTION 1:6

2. (i)

0 0 0

2 4 0

R1 $ R2

2 4 0

0 0 0

R1 ! 1

2R1

1 2 0

0 0 0

;

(ii)

0 1 3

1 2 4

R1 $ R2

1 2 4

0 1 3

R1 ! R1 ¡ 2R2

1 0 ¡2

0 1 3

;

(iii)

1 1 1

1 1 0

1 0 0

5 R2 ! R2 ¡ R1

R3 ! R3 ¡ R1

1 1 0

0 0 ¡1

0 ¡1 ¡1

R1 ! R1 + R3

R3 ! ¡R3

R2 $ R3

1 0 0

0 1 1

0 0 ¡1

5 R2 ! R2 + R3

R3 ! ¡R3

1 0 0

0 1 0

0 0 1

(iv)

2 0 0

0 0 0

¡4 0 0

5 R3 ! R3 + 2R1

R1 ! 1

2R1

1 0 0

0 0 0

3. (a)

1 1 1 2

2 3 ¡1 8

1 ¡1 ¡1 ¡8

5 R2 ! R2 ¡ 2R1

R3 ! R3 ¡ R1

1 1 1 2

0 1 ¡3 4

0 ¡2 ¡2 ¡10

R1 ! R1 ¡ R2

R3 ! R3 + 2R2

1 0 4 ¡2

0 1 ¡3 4

0 0 ¡8 ¡2

5R3 ! ¡1

8 R3

1 0 4 2

0 1 ¡3 4

0 0 1 1

R1 ! R1 ¡ 4R3

R2 ! R2 + 3R3

1 0 0 ¡3

0 1 0 19

0 0 1 1

The augmented matrix has been converted to reduced row{echelon form

and we read o® the unique solution x = ¡3; y = 19

4 ; z = 1

4 .

(b)

1 1 ¡1 2 10

3 ¡1 7 4 1

¡5 3 ¡15 ¡6 9

5 R2 ! R2 ¡ 3R1

R3 ! R3 + 5R1

1 1 ¡1 2 10

0 ¡4 10 ¡2 ¡29

0 8 ¡20 4 59

R3 ! R3 + 2R2

1 1 ¡1 2 10

0 ¡4 10 ¡2 ¡29

0 0 0 0 1

From the last matrix we see that the original system is inconsistent.

(c)

664

3 ¡1 7 0

2 ¡1 4 1

1 ¡1 1 1

6 ¡4 10 3

775

R1 $ R3

664

1 ¡1 1 1

2 ¡1 4 1

3 ¡1 7 0

6 ¡4 10 3

775

R2 ! R2 ¡ 2R1

R3 ! R3 ¡ 3R1

R4 ! R4 ¡ 6R1

664

1 ¡1 1 1

0 1 2 ¡3

0 2 4 ¡3

775

R1 ! R1 + R2

R4 ! R4 ¡ R3

R3 ! R3 ¡ 2R2

664

1 0 3 ¡1

0 1 2 ¡3

0 0 0 0

775

The augmented matrix has been converted to reduced row{echelon form

and we read o® the complete solution x = ¡1

2 ¡ 3z; y = ¡3

2 ¡ 2z, with z

arbitrary.

2 ¡1 3 a

3 1 ¡5 b

¡5 ¡5 21 c

5R2 ! R2 ¡ R1

2 ¡1 3 a

1 2 ¡8 b ¡ a

¡5 ¡5 21 c

R1 $ R2

1 2 ¡8 b ¡ a

2 ¡1 3 a

¡5 ¡5 21 c

5 R2 ! R2 ¡ 2R1

R3 ! R3 + 5R1

1 2 ¡8 b ¡ a

0 ¡5 19 ¡2b + 3a

0 5 ¡19 5b ¡ 5a + c

R3 ! R3 + R2

R2 ! ¡1

5 R2

1 2 ¡8 b ¡ a

0 1 ¡19

2b¡3a

0 0 0 3b ¡ 2a + c

R1 ! R1 ¡ 2R2

1 0 ¡2

(b+a)

0 1 ¡19

2b¡3a

0 0 0 3b ¡ 2a + c

From the last matrix we see that the original system is inconsistent if

3b¡2a+c 6= 0. If 3b¡2a+c = 0, the system is consistent and the solution

x =

(b + a)

z; y =

(2b ¡ 3a)

where z is arbitrary.

1 1 1

t 1 t

1 + t 2 3

5 R2 ! R2 ¡ tR1

R3 ! R3 ¡ (1 + t)R1

1 1 1

0 1 ¡ t 0

0 1 ¡ t 2 ¡ t

R3 ! R3 ¡ R2

1 1 1

0 1 ¡ t 0

0 0 2 ¡ t

5 = B:

Case 1. t 6= 2. No solution.

Case 2. t = 2. B =

1 0 1

0 ¡1 0

0 0 0

5 !

1 0 1

0 1 0

0 0 0

5 :

We read o® the unique solution x = 1; y = 0.

6. M2ethod 1.

664

¡3 1 1 1

1 ¡3 1 1

1 1 ¡3 1

1 1 1 ¡3

775

R1 ! R1 ¡ R4

R2 ! R2 ¡ R4

R3 ! R3 ¡ R4

664

¡4 0 0 4

0 ¡4 0 4

0 0 ¡4 4

1 1 1 ¡3

775

664

1 0 0 ¡1

0 1 0 ¡1

0 0 1 ¡1

1 1 1 ¡3

775

R4 ! R4 ¡ R3 ¡ R2 ¡ R1

664

1 0 0 ¡1

0 1 0 ¡1

0 0 1 ¡1

0 0 0 0

775

Hence the given homogeneous system has complete solution

x1 = x4; x2 = x4; x3 = x4;

with x4 arbitrary.

Method 2. Write the system as

x1 + x2 + x3 + x4 = 4x1

x1 + x2 + x3 + x4 = 4x2

x1 + x2 + x3 + x4 = 4x3

x1 + x2 + x3 + x4 = 4x4:

Then it is immediate that any solution must satisfy x1 = x2 = x3 = x4.

Conversely, if x1; x2; x3; x4 satisfy x1 = x2 = x3 = x4, we get a solution.

¸ ¡ 3 1

1 ¸ ¡ 3

R1 $ R2

1 ¸ ¡ 3

¸ ¡ 3 1

R2 ! R2 ¡ (¸ ¡ 3)R1

1 ¸ ¡ 3

0 ¡¸2 + 6¸ ¡ 8

= B:

Case 1: ¡¸2 + 6¸ ¡ 8 6= 0. That is ¡(¸ ¡ 2)(¸ ¡ 4) 6= 0 or ¸ 6= 2; 4. Here B is

row equivalent to

1 0

0 1

R2 ! 1

¡¸2+6¸¡8R2

1 ¸ ¡ 3

0 1

R1 ! R1 ¡ (¸ ¡ 3)R2

1 0

0 1

Hence we get the trivial solution x = 0; y = 0.

Case 2: ¸ = 2. Then B =

1 ¡1

0 0

and the solution is x = y, with y

arbitrary.

Case 3: ¸ = 4. Then B =

1 1

0 0

and the solution is x = ¡y, with y

arbitrary.

3 1 1 1

5 ¡1 1 ¡1

R1 !

1 1

5 ¡1 1 ¡1

R2 ! R2 ¡ 5R1

1 1

0 ¡8

3 ¡2

3 ¡8

R2 ! ¡3

1 1

0 1 1

4 1

R1 ! R1 ¡

1 0 1

4 0

0 1 1

4 1

Hence the solution of the associated homogeneous system is

x1 = ¡

x3; x2 = ¡

x3 ¡ x4;

with x3 and x4 arbitrary.

A =

6664

1 ¡ n 1 ¢ ¢ ¢ 1

1 1 ¡ n ¢ ¢ ¢ 1

...

¢ ¢ ¢

...

1 1 ¢ ¢ ¢ 1 ¡ n

7775

R1 ! R1 ¡ Rn

R2 ! R2 ¡ Rn

...

Rn¡1 ! Rn¡1 ¡ Rn

6664

¡n 0 ¢ ¢ ¢ n

0 ¡n ¢ ¢ ¢ n

...

¢ ¢ ¢

...

1 1 ¢ ¢ ¢ 1 ¡ n

7775

6664

1 0 ¢ ¢ ¢ ¡1

0 1 ¢ ¢ ¢ ¡1

...

¢ ¢ ¢

...

1 1 ¢ ¢ ¢ 1 ¡ n

7775

Rn ! Rn ¡ Rn¡1 ¢ ¢ ¢ ¡ R1

6664

1 0 ¢ ¢ ¢ ¡1

0 1 ¢ ¢ ¢ ¡1

...

¢ ¢ ¢

...

0 0 ¢ ¢ ¢ 0

7775

The last matrix is in reduced row{echelon form.

Consequently the homogeneous system with coe±cient matrix A has the

solution

x1 = xn; x2 = xn; : : : ; xn¡1 = xn;

with xn arbitrary.

Alternatively, writing the system in the form

x1 + ¢ ¢ ¢ + xn = nx1

x1 + ¢ ¢ ¢ + xn = nx2

...

x1 + ¢ ¢ ¢ + xn = nxn

shows that any solution must satisfy nx1 = nx2 = ¢ ¢ ¢ = nxn, so x1 = x2 =

¢ ¢ ¢ = xn. Conversely if x1 = xn; : : : ; xn¡1 = xn, we see that x1; : : : ; xn is a

solution.

10. Let A =

a b

c d

and assume that ad ¡ bc 6= 0.

Case 1: a 6= 0.

a b

c d

R1 ! 1

aR1

1 b

c d

R2 ! R2 ¡ cR1

1 b

0 ad¡bc

R2 ! a

ad¡bcR2

1 b

0 1

R1 ! R1 ¡ b

aR2

1 0

0 1

Case 2: a = 0. Then bc 6= 0 and hence c 6= 0.

A =

0 b

c d

R1 $ R2

c d

0 b

1 d

0 1

1 0

0 1

So in both cases, A has reduced row{echelon form equal to

1 0

0 1

11. We simplify the augmented matrix of the system using row operations:

1 2 ¡3 4

3 ¡1 5 2

4 1 a2 ¡ 14 a + 2

5 R2 ! R2 ¡ 3R1

R3 ! R3 ¡ 4R1

1 2 ¡3 4

0 ¡7 14 ¡10

0 ¡7 a2 ¡ 2 a ¡ 14

R3 ! R3 ¡ R2

R2 ! ¡1

7 R2

R1 ! R1 ¡ 2R2

1 2 ¡3 4

0 1 ¡2 10

0 0 a2 ¡ 16 a ¡ 4

5 R1 ! R1 ¡ 2R2

1 0 1 8

0 1 ¡2 10

0 0 a2 ¡ 16 a ¡ 4

5 :

Denote the last matrix by B.

Case 1: a2 ¡ 16 6= 0. i.e. a 6= §4. Then

R3 ! 1

a2¡16R3

R1 ! R1 ¡ R3

R2 ! R2 + 2R3

1 0 0 8a+25

7(a+4)

0 1 0 10a+54

7(a+4)

0 0 1 1

a+4

and we get the unique solution

x =

8a + 25

7(a + 4)

; y =

10a + 54

7(a + 4)

; z =

a + 4

Case 2: a = ¡4. Then B =

1 0 1 8

0 1 ¡2 10

0 0 0 ¡8

5, so our system is inconsistent.

Case 3: a = 4. Then B =

1 0 1 8

0 1 ¡2 10

0 0 0 0

5. We read o® that the system is

consistent, with complete solution x = 8

7 ¡ z; y = 10

7 + 2z, where z is

arbitrary.

12. We reduce the augmented array of the system to reduced row{echelon

form:

664

1 0 1 0 1

0 1 0 1 1

1 1 1 1 0

0 0 1 1 0

775

R3 ! R3 + R1

664

1 0 1 0 1

0 1 0 1 1

0 0 1 1 0

775

R3 ! R3 + R2

664

1 0 1 0 1

0 1 0 1 1

0 0 0 0 0

0 0 1 1 0

775

R1 ! R1 + R4

R3 $ R4

664

1 0 0 1 1

0 1 0 1 1

0 0 1 1 0

0 0 0 0 0

775

The last matrix is in reduced row{echelon form and we read o® the solution

of the corresponding homogeneous system:

x1 = ¡x4 ¡ x5 = x4 + x5

x2 = ¡x4 ¡ x5 = x4 + x5

x3 = ¡x4 = x4;

where x4 and x5 are arbitrary elements of Z2. Hence there are four solutions:

x1 x2 x3 x4 x5

0 0 0 0 0

1 1 0 0 1

1 1 1 1 0

0 0 1 1 1

13. (a) We reduce the augmented matrix to reduced row{echelon form:

2 1 3 4

4 1 4 1

3 1 2 0

5 R1 ! 3R1

1 3 4 2

4 1 4 1

3 1 2 0

R2 ! R2 + R1

R3 ! R3 + 2R1

1 3 4 2

0 4 3 3

0 2 0 4

5 R2 ! 4R2

1 3 4 2

0 1 2 2

0 2 0 4

R1 ! R1 + 2R2

R3 ! R3 + 3R2

1 0 3 1

0 1 2 2

0 0 1 0

5 R1 ! R1 + 2R3

R2 ! R2 + 3R3

1 0 0 1

0 1 0 2

0 0 1 0

5 :

Consequently the system has the unique solution x = 1; y = 2; z = 0.

(b) Again we reduce the augmented matrix to reduced row{echelon form:

2 1 3 4

4 1 4 1

1 1 0 3

5 R1 $ R3

1 1 0 3

4 1 4 1

2 1 3 4

R2 ! R2 + R1

R3 ! R3 + 3R1

1 1 0 3

0 2 4 4

0 4 3 3

5 R2 ! 3R2

1 1 0 3

0 1 2 2

0 4 3 3

R1 ! R1 + 4R2

R3 ! R3 + R2

1 0 3 1

0 1 2 2

0 0 0 0

5 :

We read o® the complete solution

x = 1 ¡ 3z = 1 + 2z

y = 2 ¡ 2z = 2 + 3z;

where z is an arbitrary element of Z5.

14. Suppose that (®1; : : : ; ®n) and (¯1; : : : ; ¯n) are solutions of the system

of linear equations

j=1

aijxj = bi; 1 · i · m:

Then

j=1

aij®j = bi and

j=1

aij¯j = bi

for 1 · i · m.

Let °i = (1 ¡t)®i +t¯i for 1 · i · m. Then (°1; : : : ; °n) is a solution of

the given system. For

j=1

aij°j =

j=1

aijf(1 ¡ t)®j + t¯jg

j=1

aij(1 ¡ t)®j +

j=1

aijt¯j

= (1 ¡ t)bi + tbi

= bi:

15. Suppose that (®1; : : : ; ®n) is a solution of the system of linear equations

j=1

aijxj = bi; 1 · i · m: (1)

Then the system can be rewritten as

j=1

aijxj =

j=1

aij®j ; 1 · i · m;

or equivalently

j=1

aij(xj ¡ ®j) = 0; 1 · i · m:

So we have

j=1

aijyj = 0; 1 · i · m:

where xj ¡ ®j = yj . Hence xj = ®j + yj ; 1 · j · n, where (y1; : : : ; yn) is

a solution of the associated homogeneous system. Conversely if (y1; : : : ; yn)

is a solution of the associated homogeneous system and xj = ®j + yj ; 1 ·

j · n, then reversing the argument shows that (x1; : : : ; xn) is a solution of

the system 1 .

16. We simplify the augmented matrix using row operations, working to-

wards row{echelon form:

1 1 ¡1 1 1

a 1 1 1 b

3 2 0 a 1 + a

5 R2 ! R2 ¡ aR1

R3 ! R3 ¡ 3R1

1 1 ¡1 1 1

0 1 ¡ a 1 + a 1 ¡ a b ¡ a

0 ¡1 3 a ¡ 3 a ¡ 2

R2 $ R3

R2 ! ¡R2

1 1 ¡1 1 1

0 1 ¡3 3 ¡ a 2 ¡ a

0 1 ¡ a 1 + a 1 ¡ a b ¡ a

R3 ! R3 + (a ¡ 1)R2

1 1 ¡1 1 1

0 1 ¡3 3 ¡ a 2 ¡ a

0 0 4 ¡ 2a (1 ¡ a)(a ¡ 2) ¡a2 + 2a + b ¡ 2

5 = B:

Case 1: a 6= 2. Then 4 ¡ 2a 6= 0 and

B !

1 1 ¡1 1 1

0 1 ¡3 3 ¡ a 2 ¡ a

0 0 1 a¡1

2 ¡a2+2a+b¡2

4¡2a

5 :

Hence we can solve for x; y and z in terms of the arbitrary variable w.

Case 2: a = 2. Then

B =

1 1 ¡1 1 1

0 1 ¡3 1 0

0 0 0 0 b ¡ 2

5 :

Hence there is no solution if b 6= 2. However if b = 2, then

B =

1 1 ¡1 1 1

0 1 ¡3 1 0

0 0 0 0 0

5 !

1 0 2 0 1

0 1 ¡3 1 0

0 0 0 0 0

and we get the solution x = 1 ¡ 2z; y = 3z ¡ w, where w is arbitrary.

17. (a) We ¯rst prove that 1 + 1 + 1 + 1 = 0. Observe that the elements

1 + 0; 1 + 1; 1 + a; 1 + b

are distinct elements of F by virtue of the cancellation law for addition. For

this law states that 1+x = 1+y ) x = y and hence x 6= y ) 1+x 6= 1+y.

Hence the above four elements are just the elements 0; 1; a; b in some

order. Consequently

(1 + 0) + (1 + 1) + (1 + a) + (1 + b) = 0 + 1 + a + b

(1 + 1 + 1 + 1) + (0 + 1 + a + b) = 0 + (0 + 1 + a + b);

so 1 + 1 + 1 + 1 = 0 after cancellation.

Now 1 + 1 + 1 + 1 = (1 + 1)(1 + 1), so we have x2 = 0, where x = 1 + 1.

Hence x = 0. Then a + a = a(1 + 1) = a ¢ 0 = 0.

Next a + b = 1. For a + b must be one of 0; 1; 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 = 1. Then

a + 1 = a + (a + b) = (a + a) + b = 0 + b = b:

Similarly b + 1 = a. Consequently the addition table for F is

+ 0 1 a b

0 0 1 a b

1 1 0 b a

a a b 0 1

b b a 1 0

We now ¯nd the multiplication table. First, ab must be one of 1; a; b;

however we can't have ab = a or b, so this leaves ab = 1.

Next a2 = b. For a2 must be one of 1; a; b; however a2 = a ) a = 0 or

a = 1; also

a2 = 1 ) a2 ¡ 1 = 0 ) (a ¡ 1)(a + 1) = 0 ) (a ¡ 1)2 = 0 ) a = 1;

hence a2 = b. Similarly b2 = a. Consequently the multiplication table for F

£ 0 1 a b

0 0 0 0 0

1 0 1 a b

a 0 a b 1

b 0 b 1 a

(b) We use the addition and multiplication tables for F:

A =

1 a b a

a b b 1

1 1 1 a

5 R2 ! R2 + aR1

R3 ! R3 + R1

1 a b a

0 0 a a

0 b a 0

R2 $ R3

1 a b a

0 b a 0

0 0 a a

5 R2 ! aR2

R3 ! bR3

1 a b a

0 1 b 0

0 0 1 1

R1 $ R1 + aR2

1 0 a a

0 1 b 0

0 0 1 1

5 R1 ! R1 + aR3

R2 ! R2 + bR3

1 0 0 0

0 1 0 b

0 0 1 1

5 :

The last matrix is in reduced row{echelon form.

Section 2:4

2. Suppose B =

a b

c d

e f

5 and that AB = I2. Then

¡1 0 1

0 1 0

¸ 2

a b

c d

e f

5 =

1 0

0 1

¡a + e ¡b + f

c + e d + f

Hence

¡a + e = 1

c + e = 0

; ¡b + f = 0

d + f = 1

;

e = a + 1

c = ¡e = ¡(a + 1)

;

f = b

d = 1 ¡ f = 1 ¡ b

;

B =

a b

¡a ¡ 1 1 ¡ b

a + 1 b

5 :

Next,

(BA)2B = (BA)(BA)B = B(AB)(AB) = BI2I2 = BI2 = B

4. Let pn denote the statement

An = (3n¡1)

2 A + (3¡3n)

2 I2:

Then p1 asserts that A = (3¡1)

2 A + (3¡3)

2 I2; which is true. So let n ¸ 1 and

assume pn. Then from (1),

An+1 = A ¢ An = A

(3n¡1)

2 A + (3¡3n)

2 I2

= (3n¡1)

2 A2 + (3¡3n)

2 A

= (3n¡1)

2 (4A ¡ 3I2) + (3¡3n)

2 A = (3n¡1)4+(3¡3n)

2 A + (3n¡1)(¡3)

2 I2

= (4¢3n¡3n)¡1

2 A + (3¡3n+1)

2 I2

= (3n+1¡1)

2 A + (3¡3n+1)

2 I2:

Hence pn+1 is true and the induction proceeds.

5. The equation xn+1 = axn + bxn¡1 is seen to be equivalent to

xn+1

a b

1 0

¸ ·

xn¡1

Xn = AXn¡1;

where Xn =

xn+1

and A =

a b

1 0

. Then

Xn = AnX0

if n ¸ 1. Hence by Question 3,

xn+1

(3n ¡ 1)

A +

(3 ¡ 3n)

¾·

(3n ¡ 1)

4 ¡3

1 0

· 3¡3n

2 0

0 3¡3n

¸¾·

(3n ¡ 1)2 + 3¡3n

2 (3n ¡ 1)(¡3)

3n¡1

3¡3n

Hence, equating the (2; 1) elements gives

xn =

(3n ¡ 1)

x1 +

(3 ¡ 3n)

x0 if n ¸ 1

7. Note: ¸1 + ¸2 = a + d and ¸1¸2 = ad ¡ bc.

Then

(¸1 + ¸2)kn ¡ ¸1¸2kn¡1 = (¸1 + ¸2)(¸n¡1

1 + ¸n¡2

1 ¸2 + ¢ ¢ ¢ + ¸1¸n¡2

2 + ¸n¡1

2 )

¡¸1¸2(¸n¡2

1 + ¸n¡3

1 ¸2 + ¢ ¢ ¢ + ¸1¸n¡3

2 + ¸n¡2

2 )

= (¸n

1 + ¸n¡1

1 ¸2 + ¢ ¢ ¢ + ¸1¸n¡1

2 )

+(¸n¡1

1 ¸2 + ¢ ¢ ¢ + ¸1¸n¡1

2 + ¸n

2 )

¡(¸n¡1

1 ¸2 + ¢ ¢ ¢ + ¸1¸n¡1

2 )

= ¸n

1 + ¸n¡1

1 ¸2 + ¢ ¢ ¢ + ¸1¸n¡1

2 + ¸n

2 = kn+1

If ¸1 = ¸2, we see

kn = ¸n¡1

1 + ¸n¡2

1 ¸2 + ¢ ¢ ¢ + ¸1¸n¡2

2 + ¸n¡1

= ¸n¡1

1 + ¸n¡2

1 ¸1 + ¢ ¢ ¢ + ¸1¸n¡2

1 + ¸n¡1

= n¸n¡1

If ¸1 6= ¸2, we see that

(¸1 ¡ ¸2)kn = (¸1 ¡ ¸2)(¸n¡1

1 + ¸n¡2

1 ¸2 + ¢ ¢ ¢ + ¸1¸n¡2

2 + ¸n¡1

2 )

= ¸n

1 + ¸n¡1

1 ¸2 + ¢ ¢ ¢ + ¸1¸n¡1

¡(¸n¡1

1 ¸2 + ¢ ¢ ¢ + ¸1¸n¡1

2 + ¸n

2 )

= ¸n

1 ¡ ¸n

2 :

Hence kn = ¸n

1 ¡¸n

¸1¡¸2

We have to prove

An = knA ¡ ¸1¸2kn¡1I2: ¤

n=1:

A1 = A; also k1A ¡ ¸1¸2k0I2 = k1A ¡ ¸1¸20I2

= A:

Let n ¸ 1 and assume equation ¤ holds. Then

An+1 = An ¢ A = (knA ¡ ¸1¸2kn¡1I2)A

= knA2 ¡ ¸1¸2kn¡1A:

Now A2 = (a + d)A ¡ (ad ¡ bc)I2 = (¸1 + ¸2)A ¡ ¸1¸2I2: Hence

An+1 = kn(¸1 + ¸2)A ¡ ¸1¸2I2 ¡ ¸1¸2kn¡1A

= fkn(¸1 + ¸2) ¡ ¸1¸2kn¡1gA ¡ ¸1¸2knI2

= kn+1A ¡ ¸1¸2knI2;

and the induction goes through.

8. Here ¸1; ¸2 are the roots of the polynomial x2 ¡ 2x ¡ 3 = (x ¡ 3)(x + 1).

So we can take ¸1 = 3; ¸2 = ¡1: Then

kn =

3n ¡ (¡1)n

3 ¡ (¡1)

3n + (¡1)n+1

Hence

An =

3n + (¡1)n+1

A ¡ (¡3)

3n¡1 + (¡1)n

3n + (¡1)n+1

1 2

2 1

+ 3

3n¡1 + (¡1)n

¾·

1 0

0 1

;

which is equivalent to the stated result.

9. In terms of matrices, we have

Fn+1

1 1

1 0

¸ ·

Fn¡1

for n ¸ 1:

Fn+1

1 1

1 0

¸n ·

1 1

1 0

¸n ·

Now ¸1; ¸2 are the roots of the polynomial x2 ¡ x ¡ 1 here.

Hence ¸1 = 1+p5

2 and ¸2 = 1¡

2 and

kn =

1+p5

´n¡1

1¡

´n¡1

1+p5

2 ¡

1¡

1+p5

´n¡1

1¡

´n¡1

Hence

An = knA ¡ ¸1¸2kn¡1I2

= knA + kn¡1I2

Fn+1

= (knA + kn¡1I2)

= kn

+ kn¡1

kn + kn¡1

Hence

Fn = kn =

1+p5

´n¡1

1¡

´n¡1

10. From Question 5, we know that

1 r

1 1

¸n ·

Now by Question 7, with A =

1 r

1 1

An = knA ¡ ¸1¸2kn¡1I2

= knA ¡ (1 ¡ r)kn¡1I2;

where ¸1 = 1 + pr and ¸2 = 1 ¡ pr are the roots of the polynomial

x2 ¡ 2x + (1 ¡ r) and

kn =

¸n

1 ¡ ¸n

2pr

Hence

= (knA ¡ (1 ¡ r)kn¡1I2)

µ·

kn knr

kn kn

(1 ¡ r)kn¡1 0

0 (1 ¡ r)kn¡1

¸¶·

kn ¡ (1 ¡ r)kn¡1 knr

kn kn ¡ (1 ¡ r)kn¡1

¸ ·

a(kn ¡ (1 ¡ r)kn¡1) + bknr

akn + b(kn ¡ (1 ¡ r)kn¡1)

Hence, in view of the fact that

kn¡1

¸n

1 ¡ ¸n

¸n¡1

1 ¡ ¸n¡1

¸n

1 (1 ¡ f¸2

¸1 gn)

¸n¡1

1 (1 ¡ f¸2

¸1 gn¡1) ! ¸1; as n ! 1;

we have

a(kn ¡ (1 ¡ r)kn¡1) + bknr

akn + b(kn ¡ (1 ¡ r)kn¡1)

a( kn

kn¡1 ¡ (1 ¡ r)) + b kn

kn¡1

a kn

kn¡1

+ b( kn

kn¡1 ¡ (1 ¡ r))

a(¸1 ¡ (1 ¡ r)) + b¸1r

a¸1 + b(¸1 ¡ (1 ¡ r))

a(pr + r) + b(1 + pr)r

a(1 + pr) + b(pr + r)

prfa(1 + pr) + b(1 + pr)prg

a(1 + pr) + b(pr + r)

= pr:

Section 2:7

1. [AjI2] =

1 4

¡3 1

¯¯¯¯

1 0

0 1

R2 ! R2 + 3R1

1 4

0 13

¯¯¯¯

1 0

3 1

R2 ! 1

13R2

1 4

0 1

¯¯¯¯

1 0

3=13 1=13

R1 ! R1 ¡ 4R2

1 0

0 1

¯¯¯¯

1=13 ¡4=13

3=13 1=13

Hence A is non{singular and A¡1 =

1=13 ¡4=13

3=13 1=13

Moreover

E12(¡4)E2(1=13)E21(3)A = I2;

A¡1 = E12(¡4)E2(1=13)E21(3):

Hence

A = fE21(3)g¡1fE2(1=13)g¡1fE12(¡4)g¡1 = E21(¡3)E2(13)E12(4):

2. Let D = [dij ] be an m£m diagonal matrix and let A = [ajk] be an m£n

matrix. Then

(DA)ik =

j=1

dijajk = diiaik;

as dij = 0 if i 6= j. It follows that the ith row of DA is obtained by

multiplying the ith row of A by dii.

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 (a1; : : : ; an)diag (b1; : : : ; bn) = diag (a1b1; : : : ; anbn);

as the left{hand side can be regarded as pre{multiplication of the matrix

diag (b1; : : : ; bn) by the diagonal matrix diag (a1; : : : ; an).

Finally, suppose that each of a1; : : : ; an is non{zero. Then a¡1

1 ; : : : ; a¡1

all exist and we have

diag (a1; : : : ; an)diag (a¡1

1 ; : : : ; a¡1

n ) = diag (a1a¡1

1 ; : : : ; ana¡1

n )

= diag (1; : : : ; 1) = In:

Hence diag (a1; : : : ; an) is non{singular and its inverse is diag (a¡1

1 ; : : : ; a¡1

n ).

Next suppose that ai = 0. Then diag (a1; : : : ; an) is row{equivalent to a

matix containing a zero row and is hence singular.

3. [AjI3] =

0 0 2

1 2 6

3 7 9

¯¯¯¯¯¯

1 0 0

0 1 0

0 0 1

5 R1 $ R2

1 2 6 0 1 0

0 0 2 1 0 0

3 7 9 0 0 1

R3 ! R3 ¡ 3R1

1 2 6 0 1 0

0 0 2 1 0 0

0 1 ¡9 0 ¡3 1

5 R2 $ R3

1 2 6 0 1 0

0 1 ¡9 0 ¡3 1

0 0 2 1 0 0

R3 ! 1

2R3

1 2 6 0 1 0

0 1 ¡9 0 ¡3 1

0 0 1 1=2 0 0

5 R1 ! R1 ¡ 2R2

1 0 24 0 7 ¡2

0 1 ¡9 0 ¡3 1

0 0 1 1=2 0 0

R1 ! R1 ¡ 24R3

R2 ! R2 + 9R3

1 0 0 ¡12 7 ¡2

0 1 0 9=2 ¡3 1

0 0 1 1=2 0 0

Hence A is non{singular and A¡1 =

4 ¡12 7 ¡2

9=2 ¡3 1

1=2 0 0

Also

E23(9)E13(¡24)E12(¡2)E3(1=2)E23E31(¡3)E12A = I3:

Hence

A¡1 = E23(9)E13(¡24)E12(¡2)E3(1=2)E23E31(¡3)E12;

A = E12E31(3)E23E3(2)E12(2)E13(24)E23(¡9):

A =

1 2 k

3 ¡1 1

5 3 ¡5

5 !

1 2 k

0 ¡7 1 ¡ 3k

0 ¡7 ¡5 ¡ 5k

5 !

1 2 k

0 ¡7 1 ¡ 3k

0 0 ¡6 ¡ 2k

5 = B:

Hence if ¡6¡2k 6= 0, i.e. if k 6= ¡3, we see that B can be reduced to I3

and hence A is non{singular.

If k = ¡3, then B =

1 2 ¡3

0 ¡7 10

0 0 0

5 = B and consequently A is singu-

lar, as it is row{equivalent to a matrix containing a zero row.

5. E21(2)

1 2

¡2 ¡4

1 2

0 0

. Hence, as in the previous question,

1 2

¡2 ¡4

is singular.

6. Starting from the equation A2 ¡ 2A + 13I2 = 0, we deduce

A(A ¡ 2I2) = ¡13I2 = (A ¡ 2I2)A:

Hence AB = BA = I2, where B = ¡1

13 (A ¡ 2I2). Consequently A is non{

singular and A¡1 = B.

7. We assume the equation A3 = 3A2 ¡ 3A + I3.

(ii) A4 = A3A = (3A2 ¡ 3A + I3)A = 3A3 ¡ 3A2 + A

= 3(3A2 ¡ 3A + I3) ¡ 3A2 + A = 6A2 ¡ 8A + 3I3:

(iii) A3 ¡ 3A2 + 3A = I3. Hence

A(A2 ¡ 3A + 3I3) = I3 = (A2 ¡ 3A + 3I3)A:

Hence A is non{singular and

A¡1 = A2 ¡ 3A + 3I3

4 ¡1 ¡3 1

2 4 ¡1

0 1 0

5 :

8. (i) If B3 = 0 then

(In ¡ B)(In + B + B2) = In(In + B + B2) ¡ B(In + B + B2)

= (In + B + B2) ¡ (B + B2 + B3)

= In ¡ B3 = In ¡ 0 = In:

Similarly (In + B + B2)(In ¡ B) = In.

Hence A = In ¡ B is non{singular and A¡1 = In + B + B2.

It follows that the system AX = b has the unique solution

X = A¡1b = (In + B + B2)b = b + Bb + B2b:

(ii) Let B =

0 r s

0 0 t

0 0 0

5. Then B2 =

0 0 rt

0 0 0

5 and B3 = 0. Hence

from the preceding question

(I3 ¡ B)¡1 = I3 + B + B2

1 0 0

0 1 0

0 0 1

5 +

0 r s

0 0 t

0 0 0

5 +

0 0 rt

0 0 0

1 r s + rt

0 1 t

0 0 1

5 :

9. (i) Suppose that A2 = 0. Then if A¡1 exists, we deduce that A¡1(AA) =

A¡10, 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 A2 = A and that A¡1 exists. Then

A¡1(AA) = A¡1A;

which gives A = In. Equivalently, if A2 = A and A 6= In, then A does not

have an inverse.

10. The system of linear equations

x + y ¡ z = a

z = b

2x + y + 2z = c

is equivalent to the matrix equation AX = B, where

A =

1 1 ¡1

0 0 1

2 1 2

5 ; X =

5 ; B =

5 :

By Question 7, A¡1 exists and hence the system has the unique solution

X =

4 ¡1 ¡3 1

2 4 ¡1

0 1 0

5 =

4 ¡a ¡ 3b + c

2a + 4b ¡ c

5 :

Hence x = ¡a ¡ 3b + c; y = 2a + 4b ¡ c; z = b.

12.

A = E3(2)E14E42(3) = E3(2)E14

664

1 0 0 0

0 1 0 0

0 0 1 0

0 3 0 1

775

= E3(2)

664

0 3 0 1

0 1 0 0

0 0 1 0

1 0 0 0

775

664

0 3 0 1

0 1 0 0

0 0 2 0

1 0 0 0

775

Also

A¡1 = (E3(2)E14E42(3))¡1

= (E42(3))¡1E¡1

14 (E3(2))¡1

= E42(¡3)E14E3(1=2)

= E42(¡3)E14

664

1 0 0 0

0 1 0 0

0 0 1=2 0

0 0 0 1

775

= E42(¡3)

664

0 0 0 1

0 1 0 0

0 0 1=2 0

1 0 0 0

775

664

0 0 0 1

0 1 0 0

0 0 1=2 0

1 ¡3 0 0

775

13. (All matrices in this question are over Z2.)

(a)

664

1 1 0 1

0 0 1 1

1 1 1 1

1 0 0 1

¯¯¯¯¯¯¯¯

1 0 0 0

0 1 0 0

0 0 1 0

0 0 0 1

775

664

1 1 0 1

0 0 1 1

0 0 1 0

0 1 0 0

¯¯¯¯¯¯¯¯

1 0 0 0

0 1 0 0

1 0 1 0

1 0 0 1

775

664

1 1 0 1

0 1 0 0

0 0 1 0

0 0 1 1

¯¯¯¯¯¯¯¯

1 0 0 0

1 0 0 1

1 0 1 0

0 1 0 0

775

664

1 0 0 1

0 1 0 0

0 0 1 0

0 0 0 1

¯¯¯¯¯¯¯¯

0 0 0 1

1 0 0 1

1 0 1 0

1 1 1 0

775

664

1 0 0 0

0 1 0 0

0 0 1 0

0 0 0 1

¯¯¯¯¯¯¯¯

1 1 1 1

1 0 0 1

1 0 1 0

1 1 1 0

775

Hence A is non{singular and

A¡1 =

664

1 1 1 1

1 0 0 1

1 0 1 0

1 1 1 0

775

(b) A =

664

1 1 0 1

0 1 1 1

1 0 1 0

1 1 0 1

775

R4 ! R4 + R1

664

1 1 0 1

0 1 1 1

1 0 1 0

0 0 0 0

775

, so A is singular.

14.

(a)

1 1 1

¡1 1 0

2 0 0

¯¯¯¯¯¯

1 0 0

0 1 0

0 0 1

R3 ! 1

2R3

R1 ! R1 ¡ R3

R2 ! R2 + R3

R1 $ R3

1 0 0

0 1 0

0 1 1

¯¯¯¯¯¯

0 0 1=2

0 1 1=2

1 0 ¡1=2

R3 ! R3 ¡ R2

1 0 0

0 1 0

0 0 1

¯¯¯¯¯¯

0 0 1=2

0 1 1=2

1 ¡1 ¡1

5 :

Hence A¡1 exists and

A¡1 =

0 0 1=2

0 1 1=2

1 ¡1 ¡1

5 :

(b)

2 2 4

1 0 1

0 1 0

¯¯¯¯¯¯

1 0 0

0 1 0

0 0 1

R1 ! R1 ¡ 2R2

R1 $ R2

R2 $ R3

1 0 1

0 1 0

0 2 2

¯¯¯¯¯¯

0 1 0

0 0 1

1 ¡2 0

R3 ! R3 ¡ 2R2

1 0 1

0 1 0

0 0 2

¯¯¯¯¯¯

0 1 0

0 0 1

1 ¡2 ¡2

R3 ! 1

2R3

1 0 1

0 1 0

0 0 1

¯¯¯¯¯¯

0 1 0

0 0 1

1=2 ¡1 ¡1

R1 ! R1 ¡ R3

1 0 0

0 1 0

0 0 1

¯¯¯¯¯¯

¡1=2 2 1

0 0 1

1=2 ¡1 ¡1

5 :

Hence A¡1 exists and

A¡1 =

4 ¡1=2 2 1

0 0 1

1=2 ¡1 ¡1

5 :

(c)

4 6 ¡3

0 0 7

0 0 5

5 R2 ! 1

7R2

R3 ! 1

5R3

4 6 ¡3

0 0 1

5 R3 ! R3 ¡ R2

4 6 ¡3

0 0 1

0 0 0

5 :

Hence A is singular by virtue of the zero row.

(d)

2 0 0

0 ¡5 0

0 0 7

¯¯¯¯¯¯

1 0 0

0 1 0

0 0 1

R1 ! 1

2R1

R2 ! ¡1

5 R2

R3 ! 1

7R3

1 0 0

0 1 0

0 0 1

¯¯¯¯¯¯

1=2 0 0

0 ¡1=5 0

0 0 1=7

5 :

Hence A¡1 exists and A¡1 = diag (1=2; ¡1=5; 1=7).

(Of course this was also immediate from Question 2.)

(e)

664

1 2 4 6

0 1 2 0

0 0 1 2

0 0 0 2

¯¯¯¯¯¯¯¯

1 0 0 0

0 1 0 0

0 0 1 0

0 0 0 1

775

R1 ! R1 ¡ 2R2

664

1 0 0 6

0 1 2 0

0 0 1 2

0 0 0 2

¯¯¯¯¯¯¯¯

1 ¡2 0 0

0 1 0 0

0 0 1 0

0 0 0 1

775

R2 ! R2 ¡ 2R3

664

1 0 0 6

0 1 0 ¡4

0 0 1 2

0 0 0 2

¯¯¯¯¯¯¯¯

1 ¡2 0 0

0 1 ¡2 0

0 0 1 0

0 0 0 1

775

R1 ! R1 ¡ 3R4

R2 ! R2 + 2R4

R3 ! R3 ¡ R4

R4 ! 1

2R4

664

1 0 0 0

0 1 0 0

0 0 1 0

0 0 0 1

¯¯¯¯¯¯¯¯

1 ¡2 0 ¡3

0 1 ¡2 2

0 0 1 ¡1

0 0 0 1=2

775

Hence A¡1 exists and

A¡1 =

664

1 ¡2 0 ¡3

0 1 ¡2 2

0 0 1 ¡1

0 0 0 1=2

775

(f)

1 2 3

4 5 6

5 7 9

5 R2 ! R2 ¡ 4R1

R3 ! R3 ¡ 5R1

1 2 3

0 ¡3 ¡6

5 R3 ! R3 ¡ R2

1 2 3

0 ¡3 ¡6

0 0 0

5 :

Hence A is singular by virtue of the zero row.

15. Suppose that A is non{singular. Then

AA¡1 = In = A¡1A:

Taking transposes throughout gives

(AA¡1)t = It

n = (A¡1A)t

(A¡1)tAt = In = At(A¡1)t;

so At is non{singular and (At)¡1 = (A¡1)t.

16. Let A =

a b

c d

, where ad ¡ bc = 0. Then the equation

A2 ¡ (a + d)A + (ad ¡ bc)I2 = 0

reduces to A2 ¡ (a + d)A = 0 and hence A2 = (a + d)A. From the last

equation, if A¡1 exists, we deduce that A = (a + d)I2, or

a b

c d

a + d 0

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.

17.

A =

1 a b

¡a 1 c

¡b ¡c 1

5 R2 ! R2 + aR1

R3 ! R3 + bR1

1 a b

0 1 + a2 c + ab

0 ab ¡ c 1 + b2

R2 ! 1

1+a2R2

1 a b

0 1 c+ab

1+a2

0 ab ¡ c 1 + b2

R3 ! R3 ¡ (ab ¡ c)R2

1 a b

0 1 c+ab

1+a2

0 0 1 + b2 + (c¡ab)(c+ab)

1+a2

= B:

Now

1 + b2 +

(c ¡ ab)(c + ab)

1 + a2 = 1 + b2 +

c2 ¡ (ab)2

1 + a2

1 + a2 + b2 + c2

1 + a2 6= 0:

Hence B can be reduced to I3 using four more row operations and conse-

quently A is non{singular.

18. The proposition is clearly true when n = 1. So let n ¸ 1 and assume

(P¡1AP)n = P¡1AnP. Then

(P¡1AP)n+1 = (P¡1AP)n(P¡1AP)

= (P¡1AnP)(P¡1AP)

= P¡1An(PP¡1)AP

= P¡1AnIAP

= P¡1(AnA)P

= P¡1An+1P

and the induction goes through.

19. Let A =

2=3 1=4

1=3 3=4

and P =

1 3

¡1 4

. Then P¡1 = 1

4 ¡3

1 1

We then verify that P¡1AP =

5=12 0

0 1

. Then from the previous ques-

tion,

P¡1AnP = (P¡1AP)n =

5=12 0

0 1

¸n

(5=12)n 0

0 1n

(5=12)n 0

0 1

Hence

An = P

(5=12)n 0

0 1

P¡1 =

1 3

¡1 4

¸ ·

(5=12)n 0

0 1

4 ¡3

1 1

(5=12)n 3

¡(5=12)n 4

¸ ·

4 ¡3

1 1

4(5=12)n + 3 (¡3)(5=12)n + 3

¡4(5=12)n + 4 3(5=12)n + 4

3 3

4 4

(5=12)n

4 ¡3

¡4 3

Notice that An ! 1

3 3

4 4

as n ! 1. This problem is a special case of

a more general result about Markov matrices.

20. Let A =

a b

c d

be a matrix whose elements are non{negative real

numbers satisfying

a ¸ 0; b ¸ 0; c ¸ 0; d ¸ 0; a + c = 1 = b + d:

Also let P =

b 1

c ¡1

and suppose that A 6= I2.

(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 = 1, resulting in A = I2. Hence

det P < 0 and P is non{singular.

Next,

P¡1AP = ¡1

b + c

¡1 ¡1

¡c b

¸ ·

a b

c d

¸ ·

b 1

c ¡1

= ¡1

b + c

¡a ¡ c ¡b ¡ d

¡ac + bc ¡cb + bd

¸ ·

b 1

c ¡1

= ¡1

b + c

¡1 ¡1

¡ac + bc ¡cb + bd

¸ ·

b 1

c ¡1

= ¡1

b + c

¡b ¡ c 0

(¡ac + bc)b + (¡cb + bd)c ¡ac + bc + cb ¡ bd

Now

¡acb + b2c ¡ c2b + bdc = ¡cb(a + c) + bc(b + d)

= ¡cb + bc = 0:

Also

¡(a + d ¡ 1)(b + c) = ¡ab ¡ ac ¡ db ¡ dc + b + c

= ¡ac + b(1 ¡ a) + c(1 ¡ d) ¡ bd

= ¡ac + bc + cb ¡ bd:

Hence

P¡1AP = ¡1

b + c

¡(b + c) 0

0 ¡(a + d ¡ 1)(b + c)

1 0

0 a + d ¡ 1

(ii) We next prove that if we impose the extra restriction that A 6=

0 1

1 0

then ja + d ¡ 1j < 1. This will then have the following consequence:

A = P

1 0

0 a + d ¡ 1

P¡1

An = P

1 0

0 a + d ¡ 1

¸n

P¡1

= P

1 0

0 (a + d ¡ 1)n

P¡1

! P

1 0

0 0

P¡1

b 1

c ¡1

¸ ·

1 0

0 0

¡1

b + c

¡1 ¡1

¡c b

= ¡1

b + c

b 0

c 0

¸ ·

¡1 ¡1

¡c b

= ¡1

b + c

¡b ¡b

¡c ¡c

b + c

b b

c c

;

where we have used the fact that (a + d ¡ 1)n ! 0 as n ! 1.

We ¯rst prove the inequality ja + d ¡ 1j · 1:

a + d ¡ 1 · 1 + d ¡ 1 = d · 1

a + d ¡ 1 ¸ 0 + 0 ¡ 1 = ¡1:

Next, if a+d¡1 = 1, we have a+d = 2; so a = 1 = d and hence c = 0 = b,

contradicting our assumption that A 6= I2. Also if a + d ¡ 1 = ¡1, then

a + d = 0; so a = 0 = d and hence c = 1 = b and hence A =

0 1

1 0

22. The system is inconsistent: We work towards reducing the augmented

matrix:

1 2

1 1

3 5

¯¯¯¯¯¯

5 R2 ! R2 ¡ R1

R3 ! R3 ¡ 3R1

1 2

0 ¡1

¯¯¯¯¯¯

R3 ! R3 ¡ R2

1 2

0 ¡1

0 0

¯¯¯¯¯¯

¡1

5 :

The last row reveals inconsistency.

The system in matrix form is AX = B, where

A =

1 2

1 1

3 5

5 ; X =

; B =

5 :

The normal equations are given by the matrix equation

AtAX = AtB:

Now

AtA =

1 1 3

2 1 5

¸ 2

1 2

1 1

3 5

5 =

11 18

18 30

AtB =

1 1 3

2 1 5

¸ 2

5 =

Hence the normal equations are

11x + 18y = 45

18x + 30y = 73:

These may be solved, for example, by Cramer's rule:

x =

¯¯¯¯

45 18

73 30

¯¯¯¯

11 18

18 30

¯¯¯¯

= 6

y =

¯¯¯¯

11 45

18 73

¯¯¯¯

11 18

18 30

¯¯¯¯

= ¡7

23. Substituting the coordinates of the ¯ve points into the parabola equation

gives the following equations:

a = 0

a + b + c = 0

a + 2b + 4c = ¡1

a + 3b + 9c = 4

a + 4b + 16c = 8:

The associated normal equations are given by

5 10 30

10 30 100

30 100 354

5 =

160

5 ;

which have the solution a = 1=5; b = ¡2; c = 1.

24. Suppose that A is symmetric, i.e. At = A and that AB is de¯ned. Then

(BtAB)t = BtAt(Bt)t = BtAB;

so BtAB is also symmetric.

25. Let A be m£n and B be n£m, where m > n. Then the homogeneous

system BX = 0 has a non{trivial solution X0, as the number of unknowns

is greater than the number of equations. Then

(AB)X0 = A(BX0) = A0 = 0

and the m £ m matrix AB is therefore singular, as X0 6= 0.

26. (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 At is also singular and

hence by (i) so is BtAt = (AB)t. Consequently AB is also singular

Section 3.6

1. (a) Let S be the set of vectors [x; y] satisfying x = 2y. Then S is a vector

subspace of R2. For

(i) [0; 0] 2 S as x = 2y holds with x = 0 and y = 0.

(ii) S is closed under addition. For let [x1; y1] and [x2; y2] belong to S.

Then x1 = 2y1 and x2 = 2y2. Hence

x1 + x2 = 2y1 + 2y2 = 2(y1 + y2)

and hence

[x1 + x2; y1 + y2] = [x1; y1] + [x2; y2]

belongs to S.

(iii) S is closed under scalar multiplication. For let [x; y] 2 S and t 2 R.

Then x = 2y and hence tx = 2(ty). Consequently

[tx; ty] = t[x; y] 2 S:

(b) Let S be the set of vectors [x; y] satisfying x = 2y and 2x = y. Then S is

a subspace of R2. This can be proved in the same way as (a), or alternatively

we see that x = 2y and 2x = y imply x = 4x and hence x = 0 = y. Hence

S = f[0; 0]g, the set consisting of the zero vector. This is always a subspace.

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 [1; 0] 2 S and [0; 1] 2 S, but

[1; 0] + [0; 1] = [1; 1] 62 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 [1; 0] 2 S and ¡1 2 R,

but (¡1)[1; 0] = [¡1; 0] 62 S.

2. Let X; Y; Z be vectors in Rn. Then by Lemma 3.2.1

hX + Y; X + Z; Y + Zi µ hX; Y; Zi;

as each of X + Y; X + Z; Y + Z is a linear combination of X; Y; Z.

Also

X =

(X + Y ) +

(X + Z) ¡

(Y + Z);

Y =

(X + Y ) ¡

(X + Z) +

(Y + Z);

Z = ¡1

(X + Y ) +

(X + Z) +

(Y + Z);

hX; Y; Zi µ hX + Y; X + Z; Y + Zi:

Hence

hX; Y; Zi = hX + Y; X + Z; Y + Zi:

3. Let X1 =

664

775

; X2 =

664

775

and X3 =

664

775

. We have to decide if

X1; X2; X3 are linearly independent, that is if the equation xX1 + yX2 +

zX3 = 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

2x + 2y + 3z = 0:

We reduce the coe±cient matrix to reduced row{echelon form:

664

1 0 1

0 1 1

1 1 1

2 2 3

775

664

1 0 0

0 1 0

0 0 1

0 0 0

775

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

X1 =

¡1

5 ; X2 =

4 ¡1

¡1

5 ; X3 =

4 ¡1

¡1

are linearly dependent for precisely those values of ¸ for which the equation

xX1+yX2+zX3 = 0 has a non{trivial solution. This equation is equivalent

to the system of homogeneous equations

¸x ¡ y ¡ z = 0

¡x + ¸y ¡ z = 0

¡x ¡ y + ¸z = 0:

Now the coe±cient determinant of this system is

¯¯¯¯¯¯

¸ ¡1 ¡1

¡1 ¸ ¡1

¡1 ¡1 ¸

¯¯¯¯¯¯

= (¸ + 1)2(¸ ¡ 2):

So the values of ¸ which make X1; X2; X3 linearly independent are those ¸

satisfying ¸ 6= ¡1 and ¸ 6= 2.

5. Let A be the following matrix of rationals:

A =

664

1 1 2 0 1

2 2 5 0 3

0 0 0 1 3

8 11 19 0 11

775

Then A has reduced row{echelon form

B =

664

1 0 0 0 ¡1

0 1 0 0 0

0 0 1 0 1

0 0 0 1 3

775

From B we read o® 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 ¯rst four columns of A form a basis for C(A).

From B we see that the solution is

x1 = x5

x2 = 0

x3 = ¡x5

x4 = ¡3x5;

with x5 arbitrary. Then

X =

66664

¡x5

¡3x5

77775

= x5

66664

¡1

¡3

77775

;

so [1; 0; ¡1; ¡3; 1]t is a basis for N(A).

6. In Section 1.6, problem 12, we found that the matrix

A =

664

1 0 1 0 1

0 1 0 1 1

1 1 1 1 0

0 0 1 1 0

775

has reduced row{echelon form

B =

664

1 0 0 1 1

0 1 0 1 1

0 0 1 1 0

0 0 0 0 0

775

From B we read o® the following:

(a) The three non{zero rows of B form a basis for R(A).

(b) The ¯rst three columns of A form a basis for C(A).

From B we see that the solution is

x1 = ¡x4 ¡ x5 = x4 + x5

x2 = ¡x4 ¡ x5 = x4 + x5

x3 = ¡x4 = x4;

with x4 and x5 arbitrary elements of Z2. Hence

X =

66664

x4 + x5

77775

= x4

66664

77775

66664

77775

Hence [1; 1; 1; 1; 0]t and [1; 1; 0; 0; 1]t form a basis for N(A).

7. Let A be the following matrix over Z5:

A =

664

1 1 2 0 1 3

2 1 4 0 3 2

0 0 0 1 3 0

3 0 2 4 3 2

775

We ¯nd that A has reduced row{echelon form B:

B =

664

1 0 0 0 2 4

0 1 0 0 4 4

0 0 1 0 0 0

0 0 0 1 3 0

775

From B we read o® 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 ¯rst four columns of A form a basis for C(A).

From B we see that the solution is

x1 = ¡2x5 ¡ 4x6 = 3x5 + x6

x2 = ¡4x5 ¡ 4x6 = x5 + x6

x3 = 0

x4 = ¡3x5 = 2x5;

where x5 and x6 are arbitrary elements of Z5. Hence

X = x5

6666664

7777775 +

6666664 1

7777775

;

so [3; 1; 0; 2; 1; 0]t and [1; 1; 0; 0; 0; 1]t form a basis for R(A).

8. Let F = f0; 1; a; bg be a ¯eld and let A be the following matrix over F:

A =

1 a b a

a b b 1

1 1 1 a

5 :

In Section 1.6, problem 17, we found that A had reduced row{echelon form

B =

1 0 0 0

0 1 0 b

0 0 1 1

5 :

From B we read o® 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 ¯rst three columns of A form a basis for C(A).

From B we see that the solution is

x1 = 0

x2 = ¡bx4 = bx4

x3 = ¡x4 = x4;

where x4 is an arbitrary element of F. Hence

X = x4

664

775

;

so [0; b; 1; 1]t is a basis for N(A).

9. Suppose that X1; : : : ;Xm form a basis for a subspace S. We have to

prove that

X1; X1 + X2; : : : ;X1 + ¢ ¢ ¢ + Xm

also form a basis for S.

First we prove the independence of the family: Suppose

x1X1 + x2(X1 + X2) + ¢ ¢ ¢ + xm(X1 + ¢ ¢ ¢ + Xm) = 0:

Then

(x1 + x2 + ¢ ¢ ¢ + xm)X1 + ¢ ¢ ¢ + xmXm = 0:

Then the linear independence of X1; : : : ;Xm gives

x1 + x2 + ¢ ¢ ¢ + xm = 0; : : : ; xm = 0;

form which we deduce that x1 = 0; : : : ; xm = 0.

Secondly we have to prove that every vector of S is expressible as a linear

combination of X1; X1 + X2; : : : ;X1 + ¢ ¢ ¢ + Xm. Suppose X 2 S. Then

X = a1X1 + ¢ ¢ ¢ + amXm:

We have to ¯nd x1; : : : ; xm such that

X = x1X1 + x2(X1 + X2) + ¢ ¢ ¢ + xm(X1 + ¢ ¢ ¢ + Xm)

= (x1 + x2 + ¢ ¢ ¢ + xm)X1 + ¢ ¢ ¢ + xmXm:

Then

a1X1 + ¢ ¢ ¢ + amXm = (x1 + x2 + ¢ ¢ ¢ + xm)X1 + ¢ ¢ ¢ + xmXm:

So if we can solve the system

x1 + x2 + ¢ ¢ ¢ + xm = a1; : : : ; xm = am;

we are ¯nished. Clearly these equations have the unique solution

x1 = a1 ¡ a2; : : : ; xm¡1 = am ¡ am¡1; xm = am:

10. Let A =

a b c

1 1 1

. If [a; b; c] is a multiple of [1; 1; 1], (that is,

a = b = c), then rankA = 1. For if

[a; b; c] = t[1; 1; 1];

then

R(A) = h[a; b; c]; [1; 1; 1]i = ht[1; 1; 1]; [1; 1; 1]i = h[1; 1; 1]i;

so [1; 1; 1] is a basis for R(A).

However if [a; b; c] is not a multiple of [1; 1; 1], (that is at least two

of a; b; c are distinct), then the left{to{right test shows that [a; b; c] and

[1; 1; 1] are linearly independent and hence form a basis for R(A). Conse-

quently rankA = 2 in this case.

11. Let S be a subspace of Fn with dim S = m. Also suppose that

X1; : : : ;Xm are vectors in S such that S = hX1; : : : ;Xmi. We have to

prove that X1; : : : ;Xm form a basis for S; in other words, we must prove

that X1; : : : ;Xm are linearly independent.

However if X1; : : : ;Xm were linearly dependent, then one of these vec-

tors would be a linear combination of the remaining vectors. Consequently

S would be spanned by m ¡ 1 vectors. But there exist a family of m lin-

early independent vectors in S. Then by Theorem 3.3.2, we would have the

contradiction m · m ¡ 1.

12. Let [x; y; z]t 2 S. Then x + 2y + 3z = 0. Hence x = ¡2y ¡ 3z and

5 =

4 ¡2y ¡ 3z

5 = y

4 ¡2

5 + z

4 ¡3

5 :

Hence [¡2; 1; 0]t and [¡3; 0; 1]t form a basis for S.

Next (¡1) + 2(¡1) + 3(1) = 0, so [¡1; ¡1; 1]t 2 S.

To ¯nd a basis for S which includes [¡1; ¡1; 1]t, we note that [¡2; 1; 0]t

is not a multiple of [¡1; ¡1; 1]t. Hence we have found a linearly independent

family of two vectors in S, a subspace of dimension equal to 2. Consequently

these two vectors form a basis for S.

13. Without loss of generality, suppose that X1 = X2. Then we have the

non{trivial dependency relation:

1X1 + (¡1)X2 + 0X3 + ¢ ¢ ¢ + 0Xm = 0:

14. (a) Suppose that Xm+1 is a linear combination of X1; : : : ;Xm. Then

hX1; : : : ;Xm; Xm+1i = hX1; : : : ;Xmi

and hence

dim hX1; : : : ;Xm; Xm+1i = dim hX1; : : : ;Xmi:

(b) Suppose that Xm+1 is not a linear combination of X1; : : : ;Xm. If not

all of X1; : : : ;Xm are zero, there will be a subfamily Xc1 ; : : : ;Xcr which is

a basis for hX1; : : : ;Xmi.

Then as Xm+1 is not a linear combination of Xc1 ; : : : ;Xcr , it follows that

Xc1 ; : : : ;Xcr ; Xm+1 are linearly independent. Also

hX1; : : : ;Xm; Xm+1i = hXc1 ; : : : ;Xcr ; Xm+1i:

Consequently

dim hX1; : : : ;Xm; Xm+1i = r + 1 = dim hX1; : : : ;Xmi + 1:

Our result can be rephrased in a form suitable for the second part of the

problem:

dim hX1; : : : ;Xm; Xm+1i = dim hX1; : : : ;Xmi

if and only if Xm+1 is a linear combination of X1; : : : ;Xm.

If X = [x1; : : : ; xn]t, then AX = B is equivalent to

B = x1A¤1 + ¢ ¢ ¢ + xnA¤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 2 C(A). However by the ¯rst part of this question,

B 2 C(A) if and only if dimC([AjB]) = dimC(A), that is, rank [AjB] =

rankA.

15. Let a1; : : : ; an be elements of F, not all zero. Let S denote the set of

vectors [x1; : : : ; xn]t, where x1; : : : ; xn satisfy

a1x1 + ¢ ¢ ¢ + anxn = 0:

Then S = N(A), where A is the row matrix [a1; : : : ; an]. Now rankA = 1

as A 6= 0. So by the \rank + nullity" theorem, noting that the number of

columns of A equals n, we have

dimN(A) = nullity (A) = n ¡ rankA = n ¡ 1:

16. (a) (Proof of Lemma 3.2.1) Suppose that each of X1; : : : ;Xr is a linear

combination of Y1; : : : ; Ys. Then

Xi =

j=1

aijYj ; (1 · i · r):

Now let X =

P r

i=1 xiXi be a linear combination of X1; : : : ;Xr. Then

X = x1(a11Y1 + ¢ ¢ ¢ + a1sYs)

+ ¢ ¢ ¢

+ xr(ar1Y1 + ¢ ¢ ¢ + arsYs)

= y1Y1 + ¢ ¢ ¢ + ysYs;

where yj = a1jx1+¢ ¢ ¢+arjxr. Hence X is a linear combination of Y1; : : : ; Ys.

Another way of stating Lemma 3.2.1 is

hX1; : : : ;Xri µ hY1; : : : ; Ysi; (1)

if each of X1; : : : ;Xr is a linear combination of Y1; : : : ; Ys.

(b) (Proof of Theorem 3.2.1) Suppose that each of X1; : : : ;Xr is a linear

combination of Y1; : : : ; Ys and that each of Y1; : : : ; Ys is a linear combination

of X1; : : : ;Xr. Then by (a) equation (1) above

hX1; : : : ;Xri µ hY1; : : : ; Ysi

and

hY1; : : : ; Ysi µ hX1; : : : ;Xri:

Hence

hX1; : : : ;Xri = hY1; : : : ; Ysi:

combination of X1; : : : ;Xr. Then each of X1; : : : ;Xr; Z1; : : : ;Zt is a linear

combination of X1; : : : ;Xr.

Also each of X1; : : : ;Xr is a linear combination of X1; : : : ;Xr; Z1; : : : ;Zt,

so by Theorem 3.2.1

hX1; : : : ;Xr; Z1; : : : ;Zti = hX1; : : : ;Xri:

(d) (Proof of Theorem 3.3.2) Let Y1; : : : ; Ys be vectors in hX1; : : : ;Xri

and assume that s > r. We have to prove that Y1; : : : ; Ys are linearly

dependent. So we consider the equation

x1Y1 + ¢ ¢ ¢ + xsYs = 0:

Now Yi =

P r

j=1 aijXj , for 1 · i · s. Hence

x1Y1 + ¢ ¢ ¢ + xsYs = x1(a11X1 + ¢ ¢ ¢ + a1rXr)

+ ¢ ¢ ¢

+ xr(as1X1 + ¢ ¢ ¢ + asrXr):

= y1X1 + ¢ ¢ ¢ + yrXr; (1)

where yj = a1jx1 + ¢ ¢ ¢ + asjxs. However the homogeneous system

y1 = 0; ¢ ¢ ¢ ; yr = 0

has a non{trivial solution x1; : : : ; xs, as s > r and from (1), this results in a

non{trivial solution of the equation

x1Y1 + ¢ ¢ ¢ + xsYs = 0:

Hence Y1; : : : ; Ys are linearly dependent.

17. Let R and S be subspaces of Fn, with R µ S. We ¯rst prove

dimR · dim S:

Let X1; : : : ;Xr be a basis for R. Now by Theorem 3.5.2, because X1; : : : ;Xr

form a linearly independent family lying in S, this family can be extended

to a basis X1; : : : ;Xr; : : : ;Xs for S. Then

dim S = s ¸ r = dimR:

Next suppose that dimR = dim S. Let X1; : : : ;Xr be a basis for R. Then

because X1; : : : ;Xr form a linearly independent family in S and S is a sub-

space whose dimension is r, it follows from Theorem 3.4.3 that X1; : : : ;Xr

form a basis for S. Then

S = hX1; : : : ;Xri = R:

18. Suppose that R and S are subspaces of Fn with the property that R[S

is also a subspace of Fn. We have to prove that R µ S or S µ R. We argue

by contradiction: Suppose that R 6µ S and S 6µ R. Then there exist vectors

u and v such that

u 2 R and v 62 S; v 2 S and v 62 R:

Consider the vector u+v. As we are assuming R[S is a subspace, R[S is

closed under addition. Hence u + v 2 R [ S and so u + v 2 R or u + v 2 S.

However if u + v 2 R, then v = (u + v) ¡ u 2 R, which is a contradiction;

similarly if u + v 2 S.

Hence we have derived a contradiction on the asumption that R 6µ S and

S 6µ R. Consequently at least one of these must be false. In other words

R µ S or S µ R.

19. Let X1; : : : ;Xr be a basis for S.

(i) First let

Y1 = a11X1 + ¢ ¢ ¢ + a1rXr

...

(2)

Yr = ar1X1 + ¢ ¢ ¢ + arrXr;

where A = [aij ] is non{singular. Then the above system of equations can

be solved for X1; : : : ;Xr in terms of Y1; : : : ; Yr. Consequently by Theorem

3.2.1

hY1; : : : ; Yri = hX1; : : : ;Xri = S:

It follows from problem 11 that Y1; : : : ; Yr is a basis for S.

(ii) We show that all bases for S are given by equations 2. So suppose

that Y1; : : : ; Yr forms a basis for S. Then because X1; : : : ;Xr form a basis

for S, we can express Y1; : : : ; Yr in terms of X1; : : : ;Xr as in 2, for some

matrix A = [aij ]. We show A is non{singular by demonstrating that the

linear independence of Y1; : : : ; Yr implies that the rows of A are linearly

independent.

So assume

x1[a11; : : : ; a1r] + ¢ ¢ ¢ + xr[ar1; : : : ; arr] = [0; : : : ; 0]:

Then on equating components, we have

a11x1 + ¢ ¢ ¢ + ar1xr = 0

...

a1rx1 + ¢ ¢ ¢ + arrxr = 0:

Hence

x1Y1 + ¢ ¢ ¢ + xrYr = x1(a11X1 + ¢ ¢ ¢ + a1rXr) + ¢ ¢ ¢ + xr(ar1X1 + ¢ ¢ ¢ + arrXr)

= (a11x1 + ¢ ¢ ¢ + ar1xr)X1 + ¢ ¢ ¢ + (a1rx1 + ¢ ¢ ¢ + arrxr)Xr

= 0X1 + ¢ ¢ ¢ + 0Xr = 0:

Then the linear independence of Y1; : : : ; Yr implies x1 = 0; : : : ; xr = 0.

(We mention that the last argument is reversible and provides an alter-

native proof of part (i).)

©©©©©©©©

¢@

©©©©©©©©

Section 4.1

1. We ¯rst prove that the area of a triangle P1P2P3, where the points

are in anti{clockwise orientation, is given by the formula

½¯¯¯¯

x1 x2

y1 y2

¯¯¯¯

x2 x3

y2 y3

¯¯¯¯

x3 x1

y3 y1

¯¯¯¯

Referring to the above diagram, we have

Area P1P2P3 = AreaOP1P2 + AreaOP2P3 ¡ AreaOP1P3

¯¯¯¯

x1 x2

y1 y2

¯¯¯¯

x2 x3

y2 y3

¯¯¯¯

x1 x3

y1 y3

¯¯¯¯

;

which gives the desired formula.

We now turn to the area of a quadrilateral. One possible con¯guration

occurs when the quadrilateral is convex as in ¯gure (a) below. The interior

diagonal breaks the quadrilateral into two triangles P1P2P3 and P1P3P4.

Then

Area P1P2P3P4 = Area P1P2P3 + Area P1P3P4

½¯¯¯¯

x1 x2

y1 y2

¯¯¯¯

x2 x3

y2 y3

¯¯¯¯

x3 x1

y3 y1

¯¯¯¯

` ` ` ` ` ` ` ` ` `

HH HH

(a)

` ` ` ` ` ` ` ` ` `

©©©©©©

(b) P4

½¯¯¯¯

x1 x3

y1 y3

¯¯¯¯

x3 x4

y3 y4

¯¯¯¯

x4 x1

y4 y1

¯¯¯¯

½¯¯¯¯

x1 x2

y1 y2

¯¯¯¯

x2 x3

y2 y3

¯¯¯¯

x3 x4

y3 y4

¯¯¯¯

x4 x1

y4 y1

¯¯¯¯

;

after cancellation.

Another possible con¯guration for the quadrilateral occurs when it is not

convex, as in ¯gure (b). The interior diagonal P2P4 then gives two triangles

P1P2P4 and P2P3P4 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

¯¯¯¯¯¯

a b c

x + u y + v z + w

u + a v + b w + c

¯¯¯¯¯¯

x y 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 y z

u + a v + b w + c

¯¯¯¯¯¯+

¯¯¯¯¯¯ a

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

¯¯¯¯¯¯

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

¯¯¯¯¯¯

Hence ¢ = 2

¯¯¯¯¯¯

a b c

x y z

u v w

¯¯¯¯¯¯

n2 (n + 1)2 (n + 2)2

(n + 1)2 (n + 2)2 (n + 3)2

(n + 2)2 (n + 3)2 (n + 4)2

¯¯¯¯¯¯

C3 ! C3 ¡ C2

C2 ! C2 ¡ C1

¯¯¯¯¯¯

n2 2n + 1 2n + 3

(n + 1)2 2n + 3 2n + 5

(n + 2)2 2n + 5 2n + 7

¯¯¯¯¯¯

C3 ! C3 ¡ C2

¯¯¯¯¯¯

n2 2n + 1 2

(n + 1)2 2n + 3 2

(n + 2)2 2n + 5 2

¯¯¯¯¯¯

R3 ! R3 ¡ R2

R2 ! R2 ¡ R1

¯¯¯¯¯¯

n2 2n + 1 2

2n + 1 2 0

2n + 3 2 0

¯¯¯¯¯¯

= ¡8:

4. (a)

¯¯¯¯¯¯

246 427 327

1014 543 443

¡342 721 621

¯¯¯¯¯¯

246 100 327

1014 100 443

¡342 100 621

¯¯¯¯¯¯

= 100

¯¯¯¯¯¯

246 1 327

1014 1 443

¡342 1 621

¯¯¯¯¯¯

= 100

¯¯¯¯¯¯

246 1 327

768 0 116

¡588 0 294

¯¯¯¯¯¯

= 100(¡1)

¯¯¯¯

768 116

¡588 294

¯¯¯¯

= ¡29400000:

(b)

¯¯¯¯¯¯¯¯

1 2 3 4

¡2 1 ¡4 3

3 ¡4 ¡1 2

4 3 ¡2 ¡1

¯¯¯¯¯¯¯¯

1 2 3 4

0 5 2 11

0 ¡10 ¡10 ¡10

0 ¡5 ¡14 ¡17

¯¯¯¯¯¯¯¯

¯¯¯¯¯¯

5 2 11

¡10 ¡10 ¡10

¡5 ¡14 ¡17

¯¯¯¯¯¯

= ¡10

¯¯¯¯¯¯

5 2 11

1 1 1

¡5 ¡14 ¡17

¯¯¯¯¯¯

= ¡10

¯¯¯¯¯¯

5 ¡3 6

1 0 0

¡5 ¡9 ¡12

¯¯¯¯¯¯

= ¡10(¡1)

¯¯¯¯

¡3 6

¡9 ¡12

¯¯¯¯

= 900:

5. detA =

¯¯¯¯¯¯

1 0 ¡2

3 1 4

5 2 ¡3

¯¯¯¯¯¯

1 0 0

3 1 10

5 2 7

¯¯¯¯¯¯

¯¯¯¯

1 10

2 7

¯¯¯¯

= ¡13.

Hence A is non{singular and

A¡1 =

¡13

adjA =

¡13

C11 C21 C31

C12 C22 C32

C13 C23 C33

5 =

¡13

4 ¡11 ¡4 2

29 7 ¡10

1 ¡2 1

5 :

6. (i)

¯¯¯¯¯¯

2a 2b b ¡ c

2b 2a a + c

a + b a + b b

¯¯¯¯¯¯

R1 ! R1 + R2

¯¯¯¯¯¯

2a + 2b 2b + 2a b + a

2b 2a a + c

a + b a + b b

¯¯¯¯¯¯

= (a+b)

¯¯¯¯¯¯

2 2 1

2b 2a a + c

a + b a + b b

¯¯¯¯¯¯

C1 ! C1 ¡ C2

(a+b)

¯¯¯¯¯¯

0 2 1

2(b ¡ a) 2a a + c

0 a + b b

¯¯¯¯¯¯

= 2(a + b)(a ¡ b)

¯¯¯¯

2 1

a + b b

¯¯¯¯

= ¡2(a + b)(a ¡ b)2:

(ii)

¯¯¯¯¯¯

b + c b c

c c + a a

b a a + b

¯¯¯¯¯¯

C1 ! C1 ¡ C2

¯¯¯¯¯¯

c b c

¡a c + a a

b ¡ a a a + b

¯¯¯¯¯¯

C3 ! C3 ¡ C1

¯¯¯¯¯¯

c b 0

¡a c + a 2a

b ¡ a a 2a

¯¯¯¯¯¯

= 2a

¯¯¯¯¯¯

c b 0

¡a c + a 1

b ¡ a a 1

¯¯¯¯¯¯

R3 ! R3 ¡ R2

¯¯¯¯¯¯

c b 0

¡a c + a 1

b ¡c 0

¯¯¯¯¯¯

= ¡2a

¯¯¯¯

c b

b ¡c

¯¯¯¯

= 2a(c2 + b2):

7. Suppose that the curve y = ax2 + bx + c passes through the points

(x1; y1); (x2; y2); (x3; y3), where xi 6= xj if i 6= j. Then

ax2

1 + bx1 + c = y1

ax2

2 + bx2 + c = y2

ax2

3 + bx3 + c = y3:

The coe±cient determinant is essentially a Vandermonde determinant:

¯¯¯¯¯¯

1 x1 1

2 x2 1

3 x3 1

¯¯¯¯¯¯

¯¯¯¯¯¯ x

2 x2

x1 x2 x3

1 1 1

¯¯¯¯¯¯

= ¡

¯¯¯¯¯¯

1 1 1

x1 x2 x3

1 x2

2 x2

¯¯¯¯¯¯

= ¡(x2¡x1)(x3¡x1)(x3¡x2):

Hence the coe±cient determinant is non{zero and by Cramer's rule, there

is a unique solution for a; b; c.

8. Let ¢ = detA =

¯¯¯¯¯¯

1 1 ¡1

2 3 k

1 k 3

¯¯¯¯¯¯

. Then

¢ =

C3 ! C3 + C1

C2 ! C2 ¡ C1

¯¯¯¯¯¯

1 0 0

2 1 k + 2

1 k ¡ 1 4

¯¯¯¯¯¯

¯¯¯¯

1 k + 2

k ¡ 1 4

¯¯¯¯

= 4 ¡ (k ¡ 1)(k + 2) = ¡(k2 ¡ k ¡ 6) = ¡(k + 3)(k ¡ 2):

Hence detA = 0 if and only if k = ¡3 or k = 2.

Consequently if k 6= ¡3 and k 6= 2, then detA 6= 0 and the given system

x + y ¡ z = 1

2x + 3y + kz = 3

x + ky + 3z = 2

has a unique solution. We consider the cases k = ¡3 and k = 2 separately.

k = ¡3 :

AM =

1 1 ¡1 1

2 3 ¡3 3

1 ¡3 3 2

5 R2 ! R2 ¡ 2R1

R3 ! R3 ¡ R1

1 1 ¡1 1

0 1 ¡1 1

0 ¡4 4 1

R3 ! R3 + 4R2

1 1 ¡1 1

0 1 ¡1 1

0 0 0 5

5 ;

from which we read o® inconsistency.

k = 2 :

AM =

1 1 ¡1 1

2 3 2 3

1 2 3 2

5 R2 ! R2 ¡ 2R1

R3 ! R3 ¡ R1

1 1 ¡1 1

0 1 4 1

R3 ! R3 ¡ R2

1 0 ¡5 0

0 1 4 1

0 0 0 0

5 :

We read o® the complete solution x = 5z; y = 1 ¡ 4z, where z is arbitrary.

Finally we have to determine the solution for which x2 +y2 +z2 is least.

x2 + y2 + z2 = (5z)2 + (1 ¡ 4z)2 + z2 = 42z2 ¡ 8z + 1

= 42(z2 ¡

z +

) = 42

(µ

z ¡

¶2

42 ¡

¶2

)

= 42

(µ

z ¡

¶2

882

)

We see that the least value of x2+y2+z2 is 42£ 13

882 = 13

21 and this occurs when

z = 2=21, with corresponding values x = 10=21 and y = 1¡4£ 2

21 = 13=21.

9. Let ¢ =

1 ¡2 b

a 0 2

5 2 0

¯¯¯¯¯¯

be the coe±cient determinant of the given system.

Then expanding along column 2 gives

¢ = 2

¯¯¯¯

a 2

5 0

¯¯¯¯

¡ 2

¯¯¯¯

1 b

a 2

¯¯¯¯

= ¡20 ¡ 2(2 ¡ ab)

= 2ab ¡ 24 = 2(ab ¡ 12):

Hence ¢ = 0 if and only if ab = 12. Hence if ab 6= 12, the given system has

a unique solution.

If ab = 12 we must argue with care:

AM =

1 ¡2 b 3

a 0 2 2

5 2 0 1

5 !

1 ¡2 b 3

0 2a 2 ¡ ab 2 ¡ 3a

0 12 ¡5b ¡14

1 ¡2 b 3

0 1 ¡5b

12 ¡7

0 2a 2 ¡ ab 2 ¡ 3a

5 !

1 ¡2 b 3

0 1 ¡5b

12 ¡7

0 0 12¡ab

6¡2a

1 ¡2 b 3

0 1 ¡5b

12 ¡7

0 0 0 6¡2a

5 = B:

Hence if 6 ¡ 2a 6= 0, i.e. a 6= 3, the system has no solution.

If a = 3 (and hence b = 4), then

B =

1 ¡2 4 3

0 1 ¡5

3 ¡7

0 0 0 0

5 !

1 0 ¡2=3 2=3

0 1 ¡5

3 ¡7

0 0 0 0

5 :

Consequently the complete solution of the system is x = 2

3+2

3z; y = ¡7

6 +5

3z,

where z is arbitrary. Hence there are in¯nitely many solutions.

10.

¢ =

¯¯¯¯¯¯¯¯

1 1 2 1

1 2 3 4

2 4 7 2t + 6

2 2 6 ¡ t t

¯¯¯¯¯¯¯¯

R4 ! R4 ¡ 2R1

R3 ! R3 ¡ 2R1

R2 ! R2 ¡ R1

¯¯¯¯¯¯¯¯

1 1 2 1

0 1 1 3

0 2 3 2t + 4

0 0 2 ¡ t t ¡ 2

¯¯¯¯¯¯¯¯

¯¯¯¯¯¯

1 1 3

2 3 2t + 4

0 2 ¡ t t ¡ 2

¯¯¯¯¯¯

R2 ! R2 ¡ 2R1

¯¯¯¯¯¯

1 1 3

0 1 2t ¡ 2

0 2 ¡ t t ¡ 2

¯¯¯¯¯¯

¯¯¯¯

1 2t ¡ 2

2 ¡ t t ¡ 2

¯¯¯¯

= (t ¡ 2)

¯¯¯¯

1 2t ¡ 2

¡1 1

¯¯¯¯

= (t ¡ 2)(2t ¡ 1):

Hence ¢ = 0 if and only if t = 2 or t = 1

2 . Consequently the given matrix

B is non{singular if and only if t 6= 2 and t 6= 1

2 .

11. Let A be a 3 £ 3 matrix with detA 6= 0. Then

(i)

AadjA = (detA)I3 (1)

(detA) det ( adjA) = det (detA ¢ I3) = (detA)3:

Hence, as detA 6= 0, dividing out by detA in the last equation gives

det ( adjA) = (detA)2:

(ii) . Also from equation (1)

detA

adjA = I3;

so adjA is non{singular and

( adjA)¡1 =

detA

Finally

A¡1 adj (A¡1) = (detA¡1)I3

and multiplying both sides of the last equation by A gives

adj (A¡1) = A(detA¡1)I3 =

detA

12. Let A be a real 3 £ 3 matrix satisfying AtA = I3. Then

(i) At(A ¡ I3) = AtA ¡ At = I3 ¡ At

= ¡(At ¡ I3) = ¡(At ¡ It

3 ) = ¡(A ¡ I3)t:

Taking determinants of both sides then gives

detAt det (A ¡ I3) = det (¡(A ¡ I3)t)

detAdet (A ¡ I3) = (¡1)3 det (A ¡ I3)t

= ¡det (A ¡ I3) (1):

(ii) Also detAAt = det I3, so

detAt detA = 1 = (detA)2:

Hence detA = §1.

(iii) Suppose that detA = 1. Then equation (1) gives

det (A ¡ I3) = ¡det (A ¡ I3);

so (1 + 1) det (A ¡ I3) = 0 and hence det (A ¡ I3) = 0.

13. Suppose that column 1 is a linear combination of the remaining columns:

A¤1 = x2A¤2 + ¢ ¢ ¢ + xnA¤n:

Then

detA =

¯¯¯¯¯¯¯¯¯

x2a12 + ¢ ¢ ¢ + xna1n a12 ¢ ¢ ¢ a1n

x2a22 + ¢ ¢ ¢ + xna2n a22 ¢ ¢ ¢ a2n

...

x2an2 + ¢ ¢ ¢ + xnann an2 ¢ ¢ ¢ ann

¯¯¯¯¯¯¯¯¯

Now detA is unchanged in value if we perform the operation

C1 ! C1 ¡ x2C2 ¡ ¢ ¢ ¢ ¡ xnCn :

detA =

¯¯¯¯¯¯¯¯¯

0 a12 ¢ ¢ ¢ a1n

0 a22 ¢ ¢ ¢ a2n

...

0 an2 ¢ ¢ ¢ ann

¯¯¯¯¯¯¯¯¯

= 0:

Conversely, suppose that detA = 0. Then the homogeneous system AX = 0

has a non{trivial solution X = [x1; : : : ; xn]t. So

x1A¤1 + ¢ ¢ ¢ + xnA¤n = 0:

Suppose for example that x1 6= 0. Then

A¤1 =

+ ¢ ¢ ¢ +

A¤n

and the ¯rst column of A is a linear combination of the remaining columns.

14. Consider the system

¡2x + 3y ¡ z = 1

x + 2y ¡ z = 4

¡2x ¡ y + z = ¡3

Let ¢ =

¯¯¯¯¯¯

¡2 3 ¡1

1 2 ¡1

¡2 ¡1 1

¯¯¯¯¯¯

0 7 ¡3

1 2 ¡1

0 3 ¡1

¯¯¯¯¯¯

= ¡

¯¯¯¯

7 ¡3

3 ¡1

¯¯¯¯

= ¡2 6= 0.

Hence the system has a unique solution which can be calculated using

Cramer's rule:

x =

¢1

; y =

¢2

; z =

¢3

;

where

¢1 =

¯¯¯¯¯¯

1 3 ¡1

4 2 ¡1

¡3 ¡1 1

¯¯¯¯¯¯

= ¡4;

¢2 =

¯¯¯¯¯¯

¡2 1 ¡1

1 4 ¡1

¡2 ¡3 1

¯¯¯¯¯¯

= ¡6;

¢3 =

¯¯¯¯¯¯

¡2 3 1

1 2 4

¡2 ¡1 ¡3

¯¯¯¯¯¯

= ¡8:

Hence x = ¡4

¡2 = 2; y = ¡6

¡2 = 3; z = ¡8

¡2 = 4.

15. In Remark 4.0.4, take A = In. Then we deduce

(a) detEij = ¡1;

(b) detEi(t) = t;

Now suppose that B is a non{singular n£n matrix. Then we know that B

is a product of elementary row matrices:

B = E1 ¢ ¢ ¢Em:

Consequently we have to prove that

detE1 ¢ ¢ ¢EmA = detE1 ¢ ¢ ¢Em det A:

We prove this by induction on m.

First the case m = 1. We have to prove detE1A = detE1 detA if E1 is

an elementary row matrix. This follows form Remark 4.0.4:

(a) detEijA = ¡detA = detEij detA;

(b) detEi(t)A = t detA = detEi(t) detA;

Let m ¸ 1 and assume the proposition holds for products of m elementary

row matrices. Then

detE1 ¢ ¢ ¢EmEm+1A = det (E1 ¢ ¢ ¢Em)(Em+1A)

= det (E1 ¢ ¢ ¢Em) det (Em+1A)

= det (E1 ¢ ¢ ¢Em) detEm+1 detA

= det ((E1 ¢ ¢ ¢Em)Em+1) detA

and the induction goes through.

Hence detBA = detB detA if B is non{singular.

If B is singular, problem 26, Chapter 2.7 tells us that BA is also singlular.

However singular matrices have zero determinant, so

detB = 0 detBA = 0;

so the equation detBA = detB detA holds trivially in this case.

16. ¯¯¯¯¯¯¯¯

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

¯¯¯¯¯¯¯¯

R1 ! R1 ¡ R2

R2 ! R2 ¡ R3

R3 ! R3 ¡ R4

¯¯¯¯¯¯¯¯

c ¡c 0 0

b b + c ¡b ¡ c ¡b

0 0 c ¡c

a a a + b a + b + c

¯¯¯¯¯¯¯¯

C2 ! C2 + C1

¯¯¯¯¯¯¯¯

c 0 0 0

b 2b + c ¡b ¡ c ¡b

0 0 c ¡c

a 2a a + b a + b + c

¯¯¯¯¯¯¯¯

= c

¯¯¯¯¯¯

2b + c ¡b ¡ c ¡b

0 c ¡c

2a a + b a + b + c

¯¯¯¯¯¯

C3 ! C3 + C2

¯¯¯¯¯¯

2b + c ¡b ¡ c ¡2b ¡ c

0 c 0

2a a + b 2a + 2b + c

¯¯¯¯¯¯

= c2

¯¯¯¯

2b + c ¡2b ¡ c

2a 2a + 2b + c

¯¯¯¯

= c2(2b + c)

¯¯¯¯

1 ¡1

2a 2a + 2b + c

¯¯¯¯

= c2(2b + c)(4a + 2b + c):

17. Let ¢ =

¯¯¯¯¯¯¯¯

1 + u1 u1 u1 u1

u2 1 + u2 u2 u2

u3 u3 1 + u3 u3

u4 u4 u4 1 + u4

¯¯¯¯¯¯¯¯

: Then using the operation

R1 ! R1 + R2 + R3 + R4

we have

¢ =

¯¯¯¯¯¯¯¯

t t t t

u2 1 + u2 u2 u2

u3 u3 1 + u3 u3

u4 u4 u4 1 + u4

¯¯¯¯¯¯¯¯

(where t = 1 + u1 + u2 + u3 + u4)

= (1 + u1 + u2 + u3 + u4)

¯¯¯¯¯¯¯¯

1 1 1 1

u2 1 + u2 u2 u2

u3 u3 1 + u3 u3

u4 u4 u4 1 + u4

¯¯¯¯¯¯¯¯

The last determinant equals

C2 ! C2 ¡ C1

C3 ! C3 ¡ C1

C4 ! C4 ¡ C1

¯¯¯¯¯¯¯¯

1 0 0 0

u2 1 0 0

u3 0 1 0

u4 0 0 1

¯¯¯¯¯¯¯¯

= 1:

18. Suppose that At = ¡A, that A 2 Mn£n(F), where n is odd. Then

detAt = det(¡A)

detA = (¡1)n detA = ¡det A:

Hence (1 + 1) detA = 0 and consequently detA = 0 if 1 + 1 6= 0 in F.

19.

¯¯¯¯¯¯¯¯

1 1 1 1

r 1 1 1

r r 1 1

r r r 1

¯¯¯¯¯¯¯¯

C4 ! C4 ¡ C3

C3 ! C3 ¡ C2

C2 ! C2 ¡ C1

¯¯¯¯¯¯¯¯

1 0 0 0

r 1 ¡ r 0 0

r 0 1 ¡ r 0

r 0 0 1 ¡ r

¯¯¯¯¯¯¯¯

= (1 ¡ r)3:

20.

¯¯¯¯¯¯

1 a2 ¡ bc a4

1 b2 ¡ ca b4

1 c2 ¡ ab c4

¯¯¯¯¯¯

R2 ! R2 ¡ R1

R3 ! R3 ¡ R1

¯¯¯¯¯¯

1 a2 ¡ bc a4

0 b2 ¡ ca ¡ a2 + bc b4 ¡ a4

0 c2 ¡ ab ¡ a2 + bc c4 ¡ a4

¯¯¯¯¯¯

¯¯¯¯

b2 ¡ ca ¡ a2 + bc b4 ¡ a4

c2 ¡ ab ¡ a2 + bc c4 ¡ a4

¯¯¯¯

(b ¡ a)(b + a) + c(b ¡ a) (b ¡ a)(b + a)(b2 + a2)

(c ¡ a)(c + a) + b(c ¡ a) (c ¡ a)(c + a)(c2 + a2)

¯¯¯¯

(b ¡ a)(b + a + c) (b ¡ a)(b + a)(b2 + a2)

(c ¡ a)(c + a + b) (c ¡ a)(c + a)(c2 + a2)

¯¯¯¯

= (b ¡ a)(c ¡ a)

¯¯¯¯

b + a + c (b + a)(b2 + a2)

c + a + b (c + a)(c2 + a2)

¯¯¯¯

= (b ¡ a)(c ¡ a)(a + b + c)

¯¯¯¯

1 (b + a)(b2 + a2)

1 (c + a)(c2 + a2)

¯¯¯¯

Finally

¯¯¯¯

1 (b + a)(b2 + a2)

1 (c + a)(c2 + a2)

¯¯¯¯

= (c3 + ac2 + ca2 + a3) ¡ (b3 + ab2 + ba2 + a3)

= (c3 ¡ b3) + a(c2 ¡ b2) + a2(c ¡ b)

= (c ¡ b)(c2 + cb + b2 + a(c + b) + a2)

= (c ¡ b)(c2 + cb + b2 + ac + ab + a2):

Section 5.8

(i) (¡3 + i)(14 ¡ 2i) = (¡3)(14 ¡ 2i) + i(14 ¡ 2i)

= f(¡3)14 ¡ (¡3)(2i)g + i(14) ¡ i(2i)

= (¡42 + 6i) + (14i + 2) = ¡40 + 20i:

(ii)

2 + 3i

1 ¡ 4i

(2 + 3i)(1 + 4i)

(1 ¡ 4i)(1 + 4i)

((2 + 3i) + (2 + 3i)(4i)

12 + 42

= ¡10 + 11i

= ¡10

(iii)

(1 + 2i)2

1 ¡ i

1 + 4i + (2i)2

1 ¡ i

1 + 4i ¡ 4

1 ¡ i

= ¡3 + 4i

1 ¡ i

(¡3 + 4i)(1 + i)

= ¡7 + i

= ¡

2. (i)

iz + (2 ¡ 10i)z = 3z + 2i , z(i + 2 ¡ 10i ¡ 3) = 2i

=, z(¡1 ¡ 9i) = 2i , z = ¡2i

1 + 9i

= ¡2i(1 ¡ 9i)

1 + 81

= ¡18 ¡ 2i

= ¡9 ¡ i

(ii) The coe±cient determinant is

¯¯¯¯

1 + i 2 ¡ i

1 + 2i 3 + i

¯¯¯¯

= (1 + i)(3 + i) ¡ (2 ¡ i)(1 + 2i) = ¡2 + i 6= 0:

Hence Cramer's rule applies: there is a unique solution given by

z =

¯¯¯¯

¡3i 2 ¡ i

2 + 2i 3 + i

¯¯¯¯

¡2 + i

= ¡3 ¡ 11i

¡2 + i

= ¡1 + 5i

w =

¯¯¯¯

1 + i ¡3i

1 + 2i 2 + 2i

¯¯¯¯

¡2 + i

= ¡6 + 7i

¡2 + i

19 ¡ 8i

1 + (1 + i) + ¢ ¢ ¢ + (1 + i)99 =

(1 + i)100 ¡ 1

(1 + i) ¡ 1

(1 + i)100 ¡ 1

= ¡i

(1 + i)100 ¡ 1

Now (1 + i)2 = 2i. Hence

(1 + i)100 = (2i)50 = 250i50 = 250(¡1)25 = ¡250:

Hence ¡i

(1 + i)100 ¡ 1

= ¡i(¡250 ¡ 1) = (250 + 1)i.

4. (i) Let z2 = ¡8 ¡ 6i and write z=x+iy, where x and y are real. Then

z2 = x2 ¡ y2 + 2xyi = ¡8 ¡ 6i;

so x2 ¡ y2 = ¡8 and 2xy = ¡6. Hence

y = ¡3=x; x2 ¡

¡3

¶2

= ¡8;

so x4 + 8x2 ¡ 9 = 0. This is a quadratic in x2. Hence x2 = 1 or ¡9 and

consequently x2 = 1. Hence x = 1; y = ¡3 or x = ¡1 and y = 3. Hence

z = 1 ¡ 3i or z = ¡1 + 3i.

(ii) z2 ¡ (3 + i)z + 4 + 3i = 0 has the solutions z = (3 + i § d)=2, where d is

any complex number satisfying

d2 = (3 + i)2 ¡ 4(4 + 3i) = ¡8 ¡ 6i:

Hence by part (i) we can take d = 1 ¡ 3i. Consequently

z =

3 + i § (1 ¡ 3i)

= 2 ¡ i or 1 + 2i:

(i) The number lies in the ¯rst quadrant of

the complex plane.

j4 + ij =

42 + 12 = p17:

Also Arg (4 + i) = ®, where tan ® = 1=4

and 0 < ® < ¼=2. Hence ® = tan ¡1(1=4).

¾ -

4 + i

»»»®»:

(ii) The number lies in the third quadrant of

the complex plane.

¯¯¯¯

¡3 ¡ i

¯¯¯¯

= j¡3 ¡ ij

(¡3)2 + (¡1)2 =

p9 + 1 =

p10

Also Arg (¡3¡i

2 ) = ¡¼ +®, where tan ® =

2=3

2 = 1=3 and 0 < ® < ¼=2. Hence ® =

tan ¡1(1=3).

¾ -

¡3¡i

)³®³³

(iii) The number lies in the second quadrant of

the complex plane.

j ¡ 1 + 2ij =

(¡1)2 + 22 = p5:

Also Arg (¡1+2i) = ¼¡®, where tan ® =

2 and 0 < ® < ¼=2. Hence ® = tan ¡12.

¾ -

y ¡1 + 2i

® A

(iv) The number lies in the second quadrant of

the complex plane.

¯¯¯¯¯

¡1 + ip3

¯¯¯¯¯

= j ¡ 1 + ip3j

(¡1)2 + (p3)2 =

p1 + 3 = 1:

Also Arg (¡1

2 +

2 i) = ¼ ¡ ®, where

tan ® =

2 =1

2 = p3 and 0 < ® < ¼=2.

Hence ® = ¼=3.

¾ -

¡1

2 +

2 i

® J

6. (i) Let z = (1 + i)(1 + p3i)(p3 ¡ i). Then

jzj = j1 + ijj1 + p3ijj

p3 ¡ ij

12 + 12

12 + (p3)2

(p3)2 + (¡1)2

= p2p4p4 = 4p2:

Arg z ´ Arg (1 + i) + Arg (1 + p3) + Arg (p3 ¡ i) (mod 2¼)

3 ¡

6 ´

Hence Arg z = 5

12 and the polar decomposition of z is

z = 4p2

cos

5¼

+ i sin

5¼

(ii) Let z = (1+i)5(1¡ip3)5

(p3+i)4 . Then

jzj = j(1 + i)j5j(1 ¡ ip3)j5

j(p3 + i)j4

¡p2

¢5

24 = 27=2:

Arg z ´ Arg (1 + i)5 + Arg (1 ¡

p3i)5 ¡ Arg (p3 + i)4 (mod 2¼)

´ 5Arg (1 + i) + 5Arg (1 ¡

p3i) ¡ 4Arg (p3 + i)

´ 5

+ 5

¡¼

¡ 4

6 ´ ¡13¼

12 ´

11¼

Hence Arg z = 11¼

12 and the polar decomposition of z is

z = 27=2

cos

11¼

+ i sin

11¼

7. (i) Let z = 2(cos ¼

4 + i sin ¼

4 ) and w = 3(cos ¼

6 + i sin ¼

6 ). (Both of these

numbers are already in polar form.)

(a) zw = 6(cos ( ¼

4 + ¼

6 ) + i sin ( ¼

4 + ¼

6 ))

= 6(cos 5¼

12 + i sin 5¼

12 ).

(b) z

w = 2

3 (cos ( ¼

4 ¡ ¼

6 ) + i sin ( ¼

4 ¡ ¼

6 ))

= 2

3 (cos ¼

12 + i sin ¼

12 ).

z = 3

2 (cos ( ¼

6 ¡ ¼

4 ) + i sin ( ¼

6 ¡ ¼

4 ))

= 3

2 (cos (¡¼

12 ) + i sin (¡¼

12 )).

(d) z5

w2 = 25

32 (cos ( 5¼

4 ¡ 2¼

6 ) + i sin ( 5¼

4 ¡ 2¼

6 ))

= 32

9 (cos 11¼

12 + i sin 11¼

12 ).

(a) (1 + i)2 = 2i, so

(1 + i)12 = (2i)6 = 26i6 = 64(i2)3 = 64(¡1)3 = ¡64:

(b) ( 1¡i p2

)2 = ¡i, so

1 ¡ i

¡6

Ãµ

1 ¡ i

¶2

¡3

= (¡i)¡3 = ¡1

i3 = ¡1

¡i

= ¡i:

8. (i) To solve the equation z2 = 1 + p3i, we write 1 + p3i in modulus{

argument form:

1 + p3i = 2(cos

+ i sin

Then the solutions are

zk = p2

cos

µ ¼

3 + 2k¼

+ i sin

µ ¼

3 + 2k¼

¶¶

; k = 0; 1:

Now k = 0 gives the solution

z0 = p2(cos

+ i sin

) = p2

Ãp3

Clearly z1 = ¡z0.

(ii) To solve the equation z4 = i, we write i in modulus{argument form:

i = cos

+ i sin

Then the solutions are

zk = cos

µ ¼

2 + 2k¼

+ i sin

µ ¼

2 + 2k¼

; k = 0; 1; 2; 3:

Now cos

³ ¼

2 +2k¼

= cos

¡¼

8 + k¼

, so

zk = cos

k¼

+ sin

k¼

cos

+ i sin

´k

(cos

+ i sin

)

= ik(cos

+ i sin

Geometrically, the solutions lie equi{spaced on the unit circle at arguments

;

5¼

;

+ ¼ =

9¼

;

+ 3

13¼

Also z2 = ¡z0 and z3 = ¡z1.

(iii) To solve the equation z3 = ¡8i, we rewrite the equation as

¡2i

¶3

= 1:

Then µ

¡2i

= 1; ¡1 + p3i

; or ¡1 ¡ p3i

Hence z = ¡2i; p3 + i or ¡p3 + i.

Geometrically, the solutions lie equi{spaced on the circle jzj = 2, at

arguments

;

2¼

5¼

;

+ 2

2¼

3¼

(iv) To solve z4 = 2 ¡ 2i, we write 2 ¡ 2i in modulus{argument form:

2 ¡ 2i = 23=2

cos ¡¼

+ i sin ¡¼

Hence the solutions are

zk = 23=8 cos

¡¼

4 + 2k¼

+ i sin

¡¼

4 + 2k¼

; k = 0; 1; 2; 3:

We see the solutions can also be written as

zk = 23=8ik

cos ¡¼

+ i sin ¡¼

= 23=8ik

cos

16 ¡ i sin

Geometrically, the solutions lie equi{spaced on the circle jzj = 23=8, at ar-

guments

¡¼

; ¡¼

7¼

; ¡¼

+ 2

15¼

; ¡¼

+ 3

23¼

Also z2 = ¡z0 and z3 = ¡z1.

2 + i ¡1 + 2i 2

1 + i ¡1 + i 1

1 + 2i ¡2 + i 1 + i

5 R1 ! R1 ¡ R2

R3 ! R3 ¡ R2

1 i 1

1 + i ¡1 + i 1

i ¡1 i

R2 ! R2 ¡ (1 + i)R1

R3 ! R3 ¡ iR1

1 i 1

0 0 ¡i

0 0 0

5 R2 ! iR2

1 i 1

0 0 1

0 0 0

R1 ! R1 ¡ R2

1 i 0

0 0 1

0 0 0

5 :

The last matrix is in reduced row{echelon form.

10. (i) Let p = l + im and z = x + iy. Then

pz + pz = (l ¡ im)(x + iy) + (l + im)(x ¡ iy)

= (lx + liy ¡ imx + my) + (lx ¡ liy + imx + my)

= 2(lx + my):

Hence pz + pz = 2n , lx + my = n:

(ii) Let w be the complex number which results from re°ecting the com-

plex number z in the line lx + my = n. Then because p is perpendicular to

the given line, we have

w ¡ z = tp; t 2 R: (a)

Also the midpoint w+z

2 of the segment joining w and z lies on the given line,

w + z

+ p

w + z

= n;

w + z

+ p

w + z

= n: (b)

Taking conjugates of equation (a) gives

w ¡ z = tp: (c)

Then substituting in (b), using (a) and (c), gives

2w ¡ tp

+ p

2z + tp

= n

and hence

pw + pz = n:

(iii) Let p = b ¡ a and n = jbj2 ¡ jaj2. Then

jz ¡ aj = jz ¡ bj , jz ¡ aj2 = jz ¡ bj2

, (z ¡ a)(z ¡ a) = (z ¡ b)(z ¡ b)

, zz ¡ az ¡ za + aa = zz ¡ bz ¡ zb + bb

, (b ¡ a)z + (b ¡ a)z = jbj2 ¡ jaj2

, pz + pz = n:

Suppose z lies on the circle

¯¯¯

z¡a

z¡b

¯¯¯

and let w be the re°ection 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

(a)

Substituting for z in the circle equation, using (a) gives

¸ =

¯¯¯¯¯

n¡pw

p ¡ a

n¡pw

p ¡ b

¯¯¯¯¯

¯¯¯¯

n ¡ pw ¡ pa

n ¡ pw ¡ pb

¯¯¯¯

: (b)

However

n ¡ pa = jbj2 ¡ jaj2 ¡ (b ¡ a)a

= bb ¡ aa ¡ ba + aa

= b(b ¡ a) = bp:

Similarly n ¡ pb = ap. Consequently (b) simpli¯es to

¸ =

¯¯¯¯b

ap ¡ pw

¯¯¯¯

b ¡ w

a ¡ w

¯¯¯¯

w ¡ b

w ¡ a

¯¯¯¯

;

which gives

¯¯¯

w¡a

w¡b

¯¯¯

= 1

¸.

11. Let a and b be distinct complex numbers and 0 < ® < ¼.

(i) When z1 lies on the circular arc shown, it subtends a constant angle

®. This angle is given by Arg (z1 ¡ a) ¡ Arg (z1 ¡ b). However

Arg

z1 ¡ a

z1 ¡ b

= Arg (z1 ¡ a) ¡ Arg (z1 ¡ b) + 2k¼

= ® + 2k¼:

It follows that k = 0, as 0 < ® < ¼ and ¡¼ < Arg µ · ¼. Hence

Arg

z1 ¡ a

z1 ¡ b

= ®:

Similarly if z2 lies on the circular arc shown, then

Arg

z2 ¡ a

z2 ¡ b

= ¡° = ¡(¼ ¡ ®) = ® ¡ ¼:

Replacing ® by ¼ ¡ ®, we deduce that if z4 lies on the circular arc shown,

then

Arg

z4 ¡ a

z4 ¡ b

= ¼ ¡ ®;

while if z3 lies on the circular arc shown, then

Arg

z3 ¡ a

z3 ¡ b

= ¡®:

The straight line through a and b has the equation

z = (1 ¡ t)a + tb;

where t is real. Then 0 < t < 1 describes the segment ab. Also

z ¡ a

z ¡ b

t ¡ 1

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

Arg

z ¡ a

z ¡ b

= ¼; 0;

respectively.

(ii) Case (a) Suppose z1; z2 and z3 are not collinear. Then these points

determine a circle. Now z1 and z2 partition this circle into two arcs. If z3

and z4 lie on the same arc, then

Arg

z3 ¡ z1

z3 ¡ z2

= Arg

z4 ¡ z1

z4 ¡ z2

;

whereas if z3 and z4 lie on opposite arcs, then

Arg

z3 ¡ z1

z3 ¡ z2

= ®

and

Arg

z4 ¡ z1

z4 ¡ z2

= ® ¡ ¼:

Hence in both cases

Arg

z3 ¡ z1

z3 ¡ z2

z4 ¡ z1

z4 ¡ z2

´ Arg

z3 ¡ z1

z3 ¡ z2

¡ Arg

z4 ¡ z1

z4 ¡ z2

(mod 2¼)

´ 0 or ¼:

In other words, the cross{ratio

z3 ¡ z1

z3 ¡ z2

z4 ¡ z1

z4 ¡ z2

is real.

(b) If z1; z2 and z3 are collinear, then again the cross{ratio is real.

The argument is reversible.

(iii) Assume that A; B; C; D are distinct points such that the cross{ratio

r =

z3 ¡ z1

z3 ¡ z2

z4 ¡ z1

z4 ¡ z2

is real. Now r cannot be 0 or 1. Then there are three cases:

(i) 0 < r < 1;

(ii) r < 0;

(iii) r > 1.

Case (i). Here jrj + j1 ¡ rj = 1. So

¯¯¯¯

z4 ¡ z1

z4 ¡ z2 ¢

z3 ¡ z2

z3 ¡ z1

¯¯¯¯

1 ¡

z4 ¡ z1

z4 ¡ z2 ¢

z3 ¡ z2

z3 ¡ z1

¶¯¯¯¯ = 1:

Multiplying both sides by the denominator jz4 ¡ z2jjz3 ¡ z1j gives after

simpli¯cation

jz4 ¡ z1jjz3 ¡ z2j + jz2 ¡ z1jjz4 ¡ z3j = jz4 ¡ z2jjz3 ¡ z1j;

(a) AD ¢ BC + AB ¢ CD = BD ¢ AC:

Case (ii). Here 1 + jrj = j1 ¡ rj. This leads to the equation

(b) BD ¢ AC + AD ¢ BC+ = AB ¢ CD:

Case (iii). Here 1 + j1 ¡ rj = jrj. This leads to the equation

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

jrj + j1 ¡ rj = 1; 1 + jrj = j1 ¡ rj; 1 + j1 ¡ rj = jrj;

from which we deduce that r is real.

Section 6.3

1. Let A =

4 ¡3

1 0

. Then A has characteristic equation ¸2 ¡ 4¸ + 3 = 0

or (¸ ¡ 3)(¸ ¡ 1) = 0. Hence the eigenvalues of A are ¸1 = 3 and ¸2 = 1.

¸1 = 3. The corresponding eigenvectors satisfy (A ¡ ¸1I2)X = 0, or

1 ¡3

;

or equivalently x ¡ 3y = 0. Hence

= y

and we take X1 =

Similarly for ¸2 = 1 we ¯nd the eigenvector X2 =

Hence if P = [X1jX2] =

3 1

1 1

, then P is non{singular and

P¡1AP =

3 0

0 1

Hence

A = P

3 0

0 1

P¡1

and consequently

An = P

3n 0

0 1n

P¡1

3 1

1 1

¸ ·

3n 0

0 1n

1 ¡1

¡1 3

3n+1 1

3n 1

¸ ·

1 ¡1

¡1 3

3n+1 ¡ 1 ¡3n+1 + 3

3n ¡ 1 ¡3n + 3

3n ¡ 1

A +

3 ¡ 3n

I2:

2. Let A =

3=5 4=5

2=5 1=5

. Then we ¯nd that the eigenvalues are ¸1 = 1 and

¸2 = ¡1=5, with corresponding eigenvectors

X1 =

and X2 =

¡1

Then if P = [X1jX2], P is non{singular and

P¡1AP =

1 0

0 ¡1=5

and A = P

1 0

0 ¡1=5

P¡1:

Hence

An = P

1 0

0 (¡1=5)n

P¡1

! P

1 0

0 0

P¡1

2 ¡1

1 1

¸ ·

1 0

0 0

1 1

¡1 2

2 0

1 0

¸ ·

1 1

¡1 2

2 2

1 1

2=3 2=3

1=3 1=3

3. The given system of di®erential equations is equivalent to _X = AX,

where

A =

3 ¡2

5 ¡4

and X =

The matrix P =

2 1

5 1

is a non-singular matrix of eigenvectors corre-

sponding to eigenvalues ¸1 = ¡2 and ¸2 = 1. Then

P¡1AP =

¡2 0

0 1

The substitution X = PY , where Y = [x1; y1]t, gives

¡2 0

0 1

or equivalently x_1 = ¡2x1 and y_1 = y1.

Hence x1 = x1(0)e¡2t and y1 = y1(0)et. To determine x1(0) and y1(0),

we note that

x1(0)

y1(0)

= P¡1

x(0)

y(0)

= ¡

1 ¡1

¡5 2

¸ ·

Hence x1 = 3e¡2t and y1 = 7et. Consequently

x = 2x1 + y1 = 6e¡2t + 7et and y = 5x1 + y1 = 15e¡2t + 7et:

4. Introducing the vector Xn =

, the system of recurrence relations

xn+1 = 3xn ¡ yn

yn+1 = ¡xn + 3yn;

becomes Xn+1 = AXn, where A =

3 ¡1

¡1 3

. Hence Xn = AnX0, where

X0 =

To ¯nd An we can use the eigenvalue method. We get

An =

2n + 4n 2n ¡ 4n

2n ¡ 4n 2n + 4n

Hence

Xn =

2n + 4n 2n ¡ 4n

2n ¡ 4n 2n + 4n

¸ ·

2n + 4n + 2(2n ¡ 4n)

2n ¡ 4n + 2(2n + 4n)

3 £ 2n ¡ 4n

3 £ 2n + 4n

(3 £ 2n ¡ 4n)=2

(3 £ 2n + 4n)=2

Hence xn = 1

2 (3 £ 2n ¡ 4n) and yn = 1

2(3 £ 2n + 4n).

5. Let A =

a b

c d

be a real or complex matrix with distinct eigenvalues

¸1; ¸2 and corresponding eigenvectors X1; X2. Also let P = [X1jX2].

(a) The system of recurrence relations

xn+1 = axn + byn

yn+1 = cxn + dyn

has the solution

= An

¸1 0

0 ¸2

P¡1

¶n ·

= P

¸n

1 0

0 ¸n

P¡1

= [X1jX2]

¸n

1 0

0 ¸n

¸ ·

= [X1jX2]

¸n

1 ®

¸n

2 ¯

= ¸n

1 ®X1 + ¸n

2 ¯X2;

where ·

= P¡1

(b) In matrix form, the system is _X = AX, where X =

. We substitute

X = PY , where Y = [x1; y1]t. Then

= P _Y = AX = A(PY );

= (P¡1AP)Y =

¸1 0

0 ¸2

¸ ·

Hence x_1 = ¸1x1 and y_1 = ¸2y1. Then

x1 = x1(0)e¸1t and y1 = y1(0)e¸2t:

But ·

x(0)

y(0)

= P

x1(0)

y1(0)

;

so ·

x1(0)

y1(0)

= P¡1

x(0)

y(0)

Consequently x1(0) = ® and y1(0) = ¯ and

= P

= [X1jX2]

®e¸1t

¯e¸2t

= ®e¸1tX1 + ¯e¸2tX2:

6. Let A =

a b

c d

be a real matrix with non{real eigenvalues ¸ = a + ib

and ¸ = a¡ib, with corresponding eigenvectors X = U+iV and X = U¡iV ,

where U and V are real vectors. Also let P be the real matrix de¯ned by

P = [UjV ]. Finally let a + ib = reiµ, where r > 0 and µ is real.

(a) As X is an eigenvector corresponding to the eigenvalue ¸, we have AX =

¸X and hence

A(U + iV ) = (a + ib)(U + iV )

AU + iAV = aU ¡ bV + i(bU + aV ):

Equating real and imaginary parts then gives

AU = aU ¡ bV

AV = bU + aV:

(b)

AP = A[UjV ] = [AUjAV ] = [aU¡bV jbU+aV ] = [UjV ]

a b

¡b a

= P

a b

¡b a

Hence, as P can be shown to be non{singular,

P¡1AP =

a b

¡b a

(The fact that P is non{singular is easily proved by showing the columns of

P are linearly independent: Assume xU + yV = 0, where x and y are real.

Then we ¯nd

(x + iy)(U ¡ iV ) + (x ¡ iy)(U + iV ) = 0:

Consequently x+iy = 0 as U¡iV and U+iV are eigenvectors corresponding

to distinct eigenvalues a¡ib and a+ib and are hence linearly independent.

Hence x = 0 and y = 0.)

xn+1 = axn + byn

yn+1 = cxn + dyn

has solution

= An

= P

a b

¡b a

¸n

P¡1

= P

r cos µ r sin µ

¡r sin µ r cos µ

¸n ·

= Prn

cos µ sin µ

¡sin µ cos µ

¸n ·

= rn[UjV ]

cos nµ sin nµ

¡sin nµ cos nµ

¸ ·

= rn[UjV ]

® cos nµ + ¯ sin nµ

¡® sin nµ + ¯ cos nµ

= rn f(® cos nµ + ¯ sin nµ)U + (¡® sin nµ + ¯ cos nµ)V g

= rn f(cos nµ)(®U + ¯V ) + (sin nµ)(¯U ¡ ®V )g :

(d) The system of di®erential equations

= ax + by

= cx + dy

is attacked using the substitution X = PY , where Y = [x1; y1]t. Then

= (P¡1AP)Y;

so ·

x_1

y_1

a b

¡b a

¸ ·

Equating components gives

x_1 = ax1 + by1

y_1 = ¡bx1 + ay1:

Now let z = x1 + iy1. Then

z_ = x_1 + iy_1 = (ax1 + by1) + i(¡bx1 + ay1)

= (a ¡ ib)(x1 + iy1) = (a ¡ ib)z:

Hence

z = z(0)e(a¡ib)t

x1 + iy1 = (x1(0) + iy1(0))eat(cos bt ¡ i sin bt):

Equating real and imaginary parts gives

x1 = eat fx1(0) cos bt + y1(0) sin btg

y1 = eat fy1(0) cos bt ¡ x1(0) sin btg :

Now if we de¯ne ® and ¯ by

= P¡1

x(0)

y(0)

;

we see that ® = x1(0) and ¯ = y1(0). Then

= P

= [UjV ]

eat(® cos bt + ¯ sin bt)

eat(¯ cos bt ¡ ® sin bt)

= eatf(® cos bt + ¯ sin bt)U + (¯ cos bt ¡ ® sin bt)V g

= eatfcos bt(®U + ¯V ) + sin bt(¯U ¡ ®V )g:

7. (The case of repeated eigenvalues.) Let A =

a b

c d

and suppose that

the characteristic polynomial of A, ¸2 ¡(a+d)¸+(ad¡bc), has a repeated

root ®. Also assume that A 6= ®I2.

(i)

¸2 ¡ (a + d)¸ + (ad ¡ bc) = (¸ ¡ ®)2

= ¸2 ¡ 2®¸ + ®2:

Hence a + d = 2® and ad ¡ bc = ®2 and

(a + d)2 = 4(ad ¡ bc);

a2 + 2ad + d2 = 4ad ¡ 4bc;

a2 ¡ 2ad + d2 + 4bc = 0;

(a ¡ d)2 + 4bc = 0:

(ii) Let B ¡ A ¡ ®I2. Then

B2 = (A ¡ ®I2)2 = A2 ¡ 2®A + ®2I2

= A2 ¡ (a + d)A + (ad ¡ bc)I2;

But by problem 3, chapter 2.4, A2 ¡ (a + d)A + (ad ¡ bc)I2 = 0, so

B2 = 0.

(iii) Now suppose that B 6= 0. Then BE1 6= 0 or BE2 6= 0, as BEi is the

i{th column of B. Hence BX2 6= 0, where X2 = E1 or X2 = E2.

(iv) Let X1 = BX2 and P = [X1jX2]. We prove P is non{singular by

demonstrating that X1 and X2 are linearly independent.

Assume xX1 + yX2 = 0. Then

xBX2 + yX2 = 0

B(xBX2 + yX2) = B0 = 0

xB2X2 + yBX2 = 0

x0X2 + yBX2 = 0

yBX2 = 0:

Hence y = 0 as BX2 6= 0. Hence xBX2 = 0 and so x = 0.

Finally, BX1 = B(BX2) = B2X2 = 0, so (A ¡ ®I2)X1 = 0 and

AX1 = ®X1: (2)

Also

X1 = BX2 = (A ¡ ®I2)X2 = AX2 ¡ ®X2:

Hence

AX2 = X1 + ®X2: (3)

Then, using (2) and (3), we have

AP = A[X1jX2] = [AX1jAX2]

= [®X1jX1 + ®X2]

= [X1jX2]

® 1

0 ®

Hence

AP = P

® 1

0 ®

and hence

P¡1AP =

® 1

0 ®

8. The system of di®erential equations is equivalent to the single matrix

equation _X = AX, where A =

4 ¡1

4 8

The characteristic polynomial of A is ¸2 ¡ 12¸ + 36 = (¸ ¡ 6)2, so we

can use the previous question with ® = 6. Let

B = A ¡ 6I2 =

¡2 ¡1

4 2

Then BX2 =

¡2

, if X2 =

. Also let X1 = BX2. Then if

P = [X1jX2], we have

P¡1AP =

6 1

0 6

Now make the change of variables X = PY , where Y =

. Then

= (P¡1AP)Y =

6 1

0 6

or equivalently x_1 = 6x1 + y1 and y_1 = 6y1.

Solving for y1 gives y1 = y1(0)e6t. Consequently

x_1 = 6x1 + y1(0)e6t:

Multiplying both side of this equation by e¡6t gives

(e¡6tx1) = e¡6tx_1 ¡ 6e¡6tx1 = y1(0)

e¡6tx1 = y1(0)t + c;

where c is a constant. Substituting t = 0 gives c = x1(0). Hence

e¡6tx1 = y1(0)t + x1(0)

and hence

x1 = e6t(y1(0)t + x1(0)):

However, since we are assuming x(0) = 1 = y(0), we have

x1(0)

y1(0)

= P¡1

x(0)

y(0)

¡4

0 ¡1

¡4 ¡2

¸ ·

¡4

¡1

¡6

1=4

3=2

Hence x1 = e6t( 3

2 t + 1

4 ) and y1 = 3

2e6t.

Finally, solving for x and y,

¡2 1

4 0

¸ ·

¡2 1

4 0

¸ 2

e6t( 3

2 t + 1

4 )

2e6t

(¡2)e6t( 3

2 t + 1

4 ) + 3

2e6t

4e6t( 3

2 t + 1

4 )

e6t(1 ¡ 3t)

e6t(6t + 1)

Hence x = e6t(1 ¡ 3t) and y = e6t(6t + 1).

9. Let

A =

1=2 1=2 0

1=4 1=4 1=2

5 :

(a) We ¯rst determine the characteristic polynomial chA(¸).

chA(¸) = det (¸I3 ¡ A) =

¯¯¯¯¯¯

¸ ¡ 1=2 ¡1=2 0

1=4 ¸ ¡ 1=4 ¡1=2

¡1=4 ¡1=4 ¸ ¡ 1=2

¯¯¯¯¯¯

¸ ¡

¶¯¯¯¯

¸ ¡ 1=4 ¡1=2

¡1=4 ¸ ¡ 1=2

¯¯¯¯

1=4 ¡1=2

¡1=4 ¸ ¡ 1=2

¯¯¯¯

¸ ¡

¶½µ

¸ ¡

¶µ

¸ ¡

¡1

¸ ¡

¶µ

¸2 ¡

3¸

= ¸

½µ

¸ ¡

¶µ

¸ ¡

= ¸

¸2 ¡

5¸

= ¸(¸ ¡ 1)

¸ ¡

(b) Hence the characteristic polynomial has no repeated roots and we can

use Theorem 6.2.2 to ¯nd a non{singular matrix P such that

P¡1AP = diag(1; 0;

We take P = [X1jX2jX3], where X1; X2; X3 are eigenvectors corresponding

to the respective eigenvalues 1; 0; 1

4 .

Finding X1: We have to solve (A ¡ I3)X = 0. we have

A ¡ I3 =

4 ¡1=2 1=2 0

1=4 ¡3=4 1=2

1=4 1=4 ¡1=2

5 !

1 0 ¡1

0 1 ¡1

0 0 0

5 :

Hence the eigenspace consists of vectors X = [x; y; z]t satisfying x = z and

y = z, with z arbitrary. Hence

X =

5 = z

and we can take X1 = [1; 1; 1]t.

Finding X2: We solve AX = 0. We have

A =

1=2 1=2 0

1=4 1=4 1=2

5 !

1 1 0

0 0 1

0 0 0

5 :

Hence the eigenspace consists of vectors X = [x; y; z]t satisfying x = ¡y

and z = 0, with y arbitrary. Hence

X =

4 ¡y

5 = y

4 ¡1

and we can take X2 = [¡1; 1; 0]t.

Finding X3: We solve (A ¡ 1

4 I3)X = 0. We have

A ¡

I3 =

1=4 1=2 0

1=4 0 1=2

1=4 1=4 1=4

5 !

1 0 2

0 1 ¡1

0 0 0

5 :

Hence the eigenspace consists of vectors X = [x; y; z]t satisfying x = ¡2z

and y = z, with z arbitrary. Hence

X =

4 ¡2z

5 = z

4 ¡2

and we can take X3 = [¡2; 1; 1]t.

Hence we can take P =

1 ¡1 ¡2

1 1 1

1 0 1

4 )P¡1 so An = Pdiag(1; 0; 1

4n )P¡1.

Hence

An =

1 ¡1 ¡2

1 1 1

1 0 1

1 0 0

0 0 0

0 0 1

5 1

1 1 1

0 3 ¡3

¡1 ¡1 2

1 0 ¡ 2

1 0 1

1 1 1

0 3 ¡3

¡1 ¡1 2

1 + 2

4n 1 + 2

4n 1 ¡ 4

1 ¡ 1

4n 1 ¡ 1

4n 1 + 2

1 ¡ 1

4n 1 ¡ 1

4n 1 + 2

1 1 1

5 +

3 ¢ 4n

2 2 ¡4

¡1 ¡1 2

5 :

10. Let

A =

5 2 ¡2

2 5 ¡2

¡2 ¡2 5

5 :

(a) We ¯rst determine the characteristic polynomial chA(¸).

chA(¸) =

¯¯¯¯¯¯

¸ ¡ 5 ¡2 2

¡2 ¸ ¡ 5 2

2 2 ¸ ¡ 5

5 R3 ! R3 + R2

¯¯¯¯¯¯

¸ ¡ 5 ¡2 2

¡2 ¸ ¡ 5 2

0 ¸ ¡ 3 ¸ ¡ 3

¯¯¯¯¯¯

= (¸ ¡ 3)

¯¯¯¯¯¯

¸ ¡ 5 ¡2 2

¡2 ¸ ¡ 5 2

0 1 1

¯¯¯¯¯¯

C3 ! C3 ¡ C2 = (¸ ¡ 3)

¯¯¯¯¯¯

¸ ¡ 5 ¡2 4

¡2 ¸ ¡ 5 ¡¸ + 7

0 1 0

¯¯¯¯¯¯

= ¡(¸ ¡ 3)

¯¯¯¯

¸ ¡ 5 4

¡2 ¡¸ + 7

¯¯¯¯

= ¡(¸ ¡ 3) f(¸ ¡ 5)(¡¸ + 7) + 8g

= ¡(¸ ¡ 3)(¡¸2 + 5¸ + 7¸ ¡ 35 + 8)

= ¡(¸ ¡ 3)(¡¸2 + 12¸ ¡ 27)

= ¡(¸ ¡ 3)(¡1)(¸ ¡ 3)(¸ ¡ 9)

= (¸ ¡ 3)2(¸ ¡ 9):

We have to ¯nd bases for each of the eigenspaces N(A¡9I3) and N(A¡3I3).

First we solve (A ¡ 3I3)X = 0. We have

A ¡ 3I3 =

2 2 ¡2

¡2 ¡2 2

5 !

1 1 ¡1

0 0 0

5 :

Hence the eigenspace consists of vectors X = [x; y; z]t satisfying x = ¡y+z,

with y and z arbitrary. Hence

X =

4 ¡y + z

5 = y

4 ¡1

5 + z

5 ;

so X1 = [¡1; 1; 0]t and X2 = [1; 0; 1]t form a basis for the eigenspace

corresponding to the eigenvalue 3.

Next we solve (A ¡ 9I3)X = 0. We have

A ¡ 9I3 =

4 ¡4 2 ¡2

2 ¡4 ¡2

¡2 ¡2 ¡4

5 !

1 0 1

0 1 1

0 0 0

5 :

Hence the eigenspace consists of vectors X = [x; y; z]t satisfying x = ¡z

and y = ¡z, with z arbitrary. Hence

X =

4 ¡z

¡z

5 = z

4 ¡1

¡1

and we can take X3 = [¡1; ¡1; 1]t as a basis for the eigenspace correspond-

ing to the eigenvalue 9.

Then Theorem 6.2.3 assures us that P = [X1jX2jX3] is non{singular and

P¡1AP =

3 0 0

0 3 0

0 0 9

5 :

-9 -4.5 4.5 9 13.5

4.5

13.5

-4.5

-9

-8 -4 4 8

-4

-8

Figure 1: (a): x2 ¡ 8x + 8y + 8 = 0; (b): y2 ¡ 12x + 2y + 25 = 0

Section 7.3

1. (i) x2¡8x+8y+8 = (x¡4)2+8(y¡1). So the equation x2¡8x+8y+8 = 0

becomes

1 + 8y1 = 0 (1)

if we make a translation of axes x ¡ 4 = x1; y ¡ 1 = y1.

However equation (1) can be written as a standard form

y1 = ¡

1 ;

which represents a parabola with vertex at (4; 1). (See Figure 1(a).)

(ii) y2 ¡12x+2y +25 = (y +1)2 ¡12(x¡2). Hence y2 ¡12x+2y +25 = 0

becomes

1 ¡ 12x1 = 0 (2)

if we make a translation of axes x ¡ 2 = x1; y + 1 = y1.

However equation (2) can be written as a standard form

1 = 12x1;

which represents a parabola with vertex at (2; ¡1). (See Figure 1(b).)

2. 4xy ¡ 3y2 = XtAX, where A =

0 2

2 ¡3

and X =

. The

eigenvalues of A are the roots of ¸2 + 3¸ ¡ 4 = 0, namely ¸1 = ¡4 and

¸2 = 1.

The eigenvectors corresponding to an eigenvalue ¸ are the non{zero vec-

tors [x; y]t satisfying

0 ¡ ¸ 2

2 ¡3 ¡ ¸

¸ ·

¸1 = ¡4 gives equations

4x + 2y = 0

2x + y = 0

which has the solution y = ¡2x. Hence

¡2x

= x

¡2

A corresponding unit eigenvector is [1=p5; ¡2=p5]t.

¸2 = 1 gives equations

¡x + 2y = 0

2x ¡ 4y = 0

which has the solution x = 2y. Hence

= y

A corresponding unit eigenvector is [2=p5; 1=p5]t.

Hence if

P =

1 p5

2 p5

¡2 p5

1 p5

;

then P is an orthogonal matrix. Also as det P = 1, P is a proper orthogonal

matrix and the equation

= P

represents a rotation to new x1; y1 axes whose positive directions are given

by the respective columns of P. Also

PtAP =

¡4 0

0 1

Then XtAX = ¡4x2

1 +y2

1 and the original equation 4xy ¡3y2 = 8 becomes

¡4x2

1 + y2

1 = 8, or the standard form

¡x2

= 1;

which represents an hyperbola.

The asymptotes assist in drawing the curve. They are given by the

equations

¡x2

= 0; or y1 = §2x1:

Now ·

= Pt

1 p5 ¡2 p5

2 p5

1 p5

#·

;

x1 =

x ¡ 2y

; y1 =

2x + y

Hence the asymptotes are

2x + y

= §2

x ¡ 2y

;

which reduces to y = 0 and y = 4x=3. (See Figure 2(a).)

3. 8x2 ¡ 4xy + 5y2 = XtAX, where A =

8 ¡2

¡2 5

and X =

. The

eigenvalues of A are the roots of ¸2 ¡ 13¸ + 36 = 0, namely ¸1 = 4 and

¸2 = 9. Corresponding unit eigenvectors turn out to be [1=p5; 2=p5]t and

[¡2=p5; 1=p5]t. Hence if

P =

1 p5 ¡2 p5

2 p5

1 p5

;

then P is an orthogonal matrix. Also as det P = 1, P is a proper orthogonal

matrix and the equation

= P

represents a rotation to new x1; y1 axes whose positive directions are given

by the respective columns of P. Also

PtAP =

4 0

0 9

-16 -8 8 16

-8

-16

y x

-2.85 -1.9 -0.95 0.95 1.9 2.85

0.95

1.9

2.85

-0.95

-1.9

-2.85

Figure 2: (a): 4xy ¡ 3y2 = 8; (b): 8x2 ¡ 4xy + 5y2 = 36

Then XtAX = 4x2

1 + 9y2

1 and the original equation 8x2 ¡ 4xy + 5y2 = 36

becomes 4x2

1 + 9y2

1 = 36, or the standard form

= 1;

which represents an ellipse as in Figure 2(b).

The axes of symmetry turn out to be y = 2x and x = ¡2y.

4. We give the sketch only for parts (i), (iii) and (iv). We give the working

for (ii) only. See Figures 3(a) and 4(a) and 4(b), respectively.

(ii) We have to investigate the equation

5x2 ¡ 4xy + 8y2 + 4p5x ¡ 16p5y + 4 = 0: (3)

Here 5x2 ¡ 4xy + 8y2 = XtAX, where A =

5 ¡2

¡2 8

and X =

The eigenvalues of A are the roots of ¸2 ¡13¸+36 = 0, namely ¸1 = 9 and

¸2 = 4. Corresponding unit eigenvectors turn out to be [1=p5; ¡2=p5]t and

[2=p5; 1=p5]t. Hence if

P =

1 p5

2 p5

¡2 p5

1 p5

;

then P is an orthogonal matrix. Also as det P = 1, P is a proper orthogonal

matrix and the equation

= P

-6 -3 3 6

-3

-6

y x

-4.5 -3 -1.5 1.5 3 4.5

1.5

4.5

-1.5

-3

-4.5

Figure 3: (a): 4x2 ¡ 9y2 ¡ 24x ¡ 36y ¡ 36 = 0; (b): 5x2 ¡ 4xy + 8y2 + p5x ¡ 16p5y + 4 = 0

-9 -4.5 4.5 9

4.5

-4.5

-9

-9 -4.5 4.5 9

4.5

-4.5

-9

Figure 4: (a): 4x2 +y2 ¡4xy ¡10y ¡19 = 0; (b): 77x2 +78xy ¡27y2 +

70x ¡ 30y + 29 = 0

represents a rotation to new x1; y1 axes whose positive directions are given

by the respective columns of P. Also

PtAP =

9 0

0 4

Moreover

5x2 ¡ 4xy + 8y2 = 9x2

1 + 4y2

To get the coe±cients of x1 and y1 in the transformed form of equation (3),

we have to use the rotation equations

x =

(x1 + 2y1); y =

(¡2x1 + y1):

Then equation (3) transforms to

9x2

1 + 4y2

1 + 36x1 ¡ 8y1 + 4 = 0;

or, on completing the square,

9(x1 + 2)2 + 4(y1 ¡ 1)2 = 36;

or in standard form

= 1;

where x2 = x1 + 2 and y2 = y1 ¡ 1. Thus we have an ellipse, centre

(x2; y2) = (0; 0), or (x1; y1) = (¡2; 1), or (x; y) = (0; p5).

The axes of symmetry are given by x2 = 0 and y2 = 0, or x1 + 2 = 0

and y1 ¡ 1 = 0, or

(x ¡ 2y) + 2 = 0 and

(2x + y) ¡ 1 = 0;

which reduce to x ¡ 2y + 2p5 = 0 and 2x + y ¡ p5 = 0. See Figure 3(b).

5. (i) Consider the equation

2x2 + y2 + 3xy ¡ 5x ¡ 4y + 3 = 0: (4)

¢ =

¯¯¯¯¯¯

2 3=2 ¡5=2

3=2 1 ¡2

¡5=2 ¡2 3

¯¯¯¯¯¯

= 8

¯¯¯¯¯¯

4 3 ¡5

3 2 ¡4

¡5 ¡4 6

¯¯¯¯¯¯

= 8

¯¯¯¯¯¯

1 1 ¡1

3 2 ¡4

¡2 ¡2 2

¯¯¯¯¯¯

= 0:

Let x = x1 + ®; y = y1 + ¯ and substitute in equation (4) to get

2(x1+®)2+(y1+¯)2+3(x1+®)(y1+¯)¡5(x1+®)¡4(y1+¯)+3 = 0 (5):

Then equating the coe±cients of x1 and y1 to 0 gives

4® + 3¯ ¡ 5 = 0

3® + 2¯ ¡ 4 = 0;

which has the unique solution ® = 2; ¯ = ¡1. Then equation (5) simpli¯es

2x2

1 + y2

1 + 3x1y1 = 0 = (2x1 + y1)(x1 + y1):

So relative to the x1; y1 coordinates, equation (4) describes two lines: 2x1+

y1 = 0 and x1 +y1 = 0. In terms of the original x; y coordinates, these lines

become 2(x ¡ 2) + (y +1) = 0 and (x ¡ 2) + (y + 1) = 0, i.e. 2x + y ¡ 3 = 0

and x + y ¡ 1 = 0, which intersect in the point

(x; y) = (®; ¯) = (2; ¡1):

(ii) Consider the equation

9x2 + y2 ¡ 6xy + 6x ¡ 2y + 1 = 0: (6)

Here

¢ =

¯¯¯¯¯¯

9 ¡3 3

3 1 ¡1

3 ¡1 1

¯¯¯¯¯¯

= 0;

as column 3 = ¡column 2.

Let x = x1 + ®; y = y1 + ¯ and substitute in equation (6) to get

9(x1 + ®)2 + (y1 + ¯)2 ¡ 6(x1 + ®)(y1 + ¯) + 6(x1 + ®) ¡ 2(y1 + ¯) + 1 = 0:

Then equating the coe±cients of x1 and y1 to 0 gives

18® ¡ 6¯ + 6 = 0

¡6® + 2¯ ¡ 2 = 0;

or equivalently ¡3®+¯ ¡1 = 0. Take ® = 0 and ¯ = 1. Then equation (6)

simpli¯es to

9x2

1 + y2

1 ¡ 6x1y1 = 0 = (3x1 ¡ y1)2: (7)

In terms of x; y coordinates, equation (7) becomes

(3x ¡ (y ¡ 1))2 = 0; or 3x ¡ y + 1 = 0:

(iii) Consider the equation

x2 + 4xy + 4y2 ¡ x ¡ 2y ¡ 2 = 0: (8)

Arguing as in the previous examples, we ¯nd that any translation

x = x1 + ®; y = y1 + ¯

where 2® + 4¯ ¡ 1 = 0 has the property that the coe±cients of x1 and y1

will be zero in the transformed version of equation (8). Take ¯ = 0 and

® = 1=2. Then (8) reduces to

1 + 4x1y1 + 4y2

1 ¡

= 0;

or (x1 +2y1)2 = 3=2. Hence x1 +2y1 = §3=2, with corresponding equations

x + 2y = 2 and x + 2y = ¡1:

Section 8.8

1. The given line has equations

x = 3 + t(13 ¡ 3) = 3 + 10t;

y = ¡2 + t(3 + 2) = ¡2 + 5t;

z = 7 + t(¡8 ¡ 7) = 7 ¡ 15t:

The line meets the plane y = 0 in the point (x; 0; z), where 0 = ¡2+5t, or

t = 2=5. The corresponding values for x and z are 7 and 1, respectively.

2. E = 1

2 (B + C), F = (1 ¡ t)A + tE, where

t =

AF + FE

AF=FE

(AF=FE) + 1

Hence

F =

A +

(B + C)

A +

(B + C)

(A + B + C):

3. Let A = (2; 1; 4); B = (1; ¡1; 2); C = (3; 3; 6). Then we prove

AC=

AB for some real t. We have

AC=

5 ;

AB=

4 ¡1

¡2

5 :

Hence

AC= (¡1)

AB and consequently C is on the line AB. In fact A is

between C and B, with AC = AB.

4. The points P on the line AB which satisfy AP = 2

5PB are given by

P = A + t

AB, where jt=(1 ¡ t)j = 2=5. Hence t=(1 ¡ t) = §2=5.

The equation t=(1 ¡ t) = 2=5 gives t = 2=7 and hence

P =

¡1

5 +

5 =

16=7

29=7

3=7

5 :

Hence P = (16=7; 29=7; 3=7).

The equation t=(1 ¡ t) = ¡2=5 gives t = ¡2=3 and hence

P =

¡1

5 ¡

5 =

4=3

1=3

¡13=3

5 :

Hence P = (4=3; 1=3; ¡13=3).

5. An equation for M is P = A + t

BC, which reduces to

x = 1 + 6t

y = 2 ¡ 3t

z = 3 + 7t:

An equation for N is Q = E + s

EF, which reduces to

x = 1 + 9s

y = ¡1

z = 8 + 3s:

To ¯nd if and whereMand N intersect, we set P = Q and attempt to solve

for s and t. We ¯nd the unique solution t = 1; s = 2=3, proving that the

lines meet in the point

(x; y; z) = (1 + 6; 2 ¡ 3; 3 + 7) = (7; ¡1; 10):

6. Let A = (3; 5; 6); B = (¡2; 7; 9); C = (2; 1; 7). Then

(i)

cos\ABC = (

BA ¢

BC)=(BA ¢ BC);

where

BA= [¡1; ¡2; ¡3]t and

BC= [4; ¡6; ¡2]t. Hence

cos\ABC = ¡4 + 12 + 6

p14p56

Hence \ABC = ¼=3 radians or 60±.

(ii)

cos\BAC = (

AB ¢

AC)=(AB ¢ AC);

where

AB= [1; 2; 3]t and

AC= [5; ¡4; 1]t. Hence

cos\BAC =

5 ¡ 8 + 3

p14p42

= 0:

Hence \ABC = ¼=2 radians or 90±.

(iii)

cos\ACB = (

CA ¢

CB)=(CA ¢ CB);

where

CA= [¡5; 4; ¡1]t and

CB= [¡4; 6; 2]t. Hence

cos\ACB =

20 + 24 ¡ 2

p42p56

p42

p56

Hence \ACB = ¼=6 radians or 30±.

7. By Theorem 8.5.2, the closest point P on the line AB to the origin O is

given by P = A + t

AB, where

t =

AO ¢

AB2 = ¡A¢

AB2 :

Now

A¢

AB=

4 ¡2

5 ¢

5 = ¡2:

Hence t = 2=11 and

P =

4 ¡2

5 +

5 =

4 ¡16=11

13=11

35=11

and P = (¡16=11; 13=11; 35=11).

Consequently the shortest distance OP is given by

sµ

¡16

¶2

p1650

p15 £ 11 £ 10

p150

p11

Alternatively, we can calculate the distance OP 2, where P is an arbitrary

point on the line AB and then minimize OP2:

P = A + t

AB=

4 ¡2

5 + t

5 =

4 ¡2 + 3t

1 + t

3 + t

5 :

Hence

OP2 = (¡2 + 3t)2 + (1 + t)2 + (3 + t)2

= 11t2 ¡ 4t + 14

= 11

t2 ¡

t +

= 11

Ã½

t ¡

¾2

11 ¡

121

= 11

Ã½

t ¡

¾2

150

121

Consequently

OP2 ¸ 11 £

150

121

for all t; moreover

OP2 = 11 £

150

121

when t = 2=11.

8. We ¯rst ¯nd parametric equations for N by solving the equations

x + y ¡ 2z = 1

x + 3y ¡ z = 4:

The augmented matrix is

1 1 ¡2 1

1 3 ¡1 4

;

which reduces to ·

1 0 ¡5=2 ¡1=2

0 1 1=2 3=2

Hence x = ¡1

2 + 5

2z; y = 3

2 ¡ z

2 , with z arbitrary. Taking z = 0 gives a point

A = (¡1

2 ; 3

2 ; 0), while z = 1 gives a point B = (2; 1; 1).

Hence if C = (1; 0; 1), then the closest point on N to C is given by

P = A + t

AB, where t = (

AC ¢

AB)=AB2.

Now

AC=

3=2

¡3=2

5 and

AB=

5=2

¡1=2

5 ;

t =

2 £ 5

2 + ¡3

2 £ ¡1

2 + 1 £ 1

¡5

¢2

¡1

¢2

+ 12

Hence

P =

4 ¡1=2

3=2

5 +

5=2

¡1=2

5 =

4=3

17=15

11=15

5 ;

so P = (4=3; 17=15; 11=15).

Also the shortest distance PC is given by

PC =

sµ

1 ¡

¶2

0 ¡

¶2

1 ¡

¶2

p330

9. The intersection of the planes x + y ¡ 2z = 4 and 3x ¡ 2y + z = 1 is the

line given by the equations

x =

z; y =

where z is arbitrary. Hence the line L has a direction vector [3=5; 7=5; 1]t

or the simpler [3; 7; 5]t. Then any plane of the form 3x + 7y + 5z = d will

be perpendicualr to L. The required plane has to pass through the point

(6; 0; 2), so this determines d:

3 £ 6 + 7 £ 0 + 5 £ 2 = d = 28:

10. The length of the projection of the segment AB onto the line CD is

given by the formula

CD ¢

AB j

Here

CD= [¡8; 4; ¡1]t and

AB= [4; ¡4; 3]t, so

CD ¢

AB j

= j(¡8)p£ 4 + 4 £ (¡4) + (¡1) £ 3j

(¡8)2 + 42 + (¡1)2

= jp¡ 51j 81

11. A direction vector for L is given by

BC= [¡5; ¡2; 3]t. Hence the plane

through A perpendicular to L is given by

¡5x ¡ 2y + 3z = (¡5) £ 3 + (¡2) £ (¡1) + 3 £ 2 = ¡7:

The position vector P of an arbitrary point P on L is given by P = B+t

BC,

or 2

5 =

5 + t

4 ¡5

¡2

5 ;

or equivalently x = 2 ¡ 5t; y = 1 ¡ 2t; z = 4 + 3t.

To ¯nd the intersection of line L and the given plane, we substitute the

expressions for x; y; z found in terms of t into the plane equation and solve

the resulting linear equation for t:

¡5(2 ¡ 5t) ¡ 2(1 ¡ 2t) + 3(4 + 3t) = ¡7;

which gives t = ¡7=38. Hence P =

¡111

38 ; 52

38 ; 131

and

AP =

sµ

3 ¡

111

¶2

¡1 ¡

¶2

2 ¡

131

¶2

p11134

p293 £ 38

p293

p38

12. Let P be a point inside the triangle ABC. Then the line through P and

parallel to AC will meet the segments AB and BC in D and E, respectively.

Then

P = (1 ¡ r)D + rE; 0 < r < 1;

D = (1 ¡ s)B + sA; 0 < s < 1;

E = (1 ¡ t)B + tC; 0 < t < 1:

Hence

P = (1 ¡ r) f(1 ¡ s)B + sAg + r f(1 ¡ t)B + tCg

= (1 ¡ r)sA + f(1 ¡ r)(1 ¡ s) + r(1 ¡ t)gB + rtC

= ®A + ¯B + °C;

where

® = (1 ¡ r)s; ¯ = (1 ¡ r)(1 ¡ s) + r(1 ¡ t); ° = rt:

Then 0 < ® < 1; 0 < ° < 1; 0 < ¯ < (1 ¡ r) + r = 1. Also

® + ¯ + ° = (1 ¡ r)s + (1 ¡ r)(1 ¡ s) + r(1 ¡ t) + rt = 1:

13. The line AB is given by P = A + t[3; 4; 5]t, or

x = 6 + 3t; y = ¡1 + 4t; z = 11 + 5t:

Then B is found by substituting these expressions in the plane equation

3x + 4y + 5z = 10:

We ¯nd t = ¡59=50 and consequently

B =

6 ¡

177

; ¡1 ¡

236

; 11 ¡

295

123

; ¡286

;

255

Then

AB = jj

AB jj = jjt

5 jj

= jtj

32 + 42 + 52 =

50 £

p50 =

p50

14. Let A = (¡3; 0; 2); B = (6; 1; 4); C = (¡5; 1; 0). Then the area of

triangle ABC is 1

2 jj

AB £

AC jj. Now

AB £

AC=

5 £

4 ¡2

¡2

5 =

4 ¡4

5 :

Hence jj

AB £

AC jj = p333:

15. Let A1 = (2; 1; 4); A2 = (1; ¡1; 2); A3 = (4; ¡1; 1). Then the point

P = (x; y; z) lies on the plane A1A2A3 if and only if

A1P ¢(

A1A2 £

A1A3) = 0;

or ¯¯¯¯¯¯

x ¡ 2 y ¡ 1 z ¡ 4

¡1 ¡2 ¡2

2 ¡2 ¡3

¯¯¯¯¯¯

= 2x ¡ 7y + 6z ¡ 21 = 0:

16. Non{parallel lines L and M in three dimensional space are given by

equations

P = A + sX; Q = B + tY:

(i) Suppose

PQ is orthogonal to both X and Y . Now

PQ= Q ¡ P = (B + tY ) ¡ (A + sX) =

AB +tY ¡ sX:

Hence

(

AB +tY + sX) ¢ X = 0

(

AB +tY + sX) ¢ Y = 0:

More explicitly

t(Y ¢ X) ¡ s(X ¢ X) = ¡

AB ¢X

t(Y ¢ Y ) ¡ s(X ¢ Y ) = ¡

AB ¢Y:

However the coe±cient determinant of this system of linear equations

in t and s is equal to

¯¯¯¯

Y ¢ X ¡X ¢ X

Y ¢ Y ¡X ¢ Y

¯¯¯¯

= ¡(X ¢ Y )2 + (X ¢ X)(Y ¢ Y )

= jjX £ Y jj2 6= 0;

as X 6= 0; Y 6= 0 and X and Y are not proportional (L and M are

not parallel).

(ii) P and Q can be viewed as the projections of C and D onto the line PQ,

where C and D are arbitrary points on the lines L andM, respectively.

Hence by equation (8.14) of Theorem 8.5.3, we have

PQ · CD:

Finally we derive a useful formula for PQ. Again by Theorem 8.5.3

PQ = j

AB ¢

PQ j

= j

AB ¢^nj;

¢®

¡ """"""""""""""""

"" "" ""

PPPPPPPPP

@¡

where ^n = 1

P Q

PQ is a unit vector which is orthogonal to X and Y .

Hence

^n = t(X £ Y );

where t = §1=jjX £ Y jj. Hence

PQ = j

AB ¢(X £ Y )j

jjX £ Y jj

17. We use the formula of the previous question.

Line L has the equation P = A + sX, where

X =

AC=

¡3

5 :

Line M has the equation Q = B + tY , where

Y =

BD=

5 :

Hence X £ Y = [¡6; 1; 5]t and jjX £ Y jj = p62.

Hence the shortest distance between lines AC and BD is equal to

AB ¢(X £ Y )j

jjX £ Y jj

¯¯¯¯¯¯

¡2

5 ¢

4 ¡6

¯¯¯¯¯¯

p62

18. Let E be the foot of the perpendicular from A4 to the plane A1A2A3.

Then

volA1A2A3A4 =

( area¢A1A2A3) ¢ A4E:

Now

area¢A1A2A3 =

2jj

A1A2 £

A1A3 jj:

Also A4E is the length of the projection of A1A4 onto the line A4E. See

¯gure below.)

Hence A4E = j

A1A4 ¢Xj, where X is a unit direction vector for the line

A4E. We can take

X =

A1A2 £

A1A3

A1A2 £

A1A3 jj

Hence

volA1A2A3A4 =

6jj

A1A2 £

A1A3 jjj

A1A4 ¢(

A1A2 £

A1A3)j

A1A2 £

A1A3 jj

A1A4 ¢(

A1A2 £

A1A3)j

6j(

A1A2 £

A1A3)¢

A1A4 j:

19. We have

CB= [1; 4; ¡1]t;

CD= [¡3; 3; 0]t;

AD= [3; 0; 3]t. Hence

CB £

CD= 3i + 3j + 15k;

so the vector i + j + 5k is perpendicular to the plane BCD.

Now the plane BCD has equation x+y +5z = 9, as B = (2; 2; 1) is on

the plane.

Also the line through A normal to plane BCD has equation

5 =

5 + t

5 = (1 + t)

5 :

Hence x = 1 + t; y = 1 + t; z = 5(1 + t).

[We remark that this line meets plane BCD in a point E which is given

by a value of t found by solving

(1 + t) + (1 + t) + 5(5 + 5t) = 9:

So t = ¡2=3 and E = (1=3; 1=3; 5=3).]

The distance from A to plane BCD is

j1 £ 1 + 1 £ 1 + 5 £ 5 ¡ 9j

12 + 12 + 52 =

p27

= 2p3:

To ¯nd the distance between lines AD and BC, we ¯rst note that

(a) The equation of AD is

P =

5 + t

5 =

1 + 3t

5 + 3t

5 ;

(b) The equation of BC is

Q =

5 + s

¡1

5 =

2 + s

2 + 4s

1 ¡ s

5 :

©©©©©©©©©© @

Then

PQ= [1 + s ¡ 3t; 1 + 4s; ¡4 ¡ s ¡ 3t]t and we ¯nd s and t by solving

the equations

PQ ¢

AD= 0 and

PQ ¢

BC= 0, or

(1 + s ¡ 3t)3 + (1 + 4s)0 + (¡4 ¡ s ¡ 3t)3 = 0

(1 + s ¡ 3t) + 4(1 + 4s) ¡ (¡4 ¡ s ¡ 3t) = 0:

Hence t = ¡1=2 = s.

Correspondingly, P = (¡1=2; 1; 7=2) and Q = (3=2; 0; 3=2).

Thus we have found the closest points P and Q on the respective lines

AD and BC. Finally the shortest distance between the lines is

PQ = jj

PQ jj = 3:

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