2.4 PROBLEMS

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1. Let A; B; C; D be matrices de¯ned by

A =

2

4

3 0

¡1 2

1 1

3

5 ; B =

2

4

1 5 2

¡1 1 0

¡4 1 3

3

5 ;

34 CHAPTER 2. MATRICES

C =

2

4 ¡3 ¡1

2 1

4 3

3

5 ; D =

·

4 ¡1

2 0

¸

:

Which of the following matrices are de¯ned? Compute those matrices

which are de¯ned.

A + B; A + C; AB; BA; CD; DC; D2:

[Answers: A + C; BA; CD; D2;

2

4

0 ¡1

1 3

5 4

3

5 ;

2

4

0 12

¡4 2

¡10 5

3

5,

2

4 ¡14 3

10 ¡2

22 ¡4

3

5,

·

14 ¡4

8 ¡2

¸

.]

2. Let A =

·

¡1 0 1

0 1 1

¸

. Show that if B is a 3 £ 2 such that AB = I2,

then

B =

2

4

a b

¡a ¡ 1 1 ¡ b

a + 1 b

3

5

for suitable numbers a and b. Use the associative law to show that

(BA)2B = B.

3. If A =

·

a b

c d

¸

, prove that A2 ¡ (a + d)A + (ad ¡ bc)I2 = 0.

4. If A =

·

4 ¡3

1 0

¸

, use the fact A2 = 4A ¡ 3I2 and mathematical

induction, to prove that

An =

(3n ¡ 1)

2

A +

3 ¡ 3n

2

I2 if n ¸ 1.

5. A sequence of numbers x1; x2; : : : ; xn; : : : satis¯es the recurrence rela-

tion xn+1 = axn+bxn¡1 for n ¸ 1, where a and b are constants. Prove

that ·

xn+1

xn

¸

= A

·

xn

xn¡1

¸

;

2.4. PROBLEMS 35

where A =

·

a b

1 0

¸

and hence express

·

xn+1

xn

¸

in terms of

·

x1

x0

¸

.

If a = 4 and b = ¡3, use the previous question to ¯nd a formula for

xn in terms of x1 and x0.

[Answer:

xn =

3n ¡ 1

2

x1 +

3 ¡ 3n

2

x0:]

6. Let A =

·

2a ¡a2

1 0

¸

.

(a) Prove that

An =

·

(n + 1)an ¡nan+1

nan¡1 (1 ¡ n)an

¸

if n ¸ 1.

(b) A sequence x0; x1; : : : ; xn; : : : satis¯es the recurrence relation xn+1 =

2axn ¡a2xn¡1 for n ¸ 1. Use part (a) and the previous question

to prove that xn = nan¡1x1 + (1 ¡ n)anx0 for n ¸ 1.

7. Let A =

·

a b

c d

¸

and suppose that ¸1 and ¸2 are the roots of the

quadratic polynomial x2¡(a+d)x+ad¡bc. (¸1 and ¸2 may be equal.)

Let kn be de¯ned by k0 = 0; k1 = 1 and for n ¸ 2

kn =

Xn

i=1

¸n¡i

1 ¸i¡1

2 :

Prove that

kn+1 = (¸1 + ¸2)kn ¡ ¸1¸2kn¡1;

if n ¸ 1. Also prove that

kn =

½

(¸n

1 ¡ ¸n

2 )=(¸1 ¡ ¸2) if ¸1 6= ¸2,

n¸n¡1

1 if ¸1 = ¸2.

Use mathematical induction to prove that if n ¸ 1,

An = knA ¡ ¸1¸2kn¡1I2;

[Hint: Use the equation A2 = (a + d)A ¡ (ad ¡ bc)I2.]

36 CHAPTER 2. MATRICES

8. Use Question 6 to prove that if A =

·

1 2

2 1

¸

, then

An =

3n

2

·

1 1

1 1

¸

+

(¡1)n¡1

2

·

¡1 1

1 ¡1

¸

if n ¸ 1.

9. The Fibonacci numbers are de¯ned by the equations F0 = 0; F1 = 1

and Fn+1 = Fn + Fn¡1 if n ¸ 1. Prove that

Fn =

1

p5

ÃÃ

1 + p5

2

!n

¡

Ã

1 ¡ p5

2

!n!

if n ¸ 0.

10. Let r > 1 be an integer. Let a and b be arbitrary positive integers.

Sequences xn and yn of positive integers are de¯ned in terms of a and

b by the recurrence relations

xn+1 = xn + ryn

yn+1 = xn + yn;

for n ¸ 0, where x0 = a and y0 = b.

Use Question 6 to prove that

xn

yn ! pr as n ! 1:

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