• Join over 1.2 million students every month
• Accelerate your learning by 29%
• Unlimited access from just £6.99 per month
Page
1. 1
1
2. 2
2
3. 3
3
4. 4
4
5. 5
5
6. 6
6
7. 7
7
8. 8
8
9. 9
9
10. 10
10
11. 11
11
12. 12
12
• Level: GCSE
• Subject: Maths
• Word count: 1774

# Matrix Powers

Extracts from this document...

Introduction

Robert Fox

Math SL

12/09/2007

1. Consider  the Matrix M=

Calculate Mn for n= 2, 3, 4, 5, 10, 20, 50. Describe in words any pattern you observe. Use this pattern to find a general expression for the matrix Mn in terms of n.

1. Consider the matrices P=  and S=

P2= 2 = =; S2= 2 = =

Calculate Pn and Sn for other values of n and describe any pattern you observe.

1.  Now consider matrices of the form  steps 1 and 2 contain examples of these matrices for K=1 2 and 3. Consider other values of k, and describe any pattern(s) you observe. Generalize these results in terms of K and N
1. Use technology to investigate what happens with further values of k and n. State the scope or limitations of k and n.
1. Explain why your results holds true in general.

SL type 1: Matrix Powers

1)

1. To calculate the value for matrix ‘M’ when n=2, the matrix  must be multiplied by an exponent of 2. This would be shown and calculated as,  x

Therefore the value of matrix M2 =

1. To calculate the value for matrix ‘M’ when n=3, the matrix  must be multiplied by an exponent of 3. Therefore the value of M3 =
2. To calculate the value for matrix ‘M’ when n= 4, the matrix x  must be multiplied by an exponent of 4.

Middle

5=  = 16, 2n-1=25-1= 16

P7=7==64 =2n-1=27-1=64

P10=10==512=2n-1=210-1=512

P20=20==524288=

2n-1=220-1=524288

The formula works with all values of Matrix ‘P’ therefore the equation can be assumed will work for most values of ‘n’ if not all.

Matrix “S” was a much clearer pattern. It was evident that the expression for the matrix would have to be based on the same principle. The in the numbers A and D differ from the numbers from B and C by 2. It can also be noted that the numbers in the matrix are always larger than the ones in the P matrix. I manipulated the expression for P by plugging in numbers like 6 and 5 to obtain an expression for S but these to numbers were far too great for all intents and purposes. Therefore I made it smaller and approved 3 as the number for the expression in S.

2n-1.

Tested that the formula was correctly worked:

S3=3==4=2n-1=.23-1=4

S4= 4= =82n-1=.24-1=8

According to the question the Matrices P= and S=.

This pattern can be written as:

2n-1,

For the S values the pattern can be written as:

2n-1.

3)

K= 1                 M =

K= 2                  P =

K= 3                 S =

K= 4          =

Call this matrix D:

D2=2 ==  = 2

D3=3= =4

D4=4 = = 8

Conclusion

When N=0

By use of graphic display calculator:

-P0 = 0 = =

Applying the expression:

-P0 =2-0-1=

When N=

By use of graphic display calculator:

== 8

Applying the expression:

==

Thus:

n= (-∞; ∞) ∑ R, O, Z numbers

It was seen that the graphic display calculator gave a domain error when putting in a negative value for ‘n’ however the using the general expression the patterns of difference by 2 for numbers inside the brackets. Therefore we can verify the expression acceptable by substituting a negative, integer and real numbers. K and N therefore will continue into infinity as all of their values work for M of our general expression:

2n-1

5)    In this investigation of matrix powers, one can conclude that the calculator was one of the major limitations in finding my solutions which can be said to be technology is limiting. This was shown in my attempt to solve the expression for –P2 , which gave me a domain error. This can be explained by the calculator’s inefficiency to calculate numbers larger than a certain range. This is a perfect example of usefulness of expressions created to solve such inefficiencies. When considered in terms of k and n, we can say with fairly accurate readings that the expression was proven throughout the investigation by mean of integers, negative and real numbers. In this case I can say that my results hold strong.

| Page

This student written piece of work is one of many that can be found in our GCSE Consecutive Numbers section.

## Found what you're looking for?

• Start learning 29% faster today
• 150,000+ documents available
• Just £6.99 a month

Not the one? Search for your essay title...
• Join over 1.2 million students every month
• Accelerate your learning by 29%
• Unlimited access from just £6.99 per month

# Related GCSE Consecutive Numbers essays

1. ## GCSE Maths Coursework - Maxi Product

(4.7,4.9,4.4)= 14 --> 4.7+4.9+4.4 --> 4.7x4.9x4.4=101.332 (4.7,4.7,4.6)= 14 --> 4.7+4.7+4.6 --> 4.7x4.7x4.6=101.614 I will now move on to fractional numbers as there can be no other decimal number that can give a result higher than 101.614 using three numbers. I will see in fractional numbers if I can get a number higher than 101.332 from three fractional numbers.

2. ## Investigate the Maxi Product of numbers

15 à 6+9 à 6x9 =54 (7,8)= 15 à 7+8 à 7x8 =56 I have found that 56 is the highest number so far that can be retrieved from 7 and 8 when the number is 15, in whole numbers.

1. ## In this investigation I will explore the relationship between a series of straight, non-parallel, ...

Thus, if COP(n), = n (n-1) 2 when I looked at Diagrams 1- 6 it becomes obvious that when no lines are parallel, each new line intersected exactly once with each previous line. Thus, when the nth line is added, it makes (n-1) new intersections or cross-over points and (n-2)

2. ## I am to conduct an investigation involving a number grid.

Box 1 24 25 26 X X+1 X+2 31 32 33 X+7 X+8 X+9 38 39 40 X+14 X+15 X+16 [image008.gif] [image038.gif] 24 x 40 = 960 x (x + 16) x2 + 16x [image008.gif] 38 x 26 = 988 (x + 2)

1. ## Study the topic of trios and work on from that, to discover patterns and ...

triangular, I can adapt the formula for triangular numbers to come up with a formula for trios. That formula is: (n-1)(n-2) 2 I will now test this formula: How many trios for the number 7? (7-1)(7-2) 6 x 5 = 30 30 / 2 = 15 This proves that my formula works!

2. ## The Towers of Hanoi is an ancient mathematical game. The aim of this coursework ...

So 32 is multiplied by 0.5 to get 16. Our sequence is: S= 32 + 16 + 8 4 +2 +1 To get the sum of a geometric sequence, we need to multiply by the common ratio (0.5) S = a + ar + ar2 + ar3 +... + arn-1 rS = a + ar + ar2 + ar3 +...

1. ## Binary Explained.

2 to the power of 3 is = 8 (2x2x2 = 8) 2 to the power of 4 is = 16 (2x2x2x2 = 16) 2 to the power of 5 is = 32 (2x2x2x2x2 = 32) .......and so on. Now that the table is done we can start converting from binary numbers to decimal.

2. ## Investigate calendars, and look for any patterns.

by 7, which gives: 7 14 21 28 35 This is 4 more than required each time, so subtract 4 to correct the sequence. Therefore, the expression is: n = 7n - 4 I decided to test my expression on some other columns in the calendar.

• Over 160,000 pieces
of student written work
• Annotated by
experienced teachers
• Ideas and feedback to