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• 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.

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