# Math's Portfolio SL Type 1 "Matrix Powers"

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Introduction

Newton College

Math’s Portfolio

SL Type 1

“Matrix Powers”

Mauro Gelmi

IB 2

March 2009

Introduction

Matrices are tables of numbers or any algebraic quantities that can be added or multiplied in a specific arrangement. A matrix is a block of numbers that consist of columns and rows used to present raw data, store information or to perform certain mathematical operations

The aim of this portfolio is to find a general trend in different sets of matrices, therefore find a general formula that applies for all of the matrices. The pattern will then be explained and tested to see if applies correctly to all the sets of matrices.

Method

To calculate Mnfor n = 2, 3, 4, 5, 10, 20, 50 I used a GDC.

Determinants are defined as ad-bc in a 2 2 matrix, and is denoted by det A = | A |. In Other words, det A = | A | = = ad-bc.

M2= =, det (M2) = 16 = 42

M3 = =, det (M2) = 64 = 43

M4 = =, det (M2) = 256 = 44

M5 = =, det (M2) = 1024 = 45

M10 =, det (M2) = 1048576 = 410

M20 =, det (M2) = 1.099511628 x 1012 = 420

M50 =, det (M2) = 1.2676506 x 1030 = 450

By squaring each number in the matrix by the power of Mn you get the answer for each of the matrices. So if M2 =, you then square the matrix, so you multiply it by itself. . Multiplying the matrix by itself as this: gives you the new matrix which is, in this case.

Middle

S3 = == 22, det (S3) = 1728 = 123 (here k is (12×2) +3)

S4 = == 23, det (S4) = 20736 = 124 (here k is (12×7)-3)

S5 = == 24, det (S5) = 248832 = 125 (here k is (12×20) +3)

Given these matrices, we notice that when simplifying the matrices we also use 2n-1 as the factor. We can use the general formula of k to find the numbers inside the new matrix and still a pattern between the matrices’ determinants can be noticed. The determinant of each matrix is 12 powered by the same number you powered the matrix.

If S3 =, this matrix’s determinant is 123, which is 1728, this confirms the trend I stated before. The trend in matrix P is the same in matrix Sbut 4 numbers more. The formula in matrix P is 8n, and 8 + 4 = 12 therefore the formula in matrix S is 12n.

I will make an example for k = 4, so the matrix will be V =.

We can predict according to the determinant’s pattern that the determinants will now be 16n.

V2 = == 2 , det (V2) = 256 = 162 (here k is 16)

V3 = == 22, det (V3) = 4096 = 163 (here k is 16×4)

V4 = == 23, det (V4) = 65536 = 164 (herek is 16×16)

V5 = == 24, det (V5) = 1048576 = 165(here k is 16×64)

These matrices follow the trend, because 12 + 4 = 16 so the determinants of these matrices must be 16n, where n is the power of the matrix.

Conclusion

For k = 5, the letter should be ‘Y’ according to the letter’s pattern, and according to the formula the matrix should be so you end up with Y = . Now if we power this matrix so as n = 2,3,4,5 we should see the same pattern as the former sets of matrices. The matrices determinants for this set should follow the trend and should be equal to 20n.

Y2 = == 24, det (Y2) = 400 = 202

Y3 = == 25, det (Y3) = 8000 = 203

Y4 = == 26, det (Y4) = 160000 = 204

Y5 = == 27, det (Y5) = 3200000= 205

This further value of k, confirms my previous statements about the generalized pattern of these sets of matrices, it also proves the formula and trend of the determinants in each set of matrices.

Conclusions

A general trend was found in the different sets of matrices provided by the assignment sheet. A formula was created, so that it could be applied to generate more matrices that follow the pattern found. The results were satisfactory due to the few limitations that the trend had. It can be considered a limitation that a matrix when powered to 0 gives you always the identity matrix, so n ≠ 0. Another limitation found is that is useless to power a matrix to 1, because it will remain unchanged. Another fact I found was that when k = negative numbers, the answers are exactly the same as the answers for the positive numbers.

This student written piece of work is one of many that can be found in our International Baccalaureate Maths section.

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