- Level: International Baccalaureate
- Subject: Maths
- Essay length: 1444 words
Math's Portfolio SL Type 1 "Matrix Powers"
Extracts from this essay...
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 Mn for 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)
Middle
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 (here k 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. To prove this, select a random matrix such as: V5 = , according to the pattern the determinant of this matrix should be 16 to the same power in which the matrix is powered to, in this case (5), which then 165 gives 1048576 as your answer. On the other hand, the pattern to find the numbers inside the new matrix doesn't exist here when the factor of simplification is 2n+1. This a limitation for this pattern and we cannot continue it. The pattern for matrices M, P, S and V is given by the general formula.
Conclusion
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.
Found what you're looking for?
- Start learning 29% faster today
- Over 150,000 essays available
- Just £6.99 a month