• Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month

Math's Portfolio SL Type 1 "Matrix Powers"

Extracts from this document...

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 image04.png 2 matrix, image05.png and is denoted by det A = | A |. In Other words, det A = | A | = image08.png = ad-bc.  

M2=image18.pngimage18.png =image38.png, det (M2) = 16 = 42

M3 =image18.pngimage18.pngimage18.png =image73.png, det (M2) = 64 = 43

M4 =image06.pngimage06.pngimage06.pngimage06.png =image07.png, det (M2) = 256 = 44

M5 =image06.pngimage06.pngimage06.pngimage06.pngimage06.png =image09.png, det (M2) = 1024 = 45

M10 =image10.png, det (M2) = 1048576 = 410

M20 =image11.png, det (M2) = 1.099511628 x 1012 = 420

M50 =image12.png, 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 =image06.png, you then square the matrix, so you multiply it by itself. .        Multiplying the matrix by itself as this:image13.png gives you the new matrix which is, in this caseimage14.png.

...read more.

Middle

image31.png, det (S2) = 144 = 122 (here k is 12-3)

S3 = image32.png=image33.png= 22image34.png, det (S3) = 1728 = 123 (here k is (12×2) +3)

S4 = image35.png=image36.png= 23image37.png, det (S4) = 20736 = 124 (here k is (12×7)-3)

S5 = image39.png=image40.png= 24image41.png, 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 =image33.png, 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 =image42.png.

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

V2 = image43.png=image44.png= 2image45.png , det (V2) = 256 = 162 (here k is 16)

V3 = image46.png=image47.png= 22image48.png, det (V3) = 4096 = 163 (here k is 16×4)

V4 = image49.png=image50.png= 23image51.png, det (V4) = 65536 = 164 (herek is 16×16)

V5 = image52.png=image53.png= 24image54.png, 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.

...read more.

Conclusion

k after making a hypothesis about the results of that same value.

     For k = 5, the letter should be ‘Y’ according to the letter’s pattern, and according to the formula the matrix should be image61.png so you end up with Y = image61.png. 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 = image62.png=image63.png= 24image64.png, det (Y2) = 400 = 202

Y3 = image65.png=image66.png= 25image67.png, det (Y3) = 8000 = 203

Y4 = image68.png=image69.png= 26image70.png, det (Y4) = 160000 = 204

Y5 = image71.png=image72.png= 27image74.png, 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.

...read more.

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

See related essaysSee related essays

Related International Baccalaureate Maths essays

  1. Math IB HL math portfolio type II. Deduce the formula Sn = ...

    = (S2k - Sk) + (S4k - S3k) 2[(3q-3p)-q] = (q - p) + [S4k - (3q - 3p)] 4q - 6p = q - p + S4k - 3q + 3p S4k = 6q - 8p Therefore: S3k = 3q - 3p S4k = 6q - 8p 7.

  2. Extended Essay- Math

    �k��F�1/4��1UR"L�j�Hl�_�9'm&�f��a�(c)1,1/4h5�,l\Y��Ssd+r]�-����(tm)S��nc��V(c)����u:�v�^�3��(c)�����]�� ���9""][(r)5\|E�Ò�k�Vky-��*5 ��1�_��461/4Y�� 3/4��l�W1/4Y�� 3/4{ym��� (f4ÝA�v"��2�l� _�j-�OQ� ev�)�<U�r-����@V4Ej''! !�ÛÑ z1/4Np� �Ì� �P�L5�2 1���w/�=A��-���?XE��6NGU(c)�|�ǣ���O3/4~��s?^V�|���?}� B�L�y���_�>~��'�}���xP...�iB$�I �-O�1Êk9��I�cL"��H��Y<�;*v�7'�a(r)E\�h>��=��^,�*����8q�;��^(r)�*4����?�Wq{��n�Ôog�Aߤ>��8f�2�*<")QH��"x��K(c)�� -|��]�Z�z)�Ó"MS�m�@\&>!� 7;wP�3��[�EBU`�13/4O�C�5<V8�(c)��"U ��U�3�7a�' "="��ND�����o%�סu��3/4�&����t��W'[|��$�a{4�"��>�(F"\��;ܭ� q���pߡ� ����69&MD� O,���ooÂ��VM��� M_ո�۹�׸�U>��'��7�eo 3/4��>Ѩ��N��6}�"�-�" bv�zÛ�6��?ß���ŷ�i���Lv�m����]��2�SFnH��-��D]�r��I�SX�O]�0�� ld���C��^�3ش��d$s�#�2.�h�1/2�5-6��-5-�!�v �� ���'-.��c��h��-���N�t9W 3/41/2�dum�(tm)g"�LStf+]��C9P^��%��(c)A�-W�̯��� f$�1/4�a1Q�{B"{m�qDl��� �u""...f����9�%k@��F����?g$�P0%OVK"��Rt� �--� � ?" �&�6��...Jج=�M'N=^�gUn.��S�(tm)�j ����1/4�C�R="�q��"�b�4"�Y�� )�Y�vCK�C�j�+#�;��wB>VD��� �X�a?p����"� S�(r)�4���[�NS1/2�28;�Y�1/4[�"(tm)��,��"��T1|��n�;�+��/Êj��\��\-,E:!�� �t�4.TÌ¡ e1 "�}�; [�z^�p�l��@�ok��0e g@�GGHPXNT,�...�d��e�|�*Y(r)�dT�\(tm)Y� ä°3/4�+� (�T7�$ow2���1/4�#1/2G(c)Ö��ʥ��?1/2q�� N��K�-�/M,W���g�x�FV�/�"�FQ�ⷶ-O��&�e(r)���c�x1/4�\ Q���LW@H��!�+���{�[|{��!��K�"A�i `�c m2iU-�|�Y+� Þ¨-� [[v-x�"�r�N���E'�3�pm��R '=Y�04,�!&0�+W�CÜ@o�����OS2�'S �(r)0�5��$�ɤ]pm��3�F�t� ����G�"-!�y��"�ÓV .� (r)`�עv�,...O.%��вKas-�S��ƭ�v��M�z��"`��...�3��{�9+e� ��@1/2e���"�Ly��

  1. Math IB SL BMI Portfolio

    into adult size, and that it is rather a measurement that compares one's height with one weight; it is possible that one could keep gaining weight to a certain point after reaching adult height. This assumption would be that one reaches adult height and adult weight at different times, causing

  2. Maths SL Portfolio - Parallels and Parallelograms

    A5 ? A6, = 2 - A1 ? A2 ? A4 ? A5, - A2 ? A3 ?A5 ? A6 = 2 - A1 ? A2 ? A3 ? A4 ? A5 ? A6 = 1 --> p = 6 + 7 + 2 + 2 + 1 = 18 From the earlier conclusion, p will also equal 18 if m = 4 and n = 3.

  1. Math Portfolio higher level type 1

    6 15 20 15 6 1 1 7 21 35 35 21 7 1 1 8 28 56 70 56 28 8 1 1 9 36 84 126 126 84 36 9 1 1 10 45 120 210 252 210 120 45 10 1 1 11 55 165 330 462

  2. Math Portfolio Type II Gold Medal heights

    Then following the same steps as above, a prediction for 2016 can be made. The height of 251cm seems reasonable looking at it purely mathematically. Within a timespan of 48 years (1932 to 1980) the height increased by 39 centimetres.

  1. Artificial Intelligence & Math

    However, negative aspects can also be raised, for example, that laptops prevent students from concentrating on their school work, and degrade learning (Borja, 2000). Not all students can afford laptops to buy a laptop for school, so their introduction has also raised the issue of equality and financial discrimination (Corcoran, 2002).

  2. IB Math Methods SL: Internal Assessment on Gold Medal Heights

    In accordance with the above rationale; it would be likely that if an event were held in 1944 as well; the gold medal height would be around 198 centimeters because of athletic events being a low priority in the face of war.

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