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

matrix binomilas-portfolio

Extracts from this document...












This project is about 2x2 matrices or matrix binomials; the aim of this work is to demonstrate a good and clear understanding of matrices and the operations that can be done with them.

At the end of the project we should have been able to generate different expressions for some different matrices powered by an n integer as well as a general statement to calculate the addition of specific matrices powered to an n integer.

To have a correct development of this piece of work it is essential that each question is solved in order since the result of a question will be necessary for the development of the following one.

Since there are many basic calculations that aren’t really important we will use a GDC (graphic display calculator) in order to speed up the development of the project.



Let image00.pngimage00.png andimage47.pngimage47.png. Calcúlate image11.pngimage11.png, image129.pngimage129.png; image139.pngimage139.png, image01.pngimage01.png.

With the help of a GDC (Graphic display calculator) the calculations for the matrices were done and registered below.





By considering different integer powers of X and Y find an expression forimage27.pngimage27.png, image37.pngimage37.png andimage45.pngimage45.png.

...read more.



In the case of a being 8 , image12.pngimage12.png


While in the case of a being 3 , image15.pngimage15.png


So if image06.pngimage06.pngcan be expressed like image17.pngimage17.png then image04.pngimage04.pngcan be expressed like image19.pngimage19.pngwhich can be expressed in a matrix form, like it is shown below.



To find an expression for image05.pngimage05.png we can do the same process done to generate the expression for image04.pngimage04.png , so we can star by stating that image25.pngimage25.pngisimage26.pngimage26.pngby definition, so image28.pngimage28.pngshould be equal to image29.pngimage29.png which can also be expressed asimage30.pngimage30.png), we can check this by diving each element of both of theimage28.pngimage28.png calculated previously by their respective values of image31.pngimage31.pngand the result should be equal toimage32.pngimage32.png.

In the case of b being 2 , image33.pngimage33.png


While in the case of b being 5 , image35.pngimage35.png


So if image28.pngimage28.png can be expressed like image38.pngimage38.png then image05.pngimage05.pngcan be expressed like image39.pngimage39.pngwhich can be expressed in a matrix form, like it is shown below.



The expression for image44.pngimage44.png can be found in the same way that we found the expressions for image04.pngimage04.pngand image05.pngimage05.png, we know that by definition A is aX and image25.pngimage25.pngisimage26.pngimage26.pngso image44.pngimage44.png is the same asimage46.pngimage46.png. Starting from this we can develop an expression for image44.pngimage44.png as shown below.






The expression for image48.pngimage48.png is then image54.pngimage54.png


Now consider image56.pngimage56.png

Show that image57.pngimage57.pngimage58.pngimage58.png

We know that matrix image59.pngimage59.png=image60.pngimage60.png so know we just need to add matrix A to matrix B and see if the resulting matrix is the same as image59.pngimage59.png

...read more.


We have then




Now if we use our general statement it should give use the same when we replace the values of a, b and n with the ones used to developimage110.pngimage110.png



Once again the statement worked and therefore I can consider it a general statement.


Nb b

  • The general statement developed did worked correctly and was checked two times with different values for a, b and n.
  • The relationship between the matrix image83.pngimage83.pngwith the matrices image118.pngimage118.pngand image119.pngimage119.png was found and it was expressed in matrix form in question 4.
  • We were able to generate expressions for specific matrices powered by an unknown n integer and having developed those expressions we were able to generate a general statement.


In my opinion the project was developed in a good and coherent manner, there weren’t any serious problems or difficulties during the processes and the results obtained were satisfactory, since I was able to generate the general statement and the other expressions for the different matrices powered by an n integer.

During the process I used all my knowledge about matrices which helped me to develop a good understanding of the questions in order to obtain the different answer.

However I think that the information in some parts is not presented in an organized way, this could have been improved with the use of tables to record the information in a more suitable place.

...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. Extended Essay- Math

    T�W\q��z���n*;�N�nBVd<��W]u�$)��(r)g�};�(T�hBm��h(P���Q�۴"�\�H�s���(�S�C -�"�-�>����)v�Ou�|��z,]�tn�o�"�1/4�&�(r)(r)nΣ�*�$������$-�wuuA6��S ���"\.�L����kmM�� .�U�R�-q[�ѣGk--��F���'��e�G��:-�nbÊP ���"WO�)�;���'���h�9�@$Ù4cs>5v�t�3/4�=�Y�`�}�� ;�JRf[$ vZ"�';)L��Tj������~�"8��w�"�(tm)c�I '.��*�j��BxnÙqw��ÆÆ3m�� ��u�w&���W�����-�"b PR�<��x�t_PG�g�p1/2fX\��?1/2��&��i��@� 5��o�kh�6"� ,(,����� \�Õ1/4e'H`&Z")��`4`)'�m6��jmZ4~F$P��>��c�K �]!�M��2kd� 1/4".A�á' ��0���3/4p�"5���$E�)y���6Û�5��s����5 ��T�_��d� kK���oHk6����q�� ��m1/4�����������(c)�O��Xv�:u�pL����rUk'��r�����O��T��|:w��9.�o-�G~V6[3/4�k���-;Z"�����-[�d\KO�Ry- �=�`HM#nl��Q��T"E��)�N��(tm)�s�&��-'"'2"pK$Px@:H�*�7"�'��Z"�W���� �����±ï¿½"�<W��KH`Z�...�D*���p-C��?��?��j-~ �� ��v��Ϲx�ୱ\�G��3/4ë­§3/4~�Ǵ8} T���,���w�'��� �"w6G�+��>�3�� �d3$ M"��PWB*'�0� .^;v� m�-GE�" AI�<w���2@K$�2���E$0gZ�hD��@1a�dAw-�`�� Ws~B�E�� ��EF�FÌ�">E�W�s,��&� J �����.��...�9��f'��C��'�A�z !���.G�bS&���� W�}(tm)SM$0�-�h1��&�W�����"k�+ShN��á���1/2��v"�i����V �b��":[�l�S��b�3/4}Û¶mÛ¹s�(r)]"���5��+"MÖ���|����%��hV ?�(��" �CfhhI�� ]"�#�-',uvv�0�F3/4X*��"�'���^2�&"%1-�(r)��K� oH�"0�5r��]��X >3R�4$nG:��5�Å,�r�{�c}� $5 ��ØwW�l��G`'b��'�Sγ�:H��53/4R��i�Wtx���1�n[e�...+�4"M�"���KN��2�L(r)'���1/2WJ��@t���h�eI1/43/4Õ�8F�}�?(r)L�l...4I`�'b�Æ��*A��sr��)$...1/4�" �h�bd9��>"?����'N�Y�/ �9�"��Y(c)(����hy��1/2�eN(r)(tm)c��5���-\�-`�����3/4%o@�&0 I��c��'�nÝ�P`"�{v��q� 7�Z����cÊ��M�����3/4�CL�]3/4��TTe*بF�7'M<-�D"� �"drM!*h+$�Fae�Ë?xqh��,��QL�1/4u03/4Z��c���'7�'�N��Fs�"�rrMá§)

  2. Math Portfolio: trigonometry investigation (circle trig)

    When 36 is to represent the value of x and 54 is to represent the value of y in the conjecture sinx=cosy, sin(36) would equal to sin(54) Again this should add up to 90 degrees. sinx=cosy sin(36)=cos(54) sin ?=cos(90- ?)

  1. Statistics project. Comparing and analyzing the correlation of the number of novels read per ...

    from 61-80 mark and most read a little of amount of 1 book per week. Their average mark is 63.7 and their median is 61-80. Their modal value is also 61-80. This mark with correlation to number of books read shows that number of books read has very little to do with modal mark earned.

  2. Math Portfolio - SL type 1 - matrix binomials

    It is usually represented by 'I.' When a matrix is multiplied by an identity matrix, the matrix remains unchanged. This is the property of an identity matrix. Example: This makes it easier to make an expression for (X+Y)n due to this property of identity matrix.

  1. Math IA - Matrix Binomials

    (We can also check this on a calculator). Now, using the expression, we find the value of Y5 to be the same (proving the accuracy of our expression): Y5=25-1 =24 =16 = We conclude that is indeed a valid expression. To find an expression for (X+Y)n, we must first determine the patterns by calculating the values of (X+Y), (X+Y)2, (X+Y)3, (X+Y)4.

  2. Matrix Binomials. In this Math Internal Assessment we will be dealing with matrices.

    + = 2 = Or (X + Y)1 =X+Y= n=2 (X + Y)n = (2I)n (X + Y)2 = (2I)2 X� + Y� = (2I)� + = 2� = Or (X+Y)2=(X+Y)*(X+Y)=X2+XY+YX+Y2= = = 4(I) n=3 (X + Y)n = (2I)n (X + Y)3 = (2I)3 X3+ Y3 = (2I)3 +

  1. Maths Project. Statistical Analysis of GCSE results at my secondary school summer 2010 ...

    10 46 f 46 103 Ja 9 40 f 40 102 Ja 9 46 m 46 101 Jo 9 0 m 0 100 Jo 10 0 m 0 99 Ke 11 40 m 40 98 Ki 8 40 m 40 97 Kr 9 46 m 46 96 La 8 34

  2. Math Type I SL Matrices

    a matrix one exponent higher, in this case Yn+1, therefore the following formula can be used. Yn=Yn-1 Y (original equation) (X+Y)2= (X+Y) = =2 = =2 = (X+Y)3= (X+Y)0= =3 =0 =3 =0 = = (X+Y)4= (X+Y)-1= =4 =-1 =4 = -1 = = (X+Y)0.5= 0.5 (X+Y)-0.5= -0.5 =0.5 =-0.5

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