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

Matrix Binomials IA

Extracts from this document...


Tenzin Zomkey                Maths SL Type 1

Maths Portfolio Standard Level

International Baccalaureate

Matrix Binomials

The main aim of this portfolio is to investigate the matrix binomials and observe and determine a general expression from the patterns that we obtain through the workings. Throughout the project, I shall be using solely matrices of 2 x 2 formations, and investigate the patterns I find.

1. To begin with, we consider the matrices X = image00.pngand Y =image01.png.

The values of these matrices, each raised to the power of 2, 3 and 4 are calculated, as shown below;

X2= image00.pngX image00.png= image43.pngY2 = image01.png x image01.png= image12.png

X3= image43.pngx image00.png= image02.png       and     Y3 = image12.pngx image01.png = image27.png

X4 = image02.pngx image00.png= image30.pngY4 = image27.pngx image31.png = image32.png

It can be observed that all the matrices calculated above are in the form of 2 X 2, they are all square matrices. The corresponding diagonal elements are also observed to be the same. Since the matrices of each nth power can be seen to be the value of 1 less than the nth term, the general expression for the matrix Xn in terms of n is -

Xn = image33.png

And the general expression for Yn is –

Yn = image34.png

...read more.


image48.pngB3= image49.png= image50.png

A4= image51.png= image52.pngB4= image53.png= image54.png

Note: A GCD calculator (TI 83) has been used throughout this portfolio to calculate the matrices and other calculations.

Observing the above calculations, we can detect a certain pattern in determining the values, which gives us the general expression of A and B in terms of n as;

An = image55.png               and                      Bn= image56.png


Taking n to be 4, we substitute the values in the expression of An

A4= image57.png

A4 = image58.png

A4 = image59.png

A4 = image52.png

And since A4 = image52.pngis the correct value as calculated previously, this expression is proven true and consistent.

Likewise, to find the general expression of (A + B) n , the values of (A + B) raised to the powers 2,3 and 4 are calculated;

First, we find the value of (A + B)

(A + B)= image43.png+ image44.png= image60.png

(A + B) 2 = image03.png= image04.png

(A + B) 3 = image05.png= image06.png

(A + B) 4 = image07.png= image08.png

Observing the repeating patterns in the calculations above, we can deduce the general expression of (A+B) in terms of n to be;

(A + B) n = 2 n-1 image09.png


Taking n as 3, and substituting it in the above expression –

(A + B) 3 = 2 3-1 image28.png

(A + B) 3 = 2 2image61.png

(A + B) 3 = 4 image11.png

(A + B) 3 = image06.png

...read more.


n = 2 n-1 image09.pngwhereby a=2, b= -2, and taking n=3, we calculate (A+B) raised to the third power –

(A+B)3 = 23-1 image10.png

                 = 22 image11.png

                  = 4 image11.png

                    = image06.png

So since we know from previous calculations that M3= image06.png, we can say that , M3= (A+B)3 .

Therefore, the general statement of Mnin terms of aX and bY is;

Mn= (aX+bY)n


To check the validity of this general statement, we shall take different values for a, b and n. Suppose a= 3, b=4, and n= 2 –

M2= (3X+4Y)2

M2= image13.png

M2= image14.png

M2= image15.png

M2= image16.png

And since M = (A+B) = (aX+bY),

M = image17.png

M = image18.png

So then,

M2= image15.png= image16.png

Therefore, the statement is proven true and consistent with all values of a, b and n.

5.  Using the Algebraic method, the general statement is to be verified and explained again.

Taking the expression, Mn =2 n-1 image09.png, we find:

M = image19.png

M2= image20.png= 2image21.png

M3=  image22.png

 = 2 image23.png

= 2image24.png

= 2image25.png



Substituting the above with the initial values of a and b, we find M3 -

M3= 4 image26.png

M3 = 4 image28.png

M3= 4 image29.png

M3= 4image29.png

M3= 4image11.png

M3= image06.png

Therefore, since it has been shown earlier in our work that the value M3= image06.pngis true and correct, it shows that the general statement of Mnin terms of aX and bY is true.

...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 Studies I.A

    870250000 5856.8409 Somalia 600 48.84 29304 360000 2385.3456 Spain 34,600 79.78 2760388 1197160000 6364.8484 Sudan 2,200 49.11 108042 4840000 2411.7921 Swaziland 5,100 32.23 164373 26010000 1038.7729 Switzerland 40,900 80.62 3297358 1672810000 6499.5844 Taiwan 31,900 77.5 2472250 1017610000 6006.25 Tanzania 1,300 50.71 65923 1690000 2571.5041 Timor-leste 2,400 63.5 152400 5760000 4032.25

  2. Math IA Type 1 In this task I will investigate the patterns in the ...

    In order to use the formula, where x2 and x3 are intersections of the first line with the parabola and x1 and x4 are intersections of the second line with the parabola. Thus I will now find the intersections of the second line with the parabola.

  1. Math IA - Logan's Logo

    However, when we graph, it is clear that this line is not the center line of the curve: From here, we must then add the minimum y-value (-2.8) to obtain a number that we can graph onto the axes. 3.15+-2.8=0.35.

  2. Math Studies - IA

    In 2002, 2004 and 2006 Europe won 151/2-121/2, 181/2-91/2 and 181/2-91/2 respectively. These two sets of data can then be processed into a conclusion. To compare these two different sets of data, a similar unit or measurement will have to be found. The margin of a team's win is suitable.

  1. Math IA- Type 1 The Segments of a Polygon

    The conjecture has been proved on the following page. In to order to able to test the validity of this conjecture, another triangle was produced using the geometer's sketch pad and the conjecture is validated if the ratio produced from GSP matches the value of the ratio provided by the conjecture.

  2. MATH Lacsap's Fractions IA

    The same limitations apply to the value of r, which has to be greater than zero (r ≥ 1) and an integer (â¤) as it is not possible for a negative or incomplete element to be in the pattern. Hence, the general statement will be limited to numbers from the set of natural numbers (â).

  1. Gold Medal heights IB IA- score 15

    However the coordinate (1904, 180) is a significant fluctuation as it does not follow the general trend of a positive correlation. Also, the coordinate (1936, 203) is also a significant fluctuation from the positive correlation as it has a greatly increased height from the trend.

  2. Math SL Fish Production IA

    The more variables there are, the more accurate the function model would be at fitting the data points. Below is an example of a quartic function: The x variable represents the year and the variable f(x) represents the total mass (tonnes).

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