• 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...

Introduction

Matrix Binomials

Type 1 Internal Assessment

Matrices are rectangular arrays of numbers that are arranged in rows and columns, however the regular rules of algebra do not apply.

Let X=image01.png and Y=image02.png and calculate X2, X3, X4 ; Y2, Y3, Y4.

X2=image34.pngimage01.png ×image01.png = image24.pngimage00.png

X3=image76.pngimage01.png ×image01.png × image01.png =image03.pngimage00.png

X4=image14.pngimage01.png ×image01.png × image01.png × image01.png = image05.pngimage00.png

Because matrices do not follow the algebraic rules of exponents, one can not simply distribute the exponent for each matrix value. Instead the matrix must be multiplied by itself however many times the exponent says. So for example, for X2, the matrix X must by multiplied with itself two times.

The pattern that has emerged is that with every increasing power the matrix value increases with an exponential power of 2.

21 is equal to 2 showed by the matrix image24.png. 22 is equal to 4 showed by the matrix image03.png and similarly 23 is equal to 8 represented in the matrix image05.png.

Y2=image06.pngimage02.png ×image02.png = image07.pngimage00.png

Y3=image09.pngimage02.png ×image02.png×image02.png = image10.pngimage00.png

Y4=image12.pngimage02.png ×image02.png×image02.png×image02.png = image13.pngimage00.png

The pattern is very similar to the one above except that all the negatives in the original matrix will also become negatives.

Based on the results above we can conclude that for each exponent value Xn the matrix value will result in Xn-1 and the same goes for Yn resulting in Yn-1.

...read more.

Middle

The matrix values are all multiplied by 3.

A=5X

A2= 5image34.png-5image24.png = image81.pngimage00.png

A3= 5image76.png                        5image03.png = image82.pngimage00.png

A4= 5image14.png     5 image05.png = image83.pngimage00.png

The matrix values are all multiplied by 5.

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

B= -1Y

 B2= -1image06.png--1 image16.png = image24.pngimage00.png

B3= -1image09.png                -1 image17.png        = image03.pngimage00.png

B4=-1image12.png     -1 image04.png = image05.pngimage00.png

When multiplied by -1, the matrix values become positive but remain the same. The reverse results of when matrix A was multiplied by -1.

B= Y

 B2= -½image06.png-image07.png = image08.pngimage00.png

B3= -½image09.png                -½ image10.png        = image11.pngimage00.png

B4=-½image12.png     -½ image13.png = image15.pngimage00.png

When multiplied by -1, the matrix values are divided by two (or halved) and all positive and negative the signs are switched in relation with the original matrix.

B= ½Y

 B2= ½image06.png-½ image07.png = image02.pngimage00.png

B3= ½image09.png                ½ image10.png        = image07.pngimage00.png

B4image12.png     ½ image13.png = image10.pngimage00.png

When multiplied by -1, the matrix values are divided by two (or halved) and all positive and negative the signs remain the same as in the original matrix (key difference when divides by -½).

B= 1Y

 B2= 1image06.png-1 image16.png = image16.pngimage00.png

B3= 1image09.png                1 image17.png        = image17.pngimage00.png

B4=1image12.png     1 image04.png = image04.pngimage00.png

When multiplied by 1, the matrix values remain negative. The reverse results of when matrix A was multiplied by 1.

B= 3Y

 B2= 3image06.png-3 image16.png = image18.pngimage00.png

B3= 3image09.png                3 image17.png        = image19.pngimage00.png

B4=3image12.png     3 image04.png = image20.pngimage00.png

When multiplied by 3, the matrix values remain negative.

B= 3Y

 B2= 3image06.png-3 image16.png = image18.pngimage00.png

B3= 3image09.png                3 image17.png        = image19.pngimage00.png

B4=3image12.png     3 image04.png = image20.pngimage00.png

The matrix values are all multiplied by 3.

...read more.

Conclusion

image63.png= (-9×image01.png)3  + (11×image02.png)3

image63.png= image64.png  + image59.png

image65.png= image66.png  + image62.png

image65.png= image65.png

In all these case the general statement is correct and works. However, the scope of the general statement is limited to positive exponents as well as the matrices X and Y. We have only tested a very small sample of the different matrix values and its infinite combinations for the matrices X and Y.

If we were to use different matrix values for X and Y the general statement would not apply:

Let n=3, a=9 and b=11 but let X = image24.png and let Y = image07.png

image57.png= (9×image24.png)3  + (11×image07.png)3

image57.png= image67.png  + image68.png

image60.png= image69.png  + image70.png

image60.pngimage71.png

As one can see, the general statement does not apply when the values for the matrices X and Y are changed and only works correctly when:

X=image01.png and Y=image02.png  .

To reach my general statement I used my previous findings.

I found that An = a×Xn and that Bn = b×Yn . This in turn allowed me to come to the conclusion that (A+B)n = [(a×X) + (b×Y)]n.

Using the knowledge that M=image28.png and that M=A+B and that M2=A2+B2 I could reach the following conclusion:

Mn=An+Bn

Mn=(a×X)n + (b×Y)nsubstituting A for a×X and B for b×Y

This is how I reached by general statement.

...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

    One would rarely see these in the higher life expectancy countries. Therefore, outliers are from the developing countries that are not only low in GDP but also other factors affecting it, such as war, disease, 'brain drain', economy crisis and so on.

  2. Math IA - Logan's Logo

    Notice that this value of d is the exact same y-value we calculated when we were determining variable c, by extending the data points of my curve to meet the center line. Thus, d=0.35. To recap the determined values for each of the variables, we have: a=-3.15 b= c=3.0 d=0.35

  1. Math Studies - IA

    The sources are as follows: * http://www.PGATour.co/tournaments * http://www.EuropeanTour.com/tournaments * http://www.RyderCup.org/2002 * http://www.RyderCup.org/2004 * http://www.RyderCup.org/2006 For the majors, the final scores are arranged as a total over the course of one tournament (four rounds). So to find the average of one team in a major, the players, US's for example,

  2. Math IA type 2. In this task I will be investigating Probabilities and investigating ...

    Non Deuce Deuce The possible results without deuce are 4 - 0 , 4 - 1 and 4 - 2 This would be: Now I will substitute c and d with as they represent the chances of Adam and Ben winning each point.

  1. Math IA - Matrix Binomials

    In order to find the final expression for Xn, we must multiply the general scalar value 2n-1 by matrix X: In order to test the validity of this expression, we can employ it to find X5. Using the previous method of calculations, we find that X5=.

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

    intersections because the graph was a quadratic function but in this case, there is a cubic function. Thus there will be 6 intersections between the cubic function and the 2 linear functions and D will be defined as. Where x2, x3 and x5 are intersections of one linear function and

  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. SL Math IA: Fishing Rods

    There will be two methods that will be used to solve the system of equations, seen below. Each method will be used for one of the systems being evaluated. Data Set 1 = {(1,10), (3,38), (8,149)} In the first data set, the data points will form separate equations that will be solved using a matrices equation.

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