• Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month
Page
  1. 1
    1
  2. 2
    2
  3. 3
    3
  4. 4
    4
  5. 5
    5
  6. 6
    6
  7. 7
    7
  8. 8
    8
  9. 9
    9
  10. 10
    10
  11. 11
    11
  12. 12
    12
  13. 13
    13
  14. 14
    14
  15. 15
    15
  16. 16
    16
  17. 17
    17
  18. 18
    18
  19. 19
    19
  20. 20
    20
  21. 21
    21
  22. 22
    22
  23. 23
    23
  24. 24
    24
  25. 25
    25
  26. 26
    26
  27. 27
    27
  28. 28
    28

Matrix Binomials

Extracts from this document...

Introduction

Introduction

The goal of this portfolio assessment is to find an expression for Xn, Yn, (X + Y)n, [An, Bn, (A + B)n] in both questions whilst expressing them: Mn in terms of aX and bY. The purpose of this assessment is to find out how we can interpret matrix binomials using different values and similarities to find the pattern occurring. We’ve been given a general statement to express Mn in terms of aX and bY, to do so we must substitute a into matrix X to get a new matrix ‘A’, and b into matrix Y to get the new matrix ‘B’. The task given now is to see if the pattern really did work with other numbers, and to prove the general statement.

  • Question 1

Let X =image00.png  and Y =image01.png. Calculate X2, X3, X4; Y2, Y3, Y4.

By considering integer powers of X and Y, find expressions for Xn, Yn, (X + Y)n.

Alright now to calculate X2, X3, X4; Y2, Y3, Y4, I will firstly show how these matrices are multiplied, and then I shall use my graphics calculator to do the rest. As doing so I will also look for a pattern trend in which I can use to relate to fine the expression Xn, Yn, (X + Y)n.

...read more.

Middle

image62.png

image63.pngimage56.png= image64.png

aimage00.png

a = 0.25 or image50.png

= image52.pngx image00.png

image65.png

image64.pngimage56.png

= image66.png

aimage00.png

a = 0.25 or image50.png

= image52.pngx image00.png

image67.png

image68.pngimage56.png

= image69.png

aimage00.png

a = -0.25 or -image50.png

= -image52.pngx image00.png

image71.png

The value of a (-0.25) is multiplied by the matrix given,image00.png, which equals to image72.png when multiplied by -image52.png (a). In this matrix I used a fraction instead of a decimal, simply because it’s easier to work with.

- RATIONAL NUMBER

aimage00.png

a = -0.25 or -image50.png

= -image52.pngx image00.png

image73.png

image74.pngimage74.png

= image57.png

aimage00.png

a = 0.25 or image50.png

= image52.pngx image00.png

image76.png

image59.pngimage74.png

= image77.png

aimage00.png

a = 0.25 or image50.png

= image52.pngx image00.png

image78.png

image79.pngimage80.png

= image81.png

aimage00.png

a = 0.25 or image50.png

= image52.pngx image00.png

image83.png

image81.pngimage74.png

= image84.png

aimage00.png

a = 0.25 or image50.png

= image52.pngx image00.png

image86.png

image87.pngimage74.png

= image69.png

For this type of number, the rational number, I used both negative and positive numbers. However the pattern is different, at first the sequence  has a multiple of 2 for the denominator.

aimage00.png

a = √5

= √5 x image00.png

image88.png

The value of a (√5) is multiplied by the matrix given,image00.png, which equals to image89.png when multiplied by √5  (a). In this matrix I used a root 5 instead of a decimal, simply because it’s easier to work with.

+ IRRATIONAL NUMBER

aimage00.png

a = √5

= √5 x image00.png

image90.png

image89.pngimage89.png

= image91.png

aimage00.png

a = √5

= √5 x image00.png

image92.png

image93.pngimage89.png

= image96.png

aimage00.png

a = √5

= √5ximage00.png

image97.png

image96.pngimage89.png

image98.png

aimage00.png

a = √5

= √5ximage00.png

image99.png

image98.pngimage89.png

= image101.png

aimage00.png

a = √5

= √5ximage00.png

image102.png

image103.pngimage89.png

= image104.png

aimage00.png

a = -√5

=- √5ximage00.png

image105.png

As you may have noticed, I am using a negative number now.

- IRRATIONAL NUMBER

aimage00.png

a = -√5

= -√5ximage00.png

image106.png

image107.pngimage108.png

= image91.png

aimage00.png

a = -√5

= -√5ximage00.png

image110.png

image93.pngimage108.png

= image111.png

aimage00.png

a =- √5

= -√5ximage00.png

image112.png

image113.pngimage108.png

image98.png

aimage00.png

a = -√5

= -√5ximage00.png

image114.png

image98.pngimage108.png

= image115.png

aimage00.png

a = -√5

= -√5ximage00.png

image116.png

image117.pngimage108.png

= image104.png

Whilst calculating this sequence I came to notice that every second matrix is an integer. I noticed that this pattern went from multiples of two to 10. Every second matrix is multiplied by 10.

However, knowing that the matrix A = aX

...read more.

Conclusion

image279.png.

To find the general statement which expresses Mn in terms of aX and bY.

The general statement would beimage280.png.

Now I am going to test the validity of the general statement found (Mn = An + Bn)

  • a = 3, b = 4, n = 2

image281.png=image282.png=image283.png

Or

image284.png=image285.png= image283.png

  • a = -3, b = -4, n=2

image287.png=image282.png=image283.png

Or

image288.png=image289.png= image283.png

  • a = 0.5, b = -5, n=4

image290.png=image291.png=image292.png

image293.png=image294.png=image292.png

Now I shall discuss the scopes and limitations I encountered during this portfolio. I realized that I can not have the power (n) as a negative or as a fraction or even as a root. Hence proving my point that the power can only be natural numbers. I will prove this right now giving 3 examples.

NEGATIVE

  • a = 3, b = 2, n = -2

image295.png=image296.png= this is not possible. Syntax error.

SQUARE ROOT

  • a = 3, b = 2, n = √2

image297.png=image298.png= this is not possible. Syntax error.

FRACTION

  • a = 3, b = 2, n = image299.png

image300.png=image301.png= this is not possible. Syntax error.

As you may have noticed I used the same numbers for the matrix because, no matter what number is inputted into the matrix you will get a syntax error, because the powers do not exist.

As I conclude this project I have shown all the working out, and I shown how the general statement is processed and I have also shown the different ways of which a matrix can be expressed whilst getting the same answer. The general statement is basically another way of showing how Mn = (A + B)n can be shown. This project was to strengthen out knowledge about matrix binomials and how they can be used in just simple sequences.

...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

    During the interpretation of results the weak correlation was partially explained by factors such as war, food shortage, economic crisis, epidemic of diseases, and so on. I addition, it could well be that there is a non linear correlation instead.

  2. Math IA - Logan's Logo

    Therefore, to find the value of c, I must first determine the center line of my curve - the middle line of the total height. In fact, we already determined this when we found variable a. (Recall that we divided the total height of the curve by 2 to find the amplitude).

  1. matrix power

    The first is multiplying the matrix by "n" times. The second method is simply bringing the individual digits in the matrix by the power of "n". Therefore for n = 10, 20, and 50 we can simply use the second method to determine the power of the matrix at the

  2. Math IA - Matrix Binomials

    We can find an expression for Yn in a similar fashion. , where Un=a specific term a=first term r=common ratio (multiplier between the entries of the geometric sequence) n=the number of the specific term (with relation to the rest of the sequence). For Yn: When n=1, 2, 3, 4, ...

  1. MATH IA- Filling up the petrol tank ARWA and BAO

    also consider whether the extra distance is worth the time, on that particular day. On top of that if he goes on the detour and is low on fuel and the station is closed he could use another one close by.

  2. Investigating Slopes Assessment

    However, I will do the table normally and check my results. f(x)= 3X3 X=? Tangent Gradient X= -2 Y= 36x+44 36 X= -1 Y= 9x+6 9 X= 0 Y= 0 0 X= 1 Y= 9x+(-6) 9 X= 2 Y= 36x+(-44)

  1. Lascap's Fraction Portfolio

    Like what have been described in the first question, the numerator of the sixth row is: 1 21 21 21 21 21 1 For the seventh row, the numerator can be calculated by adding 7 to the numerator of the sixth row; 21.

  2. Stellar Numbers. In this folio task, we are going to determine difference geometric shapes, ...

    = 1+21×12 S7 = 253 Now, subs X =(n-12+n-1)/2 into formula Sn=1+X×12, The general formula for p(6) Stellar number will be Sn=1+12n-12+n-1 =6n2-n+1 Again, test it by using n=5 (chose randomly), Sn=6n2-n+1 =652-5+1 =121 The value of S5 is same as the S5 value in Table on the page 5.

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