• 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

Matrix Binomials The assignment of this internal assessment fundamentally consisted of nine goals. Those nine goals are finding Xn, Yn, (X+Y)n, An, Bn, (A+B)n, proving M = A + B, proving M2 = A2 + B2, and finding Mn.

Extracts from this document...

Introduction

Internal Assessment - Matrix Binomials

        The assignment of this internal assessment fundamentally consisted of nine goals. Those nine goals are finding Xn, Yn, (X+Y)n, An, Bn, (A+B)n, proving M = A + B, proving M2 = A2 + B2, and finding Mn. With the initial matrices of X and Y, the assignment was found using calculations based off of X and Y. The assignment was also found using the value of A and B. Values for X, Y, A, and B were then used to find values for M.

All calculations were made using the GDC

        X = image00.png

        X2 = image00.pngimage00.png = image14.png

        X3 = image00.pngimage00.pngimage00.png = image14.pngimage00.png = image10.png

        X4 = image00.pngimage00.pngimage00.pngimage00.png= image10.pngimage00.png = image15.png

If X2 = Xn, then 2 = n.

Using the expression 2n-1, we find that it equals 2 when multiplied to X which is why      X2 =image14.png.

If the expression is Xn = 2n-1(X), then:

Xn = image12.pngimage24.png

Testing for n = 4 would mean: X4 = 24-1(X)

X4 = 24-1image00.png = 8 image00.png= image15.png

Testing for n = 15

X15 = (X4)(X4)(X4)(X3) = image15.pngimage15.pngimage15.pngimage10.png= image44.png

Verify using Xn = image24.png

X15 = image50.png= image53.png = image44.png

Testing for n = 9

X9 = (X4)(X4)(X) = image15.pngimage15.pngimage00.png = image18.png

Verify using Xn = image24.png

X9 = image73.png = image79.png = image18.png

Testing for n = 13

X13 = (X4)(X4)(X4)(X) = image15.pngimage15.pngimage15.pngimage00.png = image107.png

Verify using Xn = image24.png

X13 = image109.png = image115.png = image107.png

Therefore, we can conclude that Xn = 2n-1(X) which, after multiplying 2n-1image00.png, equals Xn = image24.png

Y = image02.png

Y2 = image02.pngimage02.png = image40.png

Y3 = image02.pngimage02.pngimage02.png = image40.pngimage02.png = image45.png

Y4 = image02.pngimage02.pngimage40.png = image45.pngimage02.png = image41.png

If Y2 = Yn, then 2 = n.

Using the expression 2n-1

...read more.

Middle

image00.png = image15.pngimage13.pngimage15.png = image17.png

A3 = image15.pngimage17.png = image19.png

A4 = image15.pngimage19.png = image08.png

When a = 10

A2 = 10image00.pngimage13.png 10image00.png = image25.pngimage13.pngimage25.png = image26.png

A3 = image25.pngimage26.png = image27.png

A4 = image25.pngimage27.png = image28.png

When a = 5

A2 = 5image00.pngimage13.png 5image00.png = image29.pngimage13.pngimage29.png = image30.png

A3 = image29.pngimage30.png = image31.png

A4 = image29.pngimage31.png = image32.png

When a = 3

A2 = 3image00.pngimage13.png 3image00.png = image33.pngimage13.pngimage33.png = image34.png

A3 = image33.pngimage34.png = image35.png

A4 = image33.pngimage35.png = image36.png

Because A = aX, we must remember that X = 2n-1(X). In finding An, we must find how the value of (a) changes. We can see this by looking at the results of An and how it corresponds to (a).

Since An means (aX)n, we can see that as (a)nimage13.png (X)n. Since Xn = Xn = 2n-1(X), we can consequently combine the two equations which means that An = 2n-1(X) (a)n

We can verify this by testing the previous values with the equation.

When a = 5

A2 = image30.png  A3 = image31.png A4 = image32.png

Now using the equation An = 2n-1(X) (a)n

A4 = image37.png54 = image32.png

When a = 10

A2 = image26.png A3 = image27.png A4 = image28.png

Now using the equation An = 2n-1(X) (a)n

A4 = image38.png104 = image28.png

When a = 8

A2 = image17.png A3 = image19.png A4 = image08.png

Now using the equation An = 2n-1(X) (a)n

A4 = image39.png84 = image08.png

From this evidence, we can conclude that An equals 2n-1(X) (a)n.

Next, we’ll consider B. B is a value and can be represented by B = aY. The value of (b) is always a constant, and we’ll use the value of X from the previous work. Using this information, we’ll find the values of B2, B3, B4, and Bn.

...read more.

Conclusion

n = An + Bn. However, relating this back to the original matrix form of M, we must consider that the original formula for M was M =image110.png. Therefore, Mn must equalimage132.png. However, this translates to equal the same as An + Bn because it is the same as Mn = (an)(2n-1)(X) + (bn)(2n-1)(Y). Nonetheless we must still test this with values to verify and confirm, and once again, we must take into consideration values of (a) and (b) that are both the same and different, and we can use both equations to make sure.

When a = 2 and b = 2

Using the equation Mn = (an)(2n-1)(X) + (bn)(2n-1)(Y).

M2 = (22)(2)image00.png + (22)(2)image02.png = image07.png

Verify using the Mn = image132.png

M2 = image133.png = image134.png = image07.png

When a = 8 and b = 8

Using the equation Mn = (an)(2n-1)(X) + (bn)(2n-1)(Y).

M2 = (82)(128) image00.png + (82)(128) image02.png = image75.png

Verify using the Mn = image132.png

M2 = image135.png= image75.png

When a = 8 and b = 6

Using the equation Mn = (an)(2n-1)(X) + (bn)(2n-1)(Y).

M2 = (82)(128)image00.png + (62)(32)image02.png = image136.png

Verify using the Mn = image132.png

M2 = image137.png= image136.png

From this, we can consequently conclude that Mn = image132.png because when plugged in for various values of (a) and various values of (b). The equation confirms and verifies to be true because we verified it using two equations that both equal

Mn = An + Bn.

Ultimately, the nine goals were reached based off calculations of the initial values of X and Y and all values and calculations were proven using various equations to verify and confirm

...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 IA - Matrix Binomials

    Using the previous method of calculations, we find that (X+Y)5=. (We can also check this on a calculator). Now, using the expression, we find the value of (X+Y)5 to be the same (proving the accuracy of our expression): (X+Y)5=25-1 =24 =16 = We conclude that is indeed a valid expression.

  2. Lacsap's Fractions : Internal Assessment

    = Denominator Now looking at the second row (highlighted in blue), the differences between the numerator and the denominator of the second element was recorded as 2, 4, and 6. For every consecutive row, the difference between the numerator and the denominator increases by 2.

  1. Matrix Binomials IA

    - M2 = + + = And since M2 = is the right value, it is proved that M2 = A2+B2 is true. 4. Now, in order to find the general statement expressing Mn in terms of aX and bY, we first calculate the value of Mn where n = 1, 2, 3 and 4.

  2. Mathematics Higher Level Internal Assessment Investigating the Sin Curve

    When is a positive number the translation is a horizontal translation to the right, whereas when the value of is negative the graph translates to the left by units as illustrated by Graph 3.1. When you change the value of you subtract the original value of by , and therefore

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

    This is because their determinant equals to 0(this was calculated using Microsoft Math) The general statement's limitations can also be concluded with the results above; n has to be a positive integer. n Z+ (X + Y) = = 2(I)

  2. Mathematics Internal Assessment: Finding area under a curve

    =x2+3 [0, 1], I shall divide the curve into 3 trapeziums instead of the two I have done in the example above.

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

    Another significant fluctuation would be the 1936 to 1948 time period; where the increasing height values decreased in value significantly through that time period. It is most likely that athletes in that time period were not likely to be training competitively for the Olympic level as most countries had a

  2. Type I Internal Assessment (Lascap's Triangle)

    Even though we see the numerator and the row number rise at a different rate, they both contribute to the ascending line. We observe that the line looks like a half parabola. This observation once again suggests that the general statement for Lacsap?s triangle is a quadratic equation: ax2 +

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