• 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

    For (X+Y)n: When n=1, 2, 3, 4, ... (integer powers increase), then: 1, 2, 4, 8, ... These terms represent the the scalar values multiplied to the matrix (X+Y)= to achieve an end product of (X+Y)n. Thus, we can now deduce the geometric sequence of these scalar values using the

  2. Lacsap's Fractions : Internal Assessment

    the element number and as for the equation 3, "2" in the x part of the equation is also equivalent to the element number. Hence, the x part of the equation can now be substituted by the element number, r, to form the general statement shown below: Numerator - r(n - r)

  1. Matrix Binomials IA

    previous calculations that M3 = , we can say that , M3 = (A+B)3 .

  2. Mathematics Higher Level Internal Assessment Investigating the Sin Curve

    negative number, the equation becomes and this causes the graph to translate units to the left. From Graph 3.2 it can be fe teto getal thd beCurve seen that the value of can be any number. It can either be a fraction, a whole number or even an irrational number.

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

    n Z+ (X + Y) = = 2(I) where I is the Identity matrix From this we can get the following general formula (X + Y)n = (2I)n To test the validity of this statement we will test out different values for n. n=1 (X + Y)n = (2I)n (X + Y)1 = (2I)1 X + Y = (2I)

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

    This would be attributed to Wessig?s surprise performance at the 1980 event; Wessig covertly used a new unique technique not used before which contributed to his breaking the world record by a wide margin. Given the new data expansion, it would seem appropriate to modify the model so that it fits much more with the extra data points.

  1. 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.

  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