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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.

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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.

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