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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.

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 =

X2 =  =

X3 =  =  =

X4 = =  =

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 =.

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

Xn =

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

X4 = 24-1 = 8 =

Testing for n = 15

X15 = (X4)(X4)(X4)(X3) = =

Verify using Xn =

X15 = =  =

Testing for n = 9

X9 = (X4)(X4)(X) =  =

Verify using Xn =

X9 =  =  =

Testing for n = 13

X13 = (X4)(X4)(X4)(X) =  =

Verify using Xn =

X13 =  =  =

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

Y =

Y2 =  =

Y3 =  =  =

Y4 =  =  =

If Y2 = Yn, then 2 = n.

Using the expression 2n-1

Middle

=  =

A3 =  =

A4 =  =

When a = 10

A2 = 10 10 =  =

A3 =  =

A4 =  =

When a = 5

A2 = 5 5 =  =

A3 =  =

A4 =  =

When a = 3

A2 = 3 3 =  =

A3 =  =

A4 =  =

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)n (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 =   A3 =  A4 =

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

A4 = 54 =

When a = 10

A2 =  A3 =  A4 =

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

A4 = 104 =

When a = 8

A2 =  A3 =  A4 =

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

A4 = 84 =

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.

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 =. Therefore, Mn must equal. 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) + (22)(2) =

Verify using the Mn =

M2 =  =  =

When a = 8 and b = 8

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

M2 = (82)(128)  + (82)(128)  =

Verify using the Mn =

M2 = =

When a = 8 and b = 6

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

M2 = (82)(128) + (62)(32) =

Verify using the Mn =

M2 = =

From this, we can consequently conclude that Mn =  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

This student written piece of work is one of many that can be found in our International Baccalaureate Maths section.

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