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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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, we find that it equals 2 when multiplied to Y which is why      Y2 =.

If the expression is Yn = 2n-1(Y), then:

Yn =

Testing for n = 4 would result in Y4 = 24-1(Y)

Y4 =  =

Verifying using Yn =

Y4 =  =  =

Testing for n = 15

Y15 = (Y4)(Y4)(Y4)(Y3) ==


Verify using Yn =

Y15 =  =  =

Testing for n = 9

Y9 = (Y4)(Y4)(Y) =  =

Verify using Yn =

Y9 =  =  =

Testing for n = 13

Y15 = (Y9)(Y4) = =

Verify using Yn =

Y15 =  =

From this, we can conclude that Yn = 2n-1(Y). When multiplied out, 2n-1 , this ultimately comes to Yn = .

Remember that the value of X and the value of Y are matrices. Essentially, X + Y is another matrix. Matrix (X+Y) will be called Z. Using various values we can find the values of Z2, Z3, Z4, and Zn. However, remember that Z is the value of the sum of X and Y.

Z = X + Y =  +  =

Z2 =  =

Z3 =  =  

Z4 =  =

Using information from above, we can test to see whether or not it is applicable when finding Zn. Essentially, since Xn = 2n-1(X) and Yn = 2n-1(Y), we’ll test to see if                   Zn = 2n-1(Z).

When n = 4

Z4 =  =

Verifying with Zn = 2n-1(Z),

Join now!

Z4 = 24-1 = 23 = 8 =

Although it appears that Zn = 2n-1(Z), we must check other values to determine that the equation does stay true.

When n = 15

Z15 = (Z4)(Z4)(Z4)(Z3) =  =

Verifying with Zn = 2n-1(Z),

Z15 = 215-1 = 214  = 16384 =

When n = 9

Z9 = (Z4)(Z4)(Z) =  =

Verifying with Zn = 2n-1(Z),

Z9 = 29-1 = 28 = 256 =

Testing when n = 13

Z13 = (Z4)(Z4)(Z4)(Z) =  =

With this evidence, we can come to the conclusion that Zn = 2n-1(Z). However we ...

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