Using my GDC I found

(X + Y)1 = (X + Y)3 =

(X + Y)2 = (X + Y)4 =

When I finished with these calculations, I saw another pattern that was developed. Seeing this pattern I formed an expression to solve for the matrix (X + Y)n. For the numbers of a and d the expression is 2n and for c and d it is 0. As a matrix the expression is (X + Y)n = .

To make sure all my expressions worked correctly, I calculated each matrix to the fifth power which was the next integer power in the sequence.

X5 = 2(5-1) 2(5-1) Y5 = 2(5-1) -(2(5-1)) (X + Y)5 = 25 0

2(5-1) 2(5-1) -(2(5-1)) 2(5-1) 0 25

= 24 24 = 24 -(24) =

24 24 -( 24) 24

= =

After these calculations rose to the fifth power, the pattern still continued for each matrix. Letting A = aX and B = bY, where a and b are constants; I used different values of a and b to calculate A2, A3, A4; B2, B3, B4. For the value of a, I used the number 2 and the value of b, I used 3. With these values I used my GDC to solve for A and B.

A = B =

A2 = B2 =

A3 = B3 =

A4 = B4 =

By considering integer powers of these matrices, I came up with an expression to also solve these matrices. For matrix A the expression is An = (an) 2(n-1) 2(n-1) and matrix B the expression Bn = (bn) 2(n-1) -(2(n-1)) 2(n-1) 2(n-1)

-(2(n-1)) 2(n-1)

Using my GDC I solved these matrices.

(A + B)1 = (A + B)3 =

(A + B)2 = (A + B)4 =

After solving these equations, I generated an expression to solve the matrix for

(A + B). The expression to solve the matrix is (A + B)n = (an Xn) + (bn Yn). To make sure the expression worked I plugged in number 2 and 4 for the value of n.

(A + B)2 = (22 X2) + (32 Y2) (A + B)4 = (24 X4) + (34 Y4)

Using GDC (A + B)2 = Using GDC (A + B)4 =

Considering that M = , this expression can also be derived into

M = A + B and M2 = A2 + B2. By proving the expression M = A + B, A and B needs to be substituted for aX and bY as well as keeping the constants as a variable. This will prove the expression M = through M = A + B.

M = aX + bY

= a + b

= +

=

M = A+ B

We can also prove that M2=A2+B2 to later help find a general statement for Mn, in terms of aX and bY.

M2=A2+B2

= (aX)2 + (bX)2

= a2+b2

(a2 + 2ab + b2) + (a2 + 2ab + b2) (a2 – b2) + (a2 – b2) 2a2 2a2

(a2 – b2) + (a2 – b2) (a2 + 2ab + b2) + a2 + 2ab + b2) = 2a2 2a2 +

2b2 -2b2

-2b2 2b2

2a2 + 2b2 2a2 – 2b2 2a2 + 2b2 2a2 – 2b2

2a2 – 2b2 2a2 + 2b2 = 2a2 – 2b2 2a2 + 2b2

Hence, the general statement that expresses Mn in terms of aX and bY can be expressed as:

Mn = (aX)n + (bY)n

To test the validity of my general statement by using different values of a, b, and n.

First I will use the general statement Mn = (aX)n + (bY)n and verify it by using the

n

expression Mn =

Let a = -1, b = 0.5, and n = 2

By finding if there are any limitations within this expression: Mn = (aX)n + (bY)n , I am going to change the constants ‘a’ and ‘b’ as well the power raised to ‘n’ into , decimals, fractions and negative exponents.

The first example I am going to prove that this general statement has some limitations:

Let a=0.3, b=0.32, and n=-3

M-3 = (0.3X)-3 + (0.32Y)-3

-3 -3

M-3 = +

As I plug this equation into my calculator it comes up with: (error domain). As a result this equation cannot be solved because the matrix cannot be raised to a negative power. Therefore the limitation in this expression is that ‘n’ cannot be a negative number.

In this second example I am going let ‘n’ equal to a positive integer and the constants a and b equal zero:

Let a = 0, b = 0, n = 3

M3 = (0X)3 + (0Y)3

M-3 = +

=

The limitation for the expression Mn = (aX)n + (bY)n is that ‘n’ cannot contain a negative exponent nor a decimal value or a fraction because if we multiply an exponent raised to a negative number it would make the value flip. However, both of the constants ‘a’ and ‘b’ can equal to any set of real numbers. Therefore the limitations and scope are:

To conclude this assignment, last I will explain how I arrived at my general statement using an algebraic method.

Let n = 2, a = a, b = b, X = , Y =

(aX)n + (bY)n =

a2+b2 =

2a2 2a2 + 2b2 -2b2

2a2 2a2 -2b2 2b2 =

2a2 + 2b2 2a2 – 2b2 = (a2 + 2ab + b2) + (a2 + 2ab + b2) (a2 – b2) + (a2 – b2)

2a2 – 2b2 2a2 + 2b2 (a2 – b2) + (a2 – b2) (a2 + 2ab + b2) + a2 + 2ab + b2)

2a2 + 2b2 2a2 – 2b2 2a2 + 2b2 2a2 – 2b2

2a2 – 2b2 2a2 + 2b2 = 2a2 – 2b2 2a2 + 2b2

Therefore the equation Mn = (aX)n + (bY)n equals with Mn =

Arriving at this general statement, it can help solve for the value of M in any equation.