Yn =

Likewise, the values of the matrix (X + Y), raised to the power 2, 3, and 4 is calculated to find its general expression.

The matrix: (X + Y) = + =

So, (X + Y) 2 = =

(X + Y) 3 = x =

(X + Y) 4 = x =

And from the above, we can infer that the general expression for (X + Y) n is as follows,

(X + Y) n =

Proof:

Taking n as 3, the value is substituted in the above expression –

(X + Y) 3 =

=

So since (X + Y) 3 is equal to as calculated previously, therefore the general expression (X + Y) n = is true and consistent with all values of n.

2. Consider A = aX and B = bY, where a and b are constants. In order to find the general expressions of An, Bn and (A + B)n, different values of a and b are used to calculate A2, A3, A4 and B2, B3, B4 below.

Let a = 2 and b = -2

A = aX B = bY

A = 2X = 2= B = -2Y = -2=

A2 = = B2 = =

A3 = = B3 = =

A4 = = B4 = =

Note: A GCD calculator (TI 83) has been used throughout this portfolio to calculate the matrices and other calculations.

Observing the above calculations, we can detect a certain pattern in determining the values, which gives us the general expression of A and B in terms of n as;

An = and Bn =

Proof:

Taking n to be 4, we substitute the values in the expression of An –

A4 =

A4 =

A4 =

A4 =

And since A4 = is the correct value as calculated previously, this expression is proven true and consistent.

Likewise, to find the general expression of (A + B) n , the values of (A + B) raised to the powers 2,3 and 4 are calculated;

First, we find the value of (A + B) –

(A + B) = + =

(A + B) 2 = =

(A + B) 3 = =

(A + B) 4 = =

Observing the repeating patterns in the calculations above, we can deduce the general expression of (A+B) in terms of n to be;

(A + B) n = 2 n-1

Proof:

Taking n as 3, and substituting it in the above expression –

(A + B) 3 = 2 3-1

(A + B) 3 = 2 2

(A + B) 3 = 4

(A + B) 3 =

So since (A + B) 3 is equal to as calculated previously, it is proven that this general expression for (A + B) n is true and consistent with all values of n.

3. Consider M = , to prove that M = A+B and M2 = A2+B2 , we must first calculate the value of M –

As a = 2, and b = -2,

M =

M =

To prove that M = A+B, we already know from previous workings that A = and B = -

So,

M = A+B

M= +

M =

We know that the value of M is from the previous calculation, therefore M=A+B is true.

And likewise, to prove that M2 = A2+B2 ,

M2 = = and,

A2+B2 = (A+B) 2 = A2+B2 + 2AB -------- (a/c to Algebraic Expression)

So since; A2 = =

B2 = =

2AB = 2

= 2

=

Thus, M2 = A2+B2 = A2+B2 + 2AB would be –

M2 = + +

=

And since M2 = is the right value, it is proved that M2 = A2+B2 is true.

4. Now, in order to find the general statement expressing Mn in terms of aX and bY, we first calculate the value of Mn where n = 1, 2, 3 and 4.

M =

M2 = =

M3 = =

M4 = =

Using the expression: (A + B) n = 2 n-1 whereby a=2, b= -2, and taking n=3, we calculate (A+B) raised to the third power –

(A+B)3 = 23-1

= 22

= 4

=

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

Therefore, the general statement of Mn in terms of aX and bY is;

Mn = (aX+bY)n

Proof:

To check the validity of this general statement, we shall take different values for a, b and n. Suppose a= 3, b=4, and n= 2 –

M2 = (3X+4Y)2

M2 =

M2 =

M2 =

M2 =

And since M = (A+B) = (aX+bY),

M =

M =

So then,

M2 = =

Therefore, the statement is proven true and consistent with all values of a, b and n.

5. Using the Algebraic method, the general statement is to be verified and explained again.

Taking the expression, Mn =2 n-1 , we find:

M =

M2 = = 2

M3 =

= 2

= 2

= 2

=4

Proof:

Substituting the above with the initial values of a and b, we find M3 -

M3= 4

M3 = 4

M3 = 4

M3 = 4

M3 = 4

M3 =

Therefore, since it has been shown earlier in our work that the value M3 = is true and correct, it shows that the general statement of Mn in terms of aX and bY is true.