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.