Matrix Binomials

As you might have noticed, this matrix has the same digits as Matrix A, however, the second and third numbers are negative, this and is due to the  placement.

Now to find the expression of Xn, Yn, (X + Y)n, I found a clear pattern in the matrices developed.  Each matrix was multiplied by 2. However, the power of X and Y also play a role which influence what’s on the inside of the matrix, from that I started experimenting and I came up with solution. Since all the answers are multiples of 2 and gradually increase, one is taken away from the power of X and Y.  I found that 1, 2, 4, 8, 16, 32 can also be determined as 20 21, 22, 23, 24, 25. Thus creating a pattern of 2n-1.

Example:

• = in this case n=1 because  has a power of 1, therefore the answer to this would be = = .
•  = =  n=2
• = =  n=3
• = =  n=4

Now that the pattern for matrix X has been found, you may come to realize that the highlighted green matrices are indeed multiples of two as well. So this pattern has been proved.

In order to find the pattern for matrix Y the same pattern will be used, however, negative signs will be drawn into the matrix so that it compliments.

Example:

• = in this case n=1 because  has a power of 1, therefore the answer to this would be = =.
• = =  n=2
• = =  n=3
• = =  n=4

From my observation, the pattern found suits both matrices.

Now to prove that Xn, Yn can also be equal to (X + Y)n. I will demonstrate this by using two different numbers, which aren’t given from the question in order to prove that the expression is correct and can work with other numbers given.

Matrix Xn =  and Matrix Yn =

Then (X + Y)n =

Or = 2n

• (X + Y)5 =

=

=

=

To prove this in terms of 2n

25=

=

• (X + Y)8 =

=

=

=

Proving it in terms of 2n

28=

=

As you might have noticed there is a zero involved, which in this case makes this matrix an identity kind of matrix. Such as .

Example 1:

IA= A        =

• Question 2

Let A = aX  and B = bY, where a and b are constants.

Use different values of a and b to calculate A2, A3, A4; B2, B3, B4.

By considering integer powers of A and B, find expressions for Xn, Yn, (X + Y)n.

In this part of the question I will use the GDC in order to calculate the answer of the matrix. I will choose to do so because in the previous question I have clearly shown how to multiply a matrix.

However, knowing that the matrix A = aX a pattern conceived from here will determine the expression.

Example:

a =

A2, A3, A4, A5, A6

= =

==

==

==

==

Now, using the previous matrix, a similar pattern has been derived from this matrix, however the only difference is the ‘a’. the pattern is that 1, 2, 4, 8, 16, 32 can also be determined as 20 21, 22, 23, 24, 25. Thus creating a pattern of 2n-1, therefore leaving ‘a’ as an. The new matrix evolved from this pattern will be.

Example:

• Let a = 3 and n=5, hence

= = =

Or

3= =

• Let a = -2.5 and n=3, hence

= = =

Or

-2.5= =

Now that this pattern has been proven using two types of numbers, I will not need to prove it any further.

Right now I will begin expanding on matrix B. I will use the same numbers as I did in matrix A just so that the matrices are even as well.

Now that the matrix B = bY, a pattern was conceived from here will determine the expression.

Example:

b =

B2, B3, B4, B5, B6

= =

==

==

==

==

Now, using the previous matrix, a similar pattern has been derived from this matrix, however the only difference is the ‘b’. The pattern is that 1, 2, 4, 8, 16, 32 can also be determined as 20 21, 22, 23, 24, 25. Thus creating a pattern of 2n-1, therefore leaving ‘b’ as bn. The new matrix evolved from this pattern will be.

Example:

• Let b = 7 and n=4, hence

===

Or

7= =

• Let b = -0.6 and n=3, hence

= =

Or

-0.6= =

Now to express these terms, to find the expression of (A + B)n

• Question 3

Now consider M =  

Show that M= A + B, and that M2 = A2 + B2.

Hence, find the general statement that expresses Mn in terms of aX and bY.

M = A + B

• M = aX + bY , a+ b
• +=

As you can see I used the previous matrix to show how matrix M is reached. Now to show that M2 = A2 + B2, I will use the same variables to show how M2 is attained.

M2 = A2 + B2

• +

= + =

which can also be written as .

To find the general statement which expresses Mn in terms of aX and bY.

The general statement would be.

Now I am going to test the validity of the general statement found (Mn = An + Bn)

• a = 3, b = 4, n = 2

==

Or

==

• a = -3, b = -4, n=2

==

Or

==

• a = 0.5, b = -5, n=4

==

==

Now I shall discuss the scopes and limitations I encountered during this portfolio. I realized that I can not have the power (n) as a negative or as a fraction or even as a root. Hence proving my point that the power can only be natural numbers. I will prove this right now giving 3 examples.

NEGATIVE

• a = 3, b = 2, n = -2

== this is not possible. Syntax error.

SQUARE ROOT

• a = 3, b = 2, n = √2

== this is not possible. Syntax error.

FRACTION

• a = 3, b = 2, n =

== this is not possible. Syntax error.

As you may have noticed I used the same numbers for the matrix because, no matter what number is inputted into the matrix you will get a syntax error, because the powers do not exist.

As I conclude this project I have shown all the working out, and I shown how the general statement is processed and I have also shown the different ways of which a matrix can be expressed whilst getting the same answer. The general statement is basically another way of showing how Mn = (A + B)n can be shown. This project was to strengthen out knowledge about matrix binomials and how they can be used in just simple sequences.