First, I am going to find A2 for the expression An = aX and use 4 different constants for ‘a’, where I am going to let X =

Calculation:

n=2 a=1 (without calculator) n=2 a=2 (with calculator) n=3 a=2 n=4 a=2

A2 = A2 = A2 =

= A2 = 8X A2 = 18X A2 = 32X

Now I am going to create a table for [A = aX] with all the different ‘n’ values and ‘a’ values:

Now I am going to find all the Bn by using the expression Bn = bY and use 4 different constants for ‘b’, where I am going to let Y =

By considering integer powers of A and B, find expression for An , Bn and (A+B)n

For the statement A= aX, I am going to determine a general formula by inputting different numbers for the constant ‘a’ and as well for the terms ‘n’.

An=(aX)n =

If I input a=1 and n=2 into An=(aX)n, the resulting value would be 2X:

However if I continue to input a=1 and change the terms ‘n’ to 3 and 4 a pattern begins to form:

By changing the terms n=2 up to n=4 ‘X’ increases each time from, 2X, 4X to 8X. The number of X’s that is being increased is resulted from this expression, “2n-1”. Therefore, we can convert the formula An=(aX)n to:

The same expression can be also used for the statement B=bY because both of the statements, A=aX and B=bY have the same pattern. The only difference between the two statements is that ‘X’ and ‘Y’ have different matrices. Therefore we just change, An to Bn by:

If the same values that were inputted for An=aX to Bn=bY, the resultant values will be exactly the same with ‘X’ being ‘Y’:

The expression for (A+B)n can be found by inputting different values in the expression for An=aX and Bn=bY. Therefore by inputting different values or same values for the constants ‘a’ and ‘b’ raised to the power of ‘n’

If we input the constants with the same values like a=1 and b=1 raised to the power of n=2

The resultant value for this expression contains an identity matrix when both of the values constants ‘a’ and ‘b’ are the same; however, there are some limitations. If the constants ‘a’ and ‘b’ contained different values such as (a=3 b=-12) the resultant matrix would differ, therefore the Identity Matrix cannot be used.

Let a=-3, b=5 and n=3

Therefore the expression for (A+B)n would be:

If I let a=5, b=-7 and n =3 with the expression than the expression where , will work as well.

I used a calculator to calculate this expression:

The general statement:

3

In this expression there are some limitations where ‘n’ cannot equal zero nor a negative value.

The expression can also be derived into and . By proving the expression , 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 through

Now I am going to substitute A and B for aX and bY for the expression .

Therefore the general statement that expresses Mn in terms of aX and bY could be expressed as:

Test the validity of your general statement by using different values of a,b and n

First I am going to use the general statement and then prove this general statement by using the expression

By finding if there are any limitations within this expression: , 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

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:

The limitation for the expression 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:

Use an algebraic method to explain how you arrived at your general statement.

The general statement that came from is This general statement should equal to .

To prove that this general statement equals to , I am going to expand the equation by using only variables:

Therefore the equation equals with

However, the equation would not work unless it is proven by the binomial theorem.

xn-kyk

=+2+

This calculations tells us that AB must equal to zero for this equation to work . As said before the only way the equation works is because equals to a zero matrix: .

In the end, the expression can be substituted into a different equation where (aX)n+(bY)n can be replaced as .