A4= -1 -1 =

When multiplied by -1, the matrix values become negative but remain the same.

A=-½X

A2= -½ --½ =

A3= -½ -½ =

A4= -½ -½ =

When multiplied by -½, the matrix values become negative. Also the values are divided by two (or halved).

A=½X

A2= ½ -½ =

A3= ½ ½ =

A4= ½ ½ =

When multiplied by ½, the matrix values stay positive. Also the values are divided by two (or halved).

A=1X

A2= 1 -1 =

A3= 1 1 =

A4= 1 1 =

When multiplied by 1, the matrix values remain the same.

A=3X

A2= 3 -3 =

A3= 3 3 =

A4= 3 3 =

The matrix values are all multiplied by 3.

A=5X

A2= 5 -5 =

A3= 5 5 =

A4= 5 5 =

The matrix values are all multiplied by 5.

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

B= -1Y

B2= -1 --1 =

B3= -1 -1 =

B4=-1 -1 =

When multiplied by -1, the matrix values become positive but remain the same. The reverse results of when matrix A was multiplied by -1.

B= -½Y

B2= -½ --½ =

B3= -½ -½ =

B4=-½ -½ =

When multiplied by -1, the matrix values are divided by two (or halved) and all positive and negative the signs are switched in relation with the original matrix.

B= ½Y

B2= ½ -½ =

B3= ½ ½ =

B4=½ ½ =

When multiplied by -1, the matrix values are divided by two (or halved) and all positive and negative the signs remain the same as in the original matrix (key difference when divides by -½).

B= 1Y

B2= 1 -1 =

B3= 1 1 =

B4=1 1 =

When multiplied by 1, the matrix values remain negative. The reverse results of when matrix A was multiplied by 1.

B= 3Y

B2= 3 -3 =

B3= 3 3 =

B4=3 3 =

When multiplied by 3, the matrix values remain negative.

B= 3Y

B2= 3 -3 =

B3= 3 3 =

B4=3 3 =

The matrix values are all multiplied by 3.

From these results and using the scalars -1, - ½, ½, and 1 we can conclude that:

- The scalar -1 reverses all signs of matrix values
- The scalar 1 has all signs stay the same for matrix values
- The scalar – ½ halves and reverses all signs of matrix values
- The scalar ½ halves and has all signs remain the same for matrix values
- The scalars of 3 and 5 are simply multiplied by the matrix values and do not change the signs of the matrix values.

This information can guide us to find expressions for An, Bn and (A+B)n.

An = a×Xn (whatever the exponent of A is, it will also be the exponent for X)

Bn = b×Yn (whatever the exponent of B is, it will also be the exponent for Y)

(A+B)n = [(a×X) + (b×Y)]n

Test the validity of the statement:

a=1 & b=5 & n=2

(A+B)2 = [ (1×) + (5×) ]2

(A+B)2 = [ + ]2

(A+B)2 =

(A+B)2 =

a=2 & b=3 & n=4

(A+B)4 = [ (2×) + (3×) ]4

(A+B)4 = [ + ]4

(A+B)4 =

(A+B)4 =

The pattern follows matrix M=

The top left and bottom right matrix values are positive values as the corresponding values in the matrices X and Y are also positive. Contrarily, the values in the top right and the bottom left are negative as the values in the some location on the Y matrix are also negative.

Now when the scalars are substituted for a and b in matrix M you receive the same results as when you substitute them in the statement for (A+B) without calculating the result with the exponent.

Scalars: a = 1 & b = 5

M=

M = This is the same result (A+B) when its scalars are 5 and 1

Scalars: a = 2 & b = 3

M =

M = This is the same result (A+B) when its scalars are 2 and 3

This allows us to show that M=A+B

M = A + B remembering that A=a×X & B=b×Y

= (1×) + (5×)

= (+)

=

M = A + B remembering that A=a×X & B=b×Y

= (2×) + (3×)

= +

=

Taking it a step further we can also show that M2=A2+B2

M2 = A2 + B2 remembering that A=a×X & B=b×Y

= (1×) 2 + (5×)2

= +

= +

=

M2 = A2 + B2 remembering that A=a×X & B=b×Y

= (2×)2 + (3×)2

= +

= +

=

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

We can conclude from the investigation above that since:

Mn=An+Bn

Mn=(a×X)n + (b×Y)n

Making Mn= (aX)n+(bY)n the general statement.

To test the validity of this statement let n=5, a=6 and b=7

= (6×)5 + (7×)5

= +

= +

=

To further test the validity of the statement let n=4, a=7 and b=5

= (7×)4 + (5×)4

= +

= +

=

To further test the validity of the statement let n=3, a=9 and b=11

= (9×)3 + (11×)3

= +

= +

=

To further test the validity of the statement let n=3, a=-9 and b=11

= (-9×)3 + (11×)3

= +

= +

=

In all these case the general statement is correct and works. However, the scope of the general statement is limited to positive exponents as well as the matrices X and Y. We have only tested a very small sample of the different matrix values and its infinite combinations for the matrices X and Y.

If we were to use different matrix values for X and Y the general statement would not apply:

Let n=3, a=9 and b=11 but let X = and let Y =

= (9×)3 + (11×)3

= +

= +

≠

As one can see, the general statement does not apply when the values for the matrices X and Y are changed and only works correctly when:

X= and Y= .

To reach my general statement I used my previous findings.

I found that An = a×Xn and that Bn = b×Yn . This in turn allowed me to come to the conclusion that (A+B)n = [(a×X) + (b×Y)]n.

Using the knowledge that M= and that M=A+B and that M2=A2+B2 I could reach the following conclusion:

Mn=An+Bn

Mn=(a×X)n + (b×Y)n substituting A for a×X and B for b×Y

This is how I reached by general statement.