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.