Determining X and Y
Since we know that X= and Y=, we can therefore assume that any of these matrices multiplied by themselves will give increasing exponential results.
Throughout the assignment, bolded matrices are expanded form of the original equation, and blue matrices are the final answer.
X2 = x
=
X3 = X2 =
=
X4 = X3 =
=
Since we know that X multiplied by X = X2 and X2 times X = X3, we can assume that matrix Xn multiplied by the original matrix X, gives a matrix one exponent higher, in this case Xn+1, therefore the following formula can be used.
Xn= Xn-1 x X (original equation)
Y2 = Y1 =
=
Y3 = Y2 =
=
Y4 = Y3 =
=
Since we know that Y multiplied by Y = Y2 and Y2 times Y = Y3, we can assume that matrix Yn multiplied by the original matrix Y, gives a matrix one exponent higher, in this case Yn+1, therefore the following formula can be used.
Yn=Yn-1 Y (original equation)
(X+Y)2=
(X+Y) =
=2
=
=2
=
(X+Y)3= (X+Y)0=
=3 =0
=3 =0
= =
(X+Y)4= (X+Y)-1=
=4 =-1
=4 = -1
= =
(X+Y)0.5= 0.5 (X+Y)-0.5= -0.5
=0.5 =-0.5
= =
After performing all the calculations, it is noticeable that the non-zero figures in the matrix double every time the variable “n” is increased by one. This statement applies to all the calculations, including the Xn, Yn, and (X+Y)n formulas. Because of this a coefficient of 2 can be used because it doubles. The elements in every matrix, no matter what the value of “n” is, always double in the next one in the series. By using the matrix feature on my calculator an expression of (X+Y)n = 2n-1(X+Y) was found to be valid. A limitation is that “n” has to be a whole number for the expression to work.
Determining A and B
Since we know that A=aX, and B=bY, we can give “a” and “b” respective values to test, and therefore test the theory and see if the statements are true or false.
To test the statement above, Let “a” = 2 first.
A2=(2X)2 = (2)2
= 2
=
A3=(2X)3 = (2)3
= 3
=
A4=(2X)4 = (2)4
= 4
=
Now the statement above will be tested with a negative digit. Let a = (-2).
A3=(-2X)2 = (-2)2
= 2
=
A2=(-2X)3 = (-2)3
= 3
=
A3=(-2X)4 = (-2)4
= 4
=
Decimals will be used in the next 2 examples. Let a = 0.5, and 0.25
A2=(0.5X)2 = (0.5)2
= 2
=
A3=(0.5X)3 = (0.5)3
= 3
=
A4=(0.5X)4 = (0.5)4
= 4
=
A2=(0.25X)2 = (0.25)2
= 2
=
A3=(0.25X)3 = (0.25)3
= 3
=
A4=(0.25X)4 = (0.25)4
= 4
=
These examples show that the formula works: An=anXn. From the examples, it can be concluded that positive, negative, and decimal values work for “a”. This statement has no limitations. Therefore it is entirely valid. I also checked other values on my TI-84 Calculator, and came up with the same conclusion.
Patterns that were seen during this test were that whenever “a” was equal to a positive whole number, the resulting elements in each subsequent matrix quadrupled as the exponent of “a” increased by 1. The same pattern happens when “a” is set to a negative integer, however, whenever “n” is an even number (as in ex. 2), the resulting elements are all positive, while when “n” is set equal to an odd number (as in ex. 3), the resulting elements are all negative. When “a” is set to a value less than 1 and greater than zero (a decimal or fraction), it causes the elements to become increasing smaller as the “n” value increases. There is an exception for this is when “a” is set to 0.5, the elements always add up to 0.5, regardless of the “n” value used.
To test the above statement, Let “b” = 3 first.
B2= (bY)2= (3)2
= 2
=
B3= (bY)3= (3)3
= 3
=
B4= (bY)4= (3)4
= 4
=
To test the above statement with a negative figure, Let “b” = (-3).
B2= (bY)2= (-3)2
= 2
=
B3= (bY)3= (-3)3
= 3
=
B4= (bY)4= (-3)4
= 4
=
Therefore we can assume that the formula Bn=bn×Yn-1 works in this case as well.
B2= b2xY2-1 =
= 32x = 9
=
B3= b3xY3-1 =
= 33x = 27
=
B4= b4xY4-1 =
= 34x = 81
=
B2= b2xY2-1 =
= (-3)2x = 9
=
B3= b3xY3-1 =
= (-3)3x =
(-27)=
B4= b4xY4-1 =
= (-3)4x = 81=
Decimals will be used for “b” in the following two examples.
B2=b2×Y2-1 = (0.5)2×Y1
= 0.25×
= 0.25
=
B3=b3×Y3-1 = (0.5)3×Y2
= 0.125
= 0.125
=
B4=b4×Y4-1 = (0.5)4×Y3
= 0.0625×
= 0.0625
=
B2=b2×Y2-1 = (0.25)2×Y1
= 0.0625×
= 0.0625
=
B3=b3×Y3-1 = (0.25)3×Y2
= 0.015625×
= 0.015625
=
B4=b4×Y4-1 = (0.25)4×Y3
= ×
=
=
Therefore we can conclude that the formula Bn=bnYn works. From these examples, we can conclude that this statement works with positive, negative, and decimals for the “b” figure, just as it did for “a” when tested above. There are no limitations to this statement. Therefore, the statement is valid. A also check the answers using the matrix feature on the TI-84 calculator.
Patterns that were observed during this test were that whenever “b” was equal to a positive whole number, the resulting elements in each subsequent matrix were multiplied by six as the exponent of “b” increased by 1. The same pattern happens when “b” is a negative integer, however, the elements alternate positivity and negativity, because raising by a negative integer causes positive and negative answers, depending if it is even or odd. When “b” is a value less than 1 and greater than zero (a decimal or fraction), it causes the elements to become increasing smaller as the “n” value increases. The switching of positive and negative also occurs here as the “n” value increases, just as the negative “b” value. The exception for this is when “a” is set to 0.5, the elements always add up to 0.5, regardless of the “n” value used.
Determining A and B
A and B will be combined in order to create a general statement. Let a=2, and b=3. These will be used in correlation with X and Y respectively.
If A=aX, and B=bY than (A+B)n can also be expressed as (aX + bY)n, This will be used to find the general statement, as well as prove that M = A + B and M2 = A2+ B2.
(A+B) = (aX + bY) = (2 + 3)
= ( + )
=
(A+B)2 = (aX + bY)2 = (2 + 3)2
= ( + )2
= 2
=
(A+B)3 = (aX + bY)3 = (2 + 3)3
= ( + )3
= 3
=
(A+B)4 = (aX + bY)4 = (2 + 3)4
= ( + )4
= 4
=
(A+B)-2 = (aX + bY)-2 = (2 + 3)-2
= ( + )-2
= -2
= Domain error
(A+B)0.5 = (aX + bY)0.5 = (2 + 3)0.5
= ( + )0.5
= 0.5
= Domain error
After observing the examples above, it can be concluded that only whole numbers work. A pattern that was observed is that as the exponent value “n” is increased by one, the sum of the matrix components is 6 times bigger than the sum of the previous matrix components, when “n” is a whole number. Some limitations of this formula are that it does not work with numbers less than 1 and greater than zero, as well as with any number less than zero. This was confirmed by using the matrix feature on the TI-84.
Verifying M
After performing the expressions above, they can now be used to show that M = A + B, M2 = A2+ B2 and M3 = A3+ B3.
Using constants “a” and “b”
M = + =
M2 ==
=
M3=x =
x =
=
Using “a” = 2, and “b” = 3
M = = =
M = A + B = + =
M2 = A2 + B2 = 2 + 2 = 2
= + =
M3 = A3 + B3 = 3 + 3 = 3
= + =
These examples were verified on the TI-84 calculator and are correct.
General Statement of Mn in terms of A and B:
Mn=An+Bn
This is true because if we take Mn=(A+B)n and expand the middle terms are always equal to 0. This is because Matrix A multiplied by matrix B is always equal to 0, no matter what the values of “a” and “b” are. This was confirmed using the TI-84 calculator.
This show that no matter which method is used for finding M, this will always work, and it will always prove. The general statement for M in terms of component matrices A and B is Mn=An+Bn. Since “a” was given the value of 2 throughout this assignment, and “b” was given the value of 3, these will be used in the forming and proving the general statement at first. As it is known that when the constants, “a” and “b”, are multiplied by matrices “X” and “Y” respectively, the expressions A = aX, and B = bY are formed. These will be substituted in place of “A” and “B” to make a final general statement of:
Mn = (aX)n +( bY)n Mn = anXn + bnYn
Therefore, the general statement will be tested, using different values for “a”, “b”, and “n”.
M2 = a2X2 + b2Y2 = 222 + 322
= 4 +9
= +
= =M2
Using “a” = -2, “b” = -3, and “n” = 3:
M3 = a3X3 + b3Y3 = (-2)33 + (-3)33
= -8 + (-27)
= +
= = M3
Using “a” = 2, “b” = -3, and “n” = 3:
M3 = a3X3 + b3Y3 = (2)33 + (-3)33
= 8 + (-27)
= +
= = M3
Using “a” = -2, “b” = 3, and “n” = 3:
M3 = a3X3 + b3Y3 = (2)33 + (3)33
= -8 + (27)
= +
= = M3
Using “a” = 0.5, “b” = .75, and “n” = 2:
M2 = a2X2 + b2Y2 = (0.5)22 + (.75)22
= 0.25 + 0.5625
= +
= = M2
Using “a” = 1, “b” = .75, and “n” = 2:
M2 = a2X2 + b2Y2 = (1)22 + (.75)22
= 1 + 0.5625
= +
= = M2
Using “a” = 2, “b” = 3, and “n” = -2:
M-2 = (aX + bY)-2 = (2 + 3)-2
= ( + )-2
= -2
= = Domain error
Conclusion to the General Statement
After doing all these examples and figuring out the expressions, it can be determine that this general statement works whenever “a” is positive, negative, or a decimal, “b” follows the same rules as “a”, and “n” has to be a whole number in order for the equation to be valid and to work. I have discovered that X multiplied by Y and aX multiplied by bY will always give 0. I have also discovered that equations can be modified and changed to make them easier to solve, and o get a general statement. Throughout the assignment it has been determined that a limitation/restriction that affects any expression or general statement made is that “n” cannot be anything but a whole number. Another limitation is that the matrices that are used in the formula must be similar and symmetrical to the ones in the examples. That means they should potentially be 2 by 2. When “n” is set to a negative value, or is less than 1 but greater than 0, a domain error always occurs, as shown by testing on the TI-84. “n” also cannot be 0 because the answer is always the inverse matrix. It can proven that the statement is logically correct and is accurate by using the previous expressions that were generated. It is known that “X” and “Y” can be multiplied by themselves, and can be combined to form the general statement as part of “A” and “B” respectively. We know that “A” and “B” are formed by combining the matrices “X” and “Y” with any constant. Therefore, it was proven “a” and “b” could be any real number, therefore adding to the conclusion that it is a universal statement. By using any real number in conjunction with matrices “X” and “Y”, “A” and “B” can equal almost anything, and be valid, according to rules above. Further, by knowing that “a” and “b” being combined in a matrix (page 8) equal to the same as the matrices “A” and “B” being combined to make matrix “M” (using the same values for “a” and “b” throughout), it can be confirmed that the general statement, Mn = anXn+bnYn is valid, and can be proved by using other expressions and by using examples, and provides answers that are consistent, that make sense, and that are logical.