- Level: International Baccalaureate
- Subject: Maths
- Word count: 1308
Type 1 Portfolio: Matrix Binomials
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Introduction
Type 1 Portfolio: Matrix Binomials |
Hudson Liao 12/7/2008 |
I was given the expression X = and Y = , where I calculate X2,X3,X4;Y2,Y3,Y4. Below I calculated X2 and made my way up to X4, where I also did the same with Y2 to Y4.
X2=
X3 or X2 * X1 =
X4 or X3 * X1 =
Y2 =
Y3 or Y2*Y1 =
Y4 or Y3+ * Y1 =
Xn |
X1 = 1X = 20X |
X2 = 2X = 21X |
X3 = 4X = 22X |
X4 = 8X = 23X |
Now I am going to find and expression for: [Xn, Yn, (X+Y)n], by inputting different ‘n’ values. By doing this I can find a correlation between each variable.
Expression: Xn = 2(n-1) X
This general statement was found by finding a relationship through values from X1 to X4. In the Xn table, a pattern begins to form from 1X, 2X, 4X and 8X. If we simplify these numbers by using a constant value such as 1X = 20X we can find a general statement for this expression.
Yn |
Y1= 1Y = 20Y |
Y2= 2Y = 21Y |
Y3= 4Y = 22Y |
Y4= 8Y = 23Y |
Expression: Yn = 2(n-1) Y
The same method to determine the general statement for the expression Xn = 2(n-1) X was also used for Yn = 2(n-1) Y.
(X+Y)n |
2I |
4I |
8I |
16I |
Middle
8Y
b=2
8Y
32Y
128Y
b=3
18Y
108Y
648Y
b=4
32Y
256Y
2048Y
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=1and 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 4a 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=b
Conclusion
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 .
This student written piece of work is one of many that can be found in our International Baccalaureate Maths section.
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