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Type 1 Portfolio: Matrix Binomials

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Type 1 Portfolio: Matrix Binomials

Hudson Liao


I was given the expression X = image00.pngimage00.png and Y = image73.pngimage73.png, 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= image105.pngimage105.png

X3 or X2 * X1 = image117.pngimage117.pngimage129.pngimage129.png

X4 or X3 * X1 = image01.pngimage01.pngimage15.pngimage15.png

Y2 = image27.pngimage27.png

Y3 or Y2*Y1 = image43.pngimage43.png

Y4 or Y3+ * Y1 = image61.pngimage61.png


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.


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.






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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 = image116.pngimage116.png

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

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‘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  image60.pngimage60.pngis 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 fromimage64.pngimage64.png  is  image65.pngimage65.png This general statement should equal to image66.pngimage66.png.

To prove that this general statement equals to image66.pngimage66.png, I am going to expand the equation image53.pngimage53.png by using only variables:












Therefore the equation image60.pngimage60.pngequals with image78.pngimage78.png

However, the equation image60.pngimage60.pngwould not work unless it is proven by the binomial theorem.







This calculations tells us that AB must equal to zero for this equation to work image87.pngimage87.png. As said before the only way the equation image53.pngimage53.png works is because image88.pngimage88.png equals to a zero matrix: image89.pngimage89.png.

In the end, the expression image53.pngimage53.png can be substituted into a different equation where (aX)n+(bY)n can be replaced as image90.pngimage90.png.



...read more.

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