Type 1 Portfolio: Matrix Binomials

Authors Avatar

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.


X3 or X2 * X1 =   

X4 or X3 * X1 =  

Y2 =

Y3 or Y2*Y1 =

Y4 or Y3+ * Y1 =

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.

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.

I am going to determine the expression for (X+Y)n  by letting X =  and Y = . Therefore the expression would look like:

The resultant matrix is  

        I am going to prove that this expression works with (X+Y)n :


(X+Y)n                 n=2

My expression:

(X+Y)2 = 22I =    

With (X+Y)n:

As well, since ‘X’ and ‘Y’ are singular matrices, they cannot be raised to a negative exponent since the determinant is zero.


In this section I am going use another two expression: An = aX and Bn =bY, where ‘a’ and ‘b’ are constants. I am going to use different values for both of the constants ‘a’ and ‘b’ to calculate A2 ,A3 ,A4 ; B2,B3 , B4.

Join now!

First, I am going to find A2 for the expression An = aX and use 4 different constants for ‘a’, where I am going to let X =


n=2     a=1  (without calculator)         n=2    a=2 (with calculator)            n=3    a=2                        n=4    a=2                                        

                                               A2 =                 A2 =            A2 =    
=    ...

This is a preview of the whole essay