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Matrix Binomials

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                John Vu


Matrix Binomials

        A matrix is a rectangular array of numbers arranged in rows and columns. Numbers or letters inside the brackets in matrices are called entries. Matrices can be added, subtracted, multiplied, divided, and also raised to a power. A common matrix can look like this image00.png where a, b, c, and d are the entries.

Given the matrices X = image01.png and Y = image12.png  I calculated X2, X3, X4; Y2, Y3, Y4.

Using my GDC (graphic display calculator) I evaluated these matrices.

X2 = image10.png                                                Y2 = image26.png

X3 = image29.png                                                Y3 = image40.png

X4 =  image13.png                                             Y4 = image49.png

After calculating the powers of X and Y, I observed my solutions. Looking at this I saw a trend that emerged as I increased the power of the matrix. The trend was increasing the power of the matrix by one, caused the product to double its previous solution. When X is to the second power, the entries of the solution are all 2’s; when X is to the third power the entries are all 4’s; when X is to the forth power the entries are all 8’s, and so on.

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3 = image16.png

A4 = image17.png                                                          B4 = image18.png

By considering integer powers of these matrices, I came up with an expression to also solve these matrices. For matrix A the expression is An = (an) 2(n-1)        2(n-1) and matrix B the expression Bn = (bn) 2(n-1)       -(2(n-1))                                    2(n-1)  2(n-1)

                          -(2(n-1))  2(n-1)

Using my GDC I solved these matrices.

(A + B)1 = image19.png                                        (A + B)3 = image20.png

(A + B)2 = image21.png                                (A + B)4 = image22.png

After solving these equations, I generated an expression to solve the matrix for

(A + B). The expression to solve the matrix is (A + B)n = (an   Xn) + (bn   Yn). To make sure the expression worked I plugged in number 2 and 4 for the value of n.

(A + B)2 = (22   X2) + (32   Y2)                         (A + B)4 = (24   X4) + (34   Y4)

Using GDC (A + B)2 = image21.png                       Using GDC (A + B)4 = image22.png

Considering that M = image23.png, this expression can also be derived into

M = A + B and M2 = A2 + B2. By proving the expression M = A + B, A and B needs to be substituted for aX and bY as well as keeping the constants as a variable. This will prove the expression M = image23.pngthrough M = A + B.

M = aX + bY

image23.png = a image01.png+ b image12.png

image23.png = image24.png+ image25.png

image23.png = image23.png


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Let a = 0, b = 0, n = 3

M3 = (0X)3 + (0Y)3

M-3 = image44.png       +  image45.png

= image46.png

The limitation for the expression Mn = (aX)n + (bY)n 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:



To conclude this assignment, last I will explain how I arrived at my general statement using an algebraic method.

Let n = 2, a = a, b = b, X = image01.png, Y = image12.png

(aX)n + (bY)n =image23.png

a2image10.png+b2 image26.png =  image23.pngimage23.png

2a2   2a2 +  2b2    -2b2

2a2   2a2      -2b2    2b2   = image28.png

2a2 + 2b2   2a2 – 2b2  =   (a2 + 2ab + b2) + (a2 + 2ab + b2)               (a2 – b2) + (a2 – b2)                    

2a2 – 2b2   2a2 + 2b2                (a2 – b2) + (a2 – b2)                  (a2 + 2ab + b2) + a2 + 2ab + b2)  

2a2 + 2b2   2a2 – 2b2      2a2 + 2b2   2a2 – 2b2

2a2 – 2b2   2a2 + 2b2  =  2a2 – 2b2   2a2 + 2b2

Therefore the equation Mn = (aX)n + (bY)n equals with Mn =image23.png

Arriving at this general statement, it can help solve for the value of M in any equation.

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