Matrix Binomials

Using my GDC I found

(X + Y)1 =                                         (X + Y)3 =

(X + Y)2 =                                 (X + Y)4 =

When I finished with these calculations, I saw another pattern that was developed. Seeing this pattern I formed an expression to solve for the matrix (X + Y)n. For the numbers of a and d the expression is 2n and for c and d it is 0. As a matrix the expression is (X + Y)n =  .

To make sure all my expressions worked correctly, I calculated each matrix to the fifth power which was the next integer power in the sequence.

X5 =  2(5-1)  2(5-1)                Y5 =  2(5-1)        -(2(5-1))                (X + Y)5 =  25    0                

         2(5-1)   2(5-1)                                                -(2(5-1))  2(5-1)                                0     25

=  24   24                         =  24       -(24)                        =  

    24   24                                                         -( 24)   24

=                         =                         

After these calculations rose to the fifth power, the pattern still continued for each matrix. Letting A = aX and B = bY, where a and b are constants; I used different values of a and b to calculate A2, A3, A4; B2, B3, B4. For the value of a, I used the number 2 and the value of b, I used 3. With these values I used my GDC to solve for A and B.

A =                                                  B =

A2 =                                                  B2 =

A3 =                                           B3 =

A4 =                                                           B4 =

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 =                                         (A + B)3 =

(A + B)2 =                                 (A + B)4 =                 

         

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 =                        Using GDC (A + B)4 =

Considering that M = , 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 = through M = A + B.

M = aX + bY

 = a + b

 = +

 =

M = A+ B

We can also prove that M2=A2+B2 to later help find a general statement for Mn, in terms of aX and bY.

M2=A2+B2

= (aX)2 + (bX)2

= a2+b2

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

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

2b2    -2b2

-2b2    2b2

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

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

Hence,  the general statement that expresses Mn in terms of aX and bY can be expressed as:

Mn = (aX)n + (bY)n

To test the validity of my general statement by using different values of a, b, and n.

First I will use the general statement Mn = (aX)n + (bY)n and verify it by using the

                                

  n

expression Mn = 

Let a = -1, b = 0.5, and n = 2

 

By finding if there are any limitations within this expression: Mn = (aX)n + (bY)n , I am going to change the constants ‘a’ and ‘b’ as well the power raised to ‘n’ into , decimals, fractions and negative exponents.

 

The first example I am going to prove that this general statement has some limitations:

Let a=0.3, b=0.32, and n=-3

M-3 = (0.3X)-3 + (0.32Y)-3

                     

       -3                                -3

M-3 =        +  

As I plug this equation into my calculator it comes up with: (error domain). As a result this equation cannot be solved because the matrix cannot be raised to a negative power. Therefore the limitation in this expression is that ‘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:

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

M3 = (0X)3 + (0Y)3

                         

M-3 =        +  

=

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 = , Y =                                   

(aX)n + (bY)n =  

a2+b2  =  

2a2   2a2 +  2b2    -2b2

2a2   2a2      -2b2    2b2   =

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 = 

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