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# Matrix Binomials. In this Math Internal Assessment we will be dealing with matrices.

Extracts from this document...

Introduction

 Math SL IA

Math Standard Internal Assessment

Matrix Binomials

In this Math Internal Assessment we will be dealing with matrices. A matrix is a rectangular array of numbers arranged in rows and columns. Matrices are used for many things, such as; solving of systems and equations, linear programming, business inventories, Markov chains, strategies in games, economic modelling, graph theory, assignment problems, forestry and fisheries management, cubic spline interpolation, computer graphics, flight simulation, Computer Aided Tomography, Magnetic Resonance Imaging, Fractals, Chaos, Genetics, Cryptography, and the list goes on[1].

Let X =   and Y =  Calculate X2, X3, X4; Y2, Y3, and Y4

X2=  *  = Y2=   *  =

X3=  *  = Y3=  *  =

X4=  *  = Y4=  *  =

To find an expression for Xn and Yn we must test other values for n. These values were calculated using a Texas Instrument TI-84 Plus Graphic Calculator

X7   = Y7   =

X15 = Y15 =

X20 = Y20 =

X50 =Y50 =

It should be noted how the elements Xn are equal to that of the result of 2 raised to a number.

X2 has the element 2 repeated. 2 = 21

X3 has the element 4 repeated. 4 = 22

X4 has the element 8 repeated. 8 =23

X7 has the element 64 repeated. 64 = 26

X15 has the element 16384 repeated. 16384 = 214

X20 has the element 52488 repeated. 52488 = 219

X50 has the element  repeated.  =249

It should also be noted that whatever number X

Middle

=

Or

(X+Y)2=(X+Y)*(X+Y)=X2+XY+YX+Y2=

= = 4(I)

n=3

(X + Y)n = (2I)n

(X + Y)3 = (2I)3

X3+ Y3 = (2I)3

+  = 23

=

Or

(X+Y)3=X3+3X2Y+3XY2+Y3=

Another way to approach this question is to look at how matrix binomials differ from regular binomials. Let’s look at how binomial theorem works, if (a+b)² = a²+2ab+b² then in that case the following should take place (X+Y)2=X2+2XY+Y2. However, when matrix X is multiplied by matrix Y, their product becomes the zero matrix . Anything multiplied by 0 is 0, therefore the equation will simplify to (X+Y)2=X2+Y2. The same principle can be applied to n=3, (X+Y)3=X3+Y3. It should also be noted that when matrix X is added to Y the sum is the identity matrix, I= . The results of (X+Y)2 and (X+Y)3 are all multiples of the identity matrix. We can conclude once again, (X + Y)n = Xn +Yn=2n-1*X + 2n-1*Y= (2I)n

However there are still some limitations to this general formula. Seeing that (X + Y)n = Xn+ Y n , we know from the general expression for Xnor Y n

Conclusion

a and b was less  than 0. It should be noted that since the forumula  Mn = An+Bn containsAnand Bn the same limitation exist as before where n Z+. This conclusion on limitations should be followed up with a,b

In conclusion, by considering higher powers of the matrices of X and Y, patterns were observed and this led to the formulation of a general formula for (X + Y)nXn and Y n . These formulae were then tested for their validity by substituting several different numbers for n. It can also be concluded that he matrices A and B are actually multiples of the matrices X and Y. By testing different powers of the matrices A and B patterns were observed and this led to the formulation of a general formula for(A + B)n = An+ B n. This formula was then tested for its validity by substituting different numbers for a,b, and n. Then the matrix M was then introduced. By proving that  M = A+B, we were able to generate a general formula for Mn which was expressed in the form of aX and bY. The general formula for Mn has the following limitations: a and band  n Z+.

[1] (John Owen, Robert Haese, Sandra Haese, Mark Bruce 2004)

This student written piece of work is one of many that can be found in our International Baccalaureate Maths section.

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