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# Matrix Binomials IA

Extracts from this document...

Introduction

Matrix Binomials

Type 1 Internal Assessment

Matrices are rectangular arrays of numbers that are arranged in rows and columns, however the regular rules of algebra do not apply.

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

X2= × =

X3= × ×  =

X4= × ×  ×  =

Because matrices do not follow the algebraic rules of exponents, one can not simply distribute the exponent for each matrix value. Instead the matrix must be multiplied by itself however many times the exponent says. So for example, for X2, the matrix X must by multiplied with itself two times.

The pattern that has emerged is that with every increasing power the matrix value increases with an exponential power of 2.

21 is equal to 2 showed by the matrix . 22 is equal to 4 showed by the matrix  and similarly 23 is equal to 8 represented in the matrix .

Y2= × =

Y3= ×× =

Y4= ××× =

The pattern is very similar to the one above except that all the negatives in the original matrix will also become negatives.

Based on the results above we can conclude that for each exponent value Xn the matrix value will result in Xn-1 and the same goes for Yn resulting in Yn-1.

Middle

The matrix values are all multiplied by 3.

A=5X

A2= 5-5 =

A3= 5                        5 =

A4= 5     5  =

The matrix values are all multiplied by 5.

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

B= -1Y

B2= -1--1  =

B3= -1                -1         =

B4=-1     -1  =

When multiplied by -1, the matrix values become positive but remain the same. The reverse results of when matrix A was multiplied by -1.

B= Y

B2= -½- =

B3= -½                -½         =

B4=-½     -½  =

When multiplied by -1, the matrix values are divided by two (or halved) and all positive and negative the signs are switched in relation with the original matrix.

B= ½Y

B2= ½-½  =

B3= ½                ½         =

B4     ½  =

When multiplied by -1, the matrix values are divided by two (or halved) and all positive and negative the signs remain the same as in the original matrix (key difference when divides by -½).

B= 1Y

B2= 1-1  =

B3= 1                1         =

B4=1     1  =

When multiplied by 1, the matrix values remain negative. The reverse results of when matrix A was multiplied by 1.

B= 3Y

B2= 3-3  =

B3= 3                3         =

B4=3     3  =

When multiplied by 3, the matrix values remain negative.

B= 3Y

B2= 3-3  =

B3= 3                3         =

B4=3     3  =

The matrix values are all multiplied by 3.

Conclusion

= (-9×)3  + (11×)3

=   +

=   +

=

In all these case the general statement is correct and works. However, the scope of the general statement is limited to positive exponents as well as the matrices X and Y. We have only tested a very small sample of the different matrix values and its infinite combinations for the matrices X and Y.

If we were to use different matrix values for X and Y the general statement would not apply:

Let n=3, a=9 and b=11 but let X =  and let Y =

= (9×)3  + (11×)3

=   +

=   +

As one can see, the general statement does not apply when the values for the matrices X and Y are changed and only works correctly when:

X= and Y=  .

To reach my general statement I used my previous findings.

I found that An = a×Xn and that Bn = b×Yn . This in turn allowed me to come to the conclusion that (A+B)n = [(a×X) + (b×Y)]n.

Using the knowledge that M= and that M=A+B and that M2=A2+B2 I could reach the following conclusion:

Mn=An+Bn

Mn=(a×X)n + (b×Y)nsubstituting A for a×X and B for b×Y

This is how I reached by general statement.

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

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