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# In this portfolio,my assignment is to deal with matrix binomials,and to investigate them.

Extracts from this document...

Introduction

PORTFOLIO TYPE I – MATHEMATICAL INVESTIGATION

-MATRIX BINOMIALS-

Student: Domazet Neven SL

In this portfolio,my assignment is to deal with matrix binomials,and to investigate them.

At first,my initial matrices are as follows:

and

Now,my task is to find and calculate following matrices: : X2, X3, X4; Y2,Y3 and Y4

So,since  and  we will calculate that as

We know that we can express multiplication of matrices as ,and it goes like

,so we get that , thus.  Or 2* =2* X

As we now know this,it is easy to calculate ,because or 2*X*X,which is 2*2*X or it is equal to 22*X and equal to .

is defined as X3*X=22 *X*X=22 *2*X=23 *X and our solution is =

Procedure for Y is completely the same as for X,and we do not have to go in details,so to calculate the products of , and , we will use the TI-84 Plus Texas Instruments Graphic Calculator and obtain the results in the following form ( [B] equals Y)

The results are as follows: ;  and .

Middle

This proves that  and since we know that  and   this means that . Now we can calculate

So we can now see an expression for matrix as  and substituting  we can find a general form X also same goes for Y as  Y

Now when we have found  and   as expressions our next task is to find ,and it is quite simple:

Let us first sum up X+Y matrices and it goes like

X+Y= or 2*  and it is now obvious that this is in form of 2*E,as E stands for

Now when we now this it goes like

=(2*E)=2*E=2*= because E=E

Following these general expressions, we may continue to investigate the pattern in the following equalities  and  , when and  are constants and they have different values.

It is very important to mention that X*Y= and it is obvious that X*Y=0 and 0 stands for zero matrix

Now we can calculate =(aX)=a*X*X=2 *a*X

= *A=2 *a*X*(aX)=* X2=  * a **X

= *A=**X

Conclusion

Similarly we can prove first that M2=+ and we see that this is   and when we simplified this like

2* +2  and we now see that M2=+,it  is completely the same for numbers,as we have shown for M=A+B

There is also other way to prove this,as we now that M2 =(A+B)*(A-B),and we earlier proved that A*B=0(zero matrix),now it goes easily like

M2 =(A+B)*(A-B)= +A*B+B*A+=+0+0+=+

We also already proved that =

And now if  Mn =(A+B) n  we see that

Mn =

By taking for example a=2           b=-3           n=2

==*=

=2*(+)=2*=

Through my work,I have shown that  and  belong to the set of rational number  and belongs to set of natural numbers 0.

I have also shown that  and  when multiplied give a zero matrix: .and that A*B gives zero matrix as an in example with

in which like  and XY gives zero so that

Because we know that M=A+B than referring to the law of exponentiation we see that

Mn =

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

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