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

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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:

image00.png and image01.png

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

So,since image00.png and image85.png we will calculate that as image95.png

We know that we can express multiplication of matrices as image106.png,and it goes like

image116.png,so we get that image124.png, thusimage135.png.  Or 2* image30.png=2* X

As we now know this,it is easy to calculate image02.png,becauseimage10.png or 2*X*X,which is 2*2*X or it is equal to 22*X and equal to image15.png.

image21.png is defined as X3*X=22 *X*X=22 *2*X=23 *X and our solution is image21.png=image39.png

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 image45.png, image53.pngand image57.png, 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: image77.png; image78.png and image79.png.

...read more.


This proves that image115.png and since we know that image122.png and image123.png  this means that image125.png. Now we can calculate image126.png






So we can now see an expression for matrix as image133.png and substituting image134.png we can find a general form image136.pngX also same goes for Y as  image137.pngY

Now when we have found image138.png and  image139.png as expressions our next task is to find image82.png,and it is quite simple:

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

X+Y=image140.png or 2* image141.png 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

image82.png=(2*E)image142.png=2image142.png*Eimage142.png=2image142.png*image141.png=image143.png because Eimage142.png=E

Following these general expressions, we may continue to investigate the pattern in the following equalities image28.png and image29.png , when image37.pngand image38.png are constants and they have different values.

It is very important to mention that X*Y=image144.png and it is obvious that X*Y=0 and 0 stands for zero matrix

Now we can calculate image03.png=(aX)image04.png=aimage05.png*X*X=2 *aimage05.png*X

image06.png= image03.png*A=2 *aimage05.png*X*(aX)=image07.png* X2=  image08.png* a *image09.png*X

image11.png= image06.png*A=image07.png*image12.png*X

...read more.


Similarly we can prove first that M2=image50.png+image27.png and we see that this is  image50.png and when we simplified this like

2* image51.png+2 image52.png and we now see that M2=image03.png+image54.png,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)= image03.png+A*B+B*A+image13.png=image03.png+0+0+image13.png=image03.png+image13.png

We also already proved that image55.png=image56.png

 And now if  Mn =(A+B) n  we see that

Mn =image58.png

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



 Through my work,I have shown that image37.png and image38.png belong to the set of rational number image67.png and image68.pngbelongs to set of natural numbers 0.

I have also shown that image69.png and image70.png when multiplied give a zero matrix: image71.png.and that A*B gives zero matrix as an in example with

image72.png in which like image73.png and XY gives zero so that image75.png

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

  Mn =image58.png

...read more.

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