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

Now to continue with our work we need our previous results,because they can help us to find ,  and .Results are shown in table below

Here we can see some characteristics which can help us in our work,and we will stick to them,and applying the rule to the results we will get the following matrix identities:

 ;      ;       ;     .

If we express every coefficient as ,we get  ;   ;   ; .

Next table that can help us is the one  dealing with relation between the powers

It is visible now that that

To prove it I will use an example in which .  Hence

To show that my calculations were correct I will once again use the TI-84 GDC (NOTE: [A] = Y)

Now I will find the formula for the general term  assuming that  and taking like an example that :

 and

Also

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

For B it is the same so

                                     =(bX)=b*X*X=2 *b*X

                                     = *B=2 *b*X*(bX)=* X2=* b*X

                                      = *B=**X

Or we can put some numbers,for example a=4 and b=5 and now we have

Now we can see some characteristics and substitute powers instead of numbers and get

=2**X

 We saw in earlier example that it is the same with B and it simply goes like

=2**X

We proved that X*Y=0 and commutatively it means that A*B=0 so general expression goes like

Now we have to consider matrix M=

We now from previous examples that  and .

Or A=a *  and B=b* so adding them we get

A+B=+ ==M so we proved now that

A+B=M

Or with numbers form previous example,we get that

A= and B=  and From the formula

we can calculate matrix M by substituting values for and :

And now I will calculate the sum of matrices A and B:

This proves the formula M = A + B while  and. I will take another example but this time using negative numbers as constants and to prove whether this formula is correct for those numbers as well. So in this example  and  . Then

Then we calculate  and also ;. Like in the previous example M = A + B which proves this formula also for the negative numbers.

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 ℕ0.

   

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 =