- Level: International Baccalaureate
- Subject: Maths
- Word count: 916
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 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 =
This student written piece of work is one of many that can be found in our International Baccalaureate Maths section.
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