- Level: International Baccalaureate
- Subject: Maths
- Word count: 625
Maths Portfolio Matrix Binomials
Extracts from this document...
Introduction
Maths Portfolio
SL Type 1
Matrix Binomials
In this mathematics portfolio we are instructed to investigate matrix binomials and algebraically find a general statement that combines perfectly with our matrices and equations given.
MATRIX BINOMIALS
Given that:
and 
We calculated 
Therefore:
And
We were then requested to find expressions for 
by considering the integer powers of X and Y:
These expressions were found by observing that the result of 
was always the matrix to the power of n multiplied by 2 to the power of n-1. The sequence of results gives us: 1, 2, 4 and 8, reaffirming our expressions are correct because 
.
Given that:
and 
where a and b are constants
We were asked to find 
using different values of a and b and then find the expressions for 
.
Therefore:
And:
Assuming that a = 2
Assuming that a = 3
Assuming that a = 6
Assuming that a = -4
Middle
As a result we can see that:
And since we know 
Hence: 
To find the expression for 
I did the same as for 
changing the value a to b and X to Y.
For example (assuming b = -1):
As a result we can see that:
And since we know 
Hence: 
To find 
I used the binomial theorem
From the binomial theorem we can see that the values of A and B multiply by each other on every term except the first and the last, where we find
.
However if we multiply matrix A by B we will see that the product will be a zero matrix.
This allows us to cancel every term in which A multiplies B or vice-versa, as the result will be zero.
Conclusion
= 1
...read more.
Assuming: a = -5, b = 2, n = 1
Assuming: a = 1, b = 3, n = 2
Assuming: a = 1/2, b = 2, n = 1
Assuming: a = 2, b = 2, n = -2
Assuming: a = 2, b = 2, n = 0
Not Compatible
After testing the validity of my general statement we can see that the results for both formulas were mostly compatible for all numbers of a, b and positive n, proving the validity of our statement. However when n is zero or a negative integer we find some problems with it. As we can’t power a matrix to a negative number n can’t be a negative number and when 0 we find out all matrices to the power of 0 form identity matrices that when added in the formula 
it differs from our other result which is a identity matrix as well.
To get to this formula algebraically I did:
This can be concluded by:
By knowing that AB =0:
And finally I found the general expression:
This student written piece of work is one of many that can be found in our International Baccalaureate Maths section.
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