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# Maths Portfolio Matrix Binomials

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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  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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