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
Assuming that b = -1
Assuming that b = 4
Assuming that b = 5
Assuming that b = 7
To find the expression for I observed that the final results achieved were always the value chosen for a multiplied by the matrix X and the product to the power of n.
For example (assuming a = 2):
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.
Therefore if we cancel these terms we will be only left with.
This means that
Now we were given:
And asked to prove that
So:
A B M
To find, first we need to calculate A², B² and M².
So:
Therefore:
To find the general statement that expresses in terms of aX and bY we first need to continue the sequence and find and .
So:
By analyzing all 4 values for we can see that the result can be put into multiplied by 2a+2b in a sequence.
Example:
Hence we can say that our general statement is:
Testing the validity of my general statement, to do this we had to get the same results for both expressions:
Assuming: a = 2, b = 4, n = 1
Assuming: a = -1, b = -3, n = 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:
