P3 = == 22, det (P3) = 512 = 83 (here k is 8)

P4 = == 23, det (P4) = 4096 = 84 (here k is 8×2)

P5 = == 24, det (P5) = 32768 = 85 (here k is 8×4)

Given these matrices, we can notice that when the matrix is simplified by 2n-1 you get a new matrix which the numbers can be found by using the number multiplying it as k in the general formula. A pattern between the matrices’ determinants is noticeable. The determinant of each matrix is 8 powered by the same number you powered the matrix.

If P3 =, then it’s certain that its determinant is going to be 83 which is 512.

To calculate the following matrices you use a GDC. You do the same thing to the resultant matrix of Sn, you divide each number inside the matrix by 2n-1.

S2 = == 2, det (S2) = 144 = 122 (here k is 12-3)

S3 = == 22, det (S3) = 1728 = 123 (here k is (12×2) +3)

S4 = == 23, det (S4) = 20736 = 124 (here k is (12×7)-3)

S5 = == 24, det (S5) = 248832 = 125 (here k is (12×20) +3)

Given these matrices, we notice that when simplifying the matrices we also use 2n-1 as the factor. We can use the general formula of k to find the numbers inside the new matrix and still a pattern between the matrices’ determinants can be noticed. The determinant of each matrix is 12 powered by the same number you powered the matrix.

If S3 =, this matrix’s determinant is 123, which is 1728, this confirms the trend I stated before. The trend in matrix P is the same in matrix S but 4 numbers more. The formula in matrix P is 8n, and 8 + 4 = 12 therefore the formula in matrix S is 12n.

I will make an example for k = 4, so the matrix will be V =.

We can predict according to the determinant’s pattern that the determinants will now be 16n.

V2 = == 2 , det (V2) = 256 = 162 (here k is 16)

V3 = == 22, det (V3) = 4096 = 163 (here k is 16×4)

V4 = == 23, det (V4) = 65536 = 164 (here k is 16×16)

V5 = == 24, det (V5) = 1048576 = 165(here k is 16×64)

These matrices follow the trend, because 12 + 4 = 16 so the determinants of these matrices must be 16n, where n is the power of the matrix. To prove this, select a random matrix such as:

V5 = , according to the pattern the determinant of this matrix should be 16 to the same power in which the matrix is powered to, in this case (5), which then 165 gives 1048576 as your answer.

On the other hand, the pattern to find the numbers inside the new matrix doesn’t exist here when the factor of simplification is 2n+1. This a limitation for this pattern and we cannot continue it.

The pattern for matrices M, P, S and V is given by the general formula.

As the formula for the matrix Mn is, to generalize the results in terms of k and n we can join both formulae to get one general formula for the pattern in matrices M, P, S and V. The new formula will be. We can apply this to P, S and V to prove that the formula is true.

Matrix P occurs when k is 2 so the formula will give us which happens to be the case.

If we then square the matrix, so n = 2 we get:

P2 =

Matrix S occurs when k is 3 so the formula will give us which happens to be true.

If we then square the matrix, so n = 2 we get:

S2 =

Matrix V occurs when k is 4 so the formula will give us which happens to follow the trend.

If we then square the matrix, so n = 2 we get:

V2 =

Furthermore, the determinants in each set of matrices follow a common pattern between them. In the set of matrices Mn, Pn, Sn, Vn the determinants have a pattern in each of them. In the set of matrices Mn, the determinants are always equal to 4n. In Pn, the determinants are given by the formula 8n. In the set of matrices Sn, the determinants can be found by replacing ‘n’ in 12n. Finally in Vn, the determinants are given by the formula 16n.

If you then consider this individual pattern of each set of matrices at a bigger scale you can notice another trend that applies to these set of matrices as a whole.

This pattern is easy to see if we display the information in a sequence.

det (M2) = 42, det (M3) = 43, det (M4) =44, det (M5) = 45 … so det (Mn) = 4n

+4 +4 +4 +4 +4

det (P2) = 82, det (P3) = 83, det (P4) =84, det (P5) = 85 … so det (Pn) = 8n

+4 +4 +4 +4 +4

det (S2) = 122, det (S3) = 123, det (S4) =124, det (S5) = 125 … so det (Sn) = 12n

+4 +4 +4 +4 +4

det (V2) = 162, det (V3) = 163, det (V4) =164, det (V5) = 165 … so det (Vn) = 16n

Every time k increases by one, the matrix’s determinants formula increases by 4.

I also noticed that the letters that represent each matrix increase by 4 as well. This is why I select the letter ‘V’ to represent the matrix formed when k = 4.

To prove the validity of my statements I will consider a further value of k after making a hypothesis about the results of that same value.

For k = 5, the letter should be ‘Y’ according to the letter’s pattern, and according to the formula the matrix should be so you end up with Y = . Now if we power this matrix so as n = 2,3,4,5 we should see the same pattern as the former sets of matrices. The matrices determinants for this set should follow the trend and should be equal to 20n.

Y2 = == 24, det (Y2) = 400 = 202

Y3 = == 25, det (Y3) = 8000 = 203

Y4 = == 26, det (Y4) = 160000 = 204

Y5 = == 27, det (Y5) = 3200000= 205

This further value of k, confirms my previous statements about the generalized pattern of these sets of matrices, it also proves the formula and trend of the determinants in each set of matrices.

Conclusions

A general trend was found in the different sets of matrices provided by the assignment sheet. A formula was created, so that it could be applied to generate more matrices that follow the pattern found. The results were satisfactory due to the few limitations that the trend had. It can be considered a limitation that a matrix when powered to 0 gives you always the identity matrix, so n ≠ 0. Another limitation found is that is useless to power a matrix to 1, because it will remain unchanged. Another fact I found was that when k = negative numbers, the answers are exactly the same as the answers for the positive numbers.