Maths - Matrices?

Question

Matrices-calc project please please help?
any form of help is appreciated!
note: i dont know how to make matrices on computer so each of these is a 2 by 2 matrix...just in one set of brackets. numbers before comma are top row, after comma are bottom row

1. Consider matrix M= [2 0 , 0 2] calculate M^n for n= 2, 3, 4, 5, 10, 20, 50. Describe in workds any pattern observed. use pattern to find general expression for matrix M^n in terms of n

2. Consider matrices P=[3 1, 1 3] and S= [4 2, 2 4]
P^2=[3 1, 1 3]^2 =[10 6, 6 10] = 2[5 3, 3 5] ; S^2= same type of thing as P
Calculate P^n and S^n for other values of n and describe any observed patterns

3. Consider [k+1 k-1, k-1 k+1]
steps 1 and 2 contain examples of these matrices for k=1, 2, and 3. consider other values of k and describe any patterns you observe. generalize these results in terms of k and n

4. use technology to investigate what happens when further values of k & n. state scope of limitations of k and n

Answer 1

Hi,

1. Consider matrix M= [2 0 , 0 2] calculate M^n for n= 2, 3, 4, 5, 10, 20, 50. Describe in words any pattern observed. use pattern to find general expression for matrix M^n in terms of n

[2...0] is matrix M
[0...2]

[4...0] is matrix M²
[0...4]

[8...0] is matrix M³
[0...8]

[16...0] is matrix M^4
[0...16]

[32...0] is matrix M^5
[0...32]

[1024...0] is matrix M^10
[0...1024]

[1048576...0] is matrix M^20
[0...1048576]

M^50 follows the same pattern, but is even larger.

General pattern:

[m^n....0] = M^n
[0....m^n]


2. Consider matrices P=[3 1, 1 3] and S= [4 2, 2 4]
P^2=[3 1, 1 3]^2 =[10 6, 6 10] = 2[5 3, 3 5] ; S^2= same type of thing as P
Calculate P^n and S^n for other values of n and describe any observed patterns

[3...1] is matrix P
[1...3]

[10...6] or 2[5....3] is matrix P²
[6...10]......[3....5]

[36...28] or 4[9....7] is matrix P³
[28...36]......[7....9]

[136...120] or 8[17....15] is matrix P^4
[120...136]......[15....17]

[528...496] or 16[33....31] is matrix P^5
[496...528]......[31....33]

2^(n-1)[2^n.+.1...2^n.-.1] is matrix P^n
............[2^n.-.1..2^n.+.1]

Now matrix S
[4...2] or 2[2....1] is matrix S
[2...4].......[1....2]

[20...16] or 4[5....4] is matrix S²
[16...20].......[4....5]

[112...104] or 8[14....13] is matrix S³
[104...112].......[13....14]

[656...640] or 16[41....40] is matrix S^4
[640...656].......[40....41]

[3904...3872] or 32[122....121] is matrix S^5
[3872...3904].......[121....122]

2^n[(3^n)/2 + .5....(3^n)/2 - .5] is matrix S^n
....[(3^n)/2 - .5....(3^n)/2 + .5]



3. Consider [k+1 k-1, k-1 k+1]
steps 1 and 2 contain examples of these matrices for k=1, 2, and 3. consider other values of k and describe any patterns you observe. generalize these results in terms of k and n

If the matrix A is

[k+1..k-1]
[k-1..k+1], then A² = [A][A] =

[(k+1)².+.((k-1)²...2(k+1)(k-1)]
[2(k+1)(k-1)...(k+1)².+.((k-1)²] =

[2k²+2...2k²-2]
[2k²-2...2k²+2] =

2*[k²+1...k²-1]
....[k²-1...k²+1]

If we ignore the constant 2 for now, lets multiply

[k²+1...k²-1].*.[k+1..k-1]
[k²-1...k²+1].*.[k-1..k+1] =

[(k²+1)(k+1)...(k²-1)(k-1)]
[(k²-1)(k+1)...(k²+1)(k-1)] =

[2k³+2...2k³-2]
[2k³-2...2k³+2] =

2[k³+1...k³-1]
...[k³-1...k³+1] If we multiply this by the earlier constant of 2, then A³ is:

4[k³+1...k³-1]
...[k³-1...k³+1]

Continuing this pattern A^4 is

8[k^4+1...k^4-1]
...[k^4-1...k^4+1]

So, in general A^n =

2^(n-1)[k^(n)+1.k^(n)-1]
..........[k^(n)-1.k^(n)+1]

4. use technology to investigate what happens when further values of k & n. state scope of limitations of k and n

Eventually the numbers are so large, they are written in scientific notation. This makes the numbers difficult to work with, but there is no reason the patterns would not continue.

I hope that helps!! :-)

(Sorry it took so long... grandkids come first)

Answer 2

1. This is easy if you observe that M = 2I, where I = [1 0, 0 1], the identity matrix. Therefore M^2 = (2I)(2I) = (2^2)I, ..., M^n = (2^n)I.

3. We consider powers of A = [k+1 k-1, k-1 k+1]. For A2 we have [u v, v u], where u = (k+1)^2+(k-1)^2 = 2(k^2 + 1) and v = (k^2-1) + (k^2-1) = 2(k^2 - 1). Therefore, A^2 = 2[k^2+1 k^2-1,k^2-1 k^2+1].

For A^3 we get 4[k^3+1 k^3-1, k^3-1 k^3+1]. Continuing by mathematical induction, we find A^n = (2^(n-1))[k^n+1 k^n-1, k^n-1 k^n+1].

By choosing appropriate values for k, we have formulas for the powers of P and S of part 2.

Answer 3

I'll answer 1 for a start since this will take a while to document for all 4 questions.

1. M= [2 0, 0 2]
now we know I = [1 0, 0 1] so
M= 2I
now M^2 = M*M = 2I * 2I
we know that I*I = I => this means

M^2 = 4 I

for M^3 = M^2 *M or 4I * 2I = 8I

so the pattern we are seeing is that M^n = 2^n*I or
M^n = [2^n 0, 0 2^n]

Answer 4

AB = (3x4) X (4x3 )= 3 x 3 /// use outside numbers to work out order BA = (4x3) X (3 x 4 ) = 4 x 4 //// use outside numbers to work out order

Answer 5

okayyy so Pi R Squared. youre brilliantt.. please continue? slashh finish after church :) your insight has helped us a a lot.

this is what we have so far:
for 4: the limitations HINT: try testing fractions and negatives (eigenvalues...)
and what is everyone saying for 5? are we expected to restate our general expression?

Answer 6

well remeber when subtracting two matrices and there is a negative number in one, that two negatives make a positive *hint* *hint*