# Determinate of a Matrix

#### agary12

1. The problem statement, all variables and given/known data

In my other post I figured out that when
(2 0)
(0 2) = m

(2^n 0)
(0 2^n) = m^n

Now I have to find the determinate of each matrix M^N with powers from 1-5, 10 and 20.

2. Relevant equations

3. The attempt at a solution
det (M) is 4 which is also 2^2
det (M^2) is 16 which is also 2^4
det (M^3) is 64 which is also 2^6
det (M^4) is 256 which is also 2^8
det (M^5) is 1024 which is also 2^10
det (M^10) is 1048576 which is also 2^20

another way of writing this would be
det(M) the answer is 4 or 2^(2)
det(M^2) the answer is 16 or 2^(2+2)
det(M^3) the answer is 64 or 2^(2+4)
det(M^4) the answer is 256 or 2^(2+6)
det(M^5) the answer is 1024 or 2^(2+8)
det(M^10) the answer is 1048576 or 2^(2+18)

So basically it starts with det (M) being 4 and then each time you increase n by one the power of the answer goes up by two? How do I "generalize the pattern into an expression for det(M^n) in terms of n?

#### tiny-tim

Homework Helper
Hi agary12! Thanks for the PM det (M) is 4 which is also 2^2
det (M^2) is 16 which is also 2^4
det (M^3) is 64 which is also 2^6
det (M^4) is 256 which is also 2^8
det (M^5) is 1024 which is also 2^10
det (M^10) is 1048576 which is also 2^20

another way of writing this would be
det(M) the answer is 4 or 2^(2)
det(M^2) the answer is 16 or 2^(2+2)
det(M^3) the answer is 64 or 2^(2+4)
det(M^4) the answer is 256 or 2^(2+6)
det(M^5) the answer is 1024 or 2^(2+8)
det(M^10) the answer is 1048576 or 2^(2+18)

So basically it starts with det (M) being 4 and then each time you increase n by one the power of the answer goes up by two? How do I "generalize the pattern into an expression for det(M^n) in terms of n?
oooh … stop treating it like a "brainteaser" problem …

you have the pattern …
In my other post I figured out that

(2^n 0)
(0 2^n) = m^n
… so what is the determinant
|2^n 0|
|0 2^n| ? #### agary12

|2^n 0|
|0 2^n|

So the determinate for this is
(2^n) * (2^N) - 0 = 2^(n+n)

is the pattern

|2^(n+n) 0|
|0 2^(n+n)|

#### tiny-tim

Homework Helper
simplify!

|2^n 0|
|0 2^n|

So the determinate for this is
(2^n) * (2^N) - 0 = 2^(n+n)
grrr! why do you do this?

you had 0n before, instead of just 0, and now you have n+n …

it's 2n !!!

apart from that, yes, fine, the determinant is 22n ! #### agary12

So what would the final det(m^n) generalization look like?

This is what I was trying to say:
(2^n 0^n)
(0^n 2^n)
The above matrix shows the expression for the matrix M in terms of n. Now I was saying that to find the determinate of that expression you have to do this:
(2^n * 2^n) - (0^n * 0^n) = det(m^n)
So first I multiply the 2^n by the other 2^n which is (2*2)^(n+n). The 0*0=0 so the det(M^n) would equal 4^2n (whoops, I forgot to multiply the 2 by 2 and just added the exponents).

So:
det(m^n)= 4^2n

Sorry that I keep doing weird things with the math, I'm just used to them trying to trick me with problems.

#### tiny-tim

Homework Helper
So:
det(m^n)= 4^2n
oooh, you made it so complicated!

det(mn)= 22n = 4n
Sorry that I keep doing weird things with the math, I'm just used to them trying to trick me with problems.
so you like to get in first? #### agary12

Haha, I didn't even realize I did that first thing you mentioned. Once again, I appreciate your time and effort to help me.

If I'm trying to do the same things I've done on the last 2 problems with a new matrix it changes the pattern and makes the "patters a bit more complex"

They give me
(3 1)
(1 3)

(3 1)^2 (10 6)
(1 3) = (6 10)

(3 1)^3 (36 28)
(1 3) = (28 36)

(3 1)^4 (136 120)
(1 3) = (120 136)

(3 1)^5 (528 496)
(1 3) = (496 528)

(3 1)^10 (524800 523766)
(1 3) = (523766 524800)

(3 1)^20 (5.497563382*10^11 5.497552896*10^11)
(1 3) = (5.497552896*10^11 5.497563382*10^11)

(3 1)^50 (6.338253001 * 10^29 6.338253001 * 10^29)
(1 3) = (6.338253001 * 10^29 6.338253001 * 10^29)

I don't think I can do what I did last time, that is:
(3^n 1^n)
(1^n 3^n)

Can you help me discover the pattern?

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#### tiny-tim

Homework Helper
ok, take the first five …

3 10 36 136 528 …

factor them … what do you get? #### agary12

3 - 1*3
10 - 2*5
36 - 2*2*3*3
136 - 2*2*17
528 - 2*2*2 *2*2*11

This is what you meant when you said to factor them right?

3 - (1*3)
10 - (2*5)
36 - (2*18), (6*6), (3*12), (4*9)
136 - (2*68), (4 * 34), (8 * 17)
528 - (2*264), (3*176), (4*133), (6* 88), (8*66) (11*48) (12*48), (16 *33) (22*24)

I don't really see any patters yet, did I do something incorrectly?

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#### agary12

My friend said that this was helpful, but it seems fairly complicated to me. Do you know of any other ways to do this or is this the only way to come to the right conclusion?

First, lets call any such matrix as
(a,b)
(b,a)
to generalize the numbers. Then, every power n of such a matrix will always have a form,
(X,Y)
(Y,X)
where the values of X,Y will be different for each power, and also show a pattern that I will explain in a moment.
Also, X+Y = (a+b)^n will always hold. This means that in case of the example you have given, any power of P will be of the form
(X,Y)
(Y,X)
and X+Y will be a power of 4 (as your a,b are 3,1).

Secondly, the element X will be a^n, and all those terms of the binomial expansion of (a+b)^n, which are contain the element a raised to the power n, n-2, n-4, n-6 etc. till either a^1, or a^0 depending on n being odd, or even.

Similarly, the element Y will contain all the alternating powers of a, commencing from n-1, till power 1 or 0 depending on n being odd or even.

So, for any n, the term X is a^n+C(n,2)a^(n-2)b^2+C(n,4)a^(n-4)b^4+C(... and the term Y is C(n,1)a^(n-1)b+C(n,3)a^(n-3)b^3...

or in simple English, the terms are sums of alternating terms of Binomial expansion of (a+b)^n, expressed as {(X,Y),(Y,X)}.

An example is say,
p=
(a,b)
(b,a)
Then p^5 will be of the form
(X,Y)
(Y,X)

where,

X= a^5+C(5,2)a^3b^2+C(5,4)ab^4 = a^5+10a^3b^2+5ab^4
Y= C(5,1)a^4b+C(5,3)a^2b^3+b^5 = 5a^4b+10a^2b^3+b^5

If we take specific example of P=
(3,1)
(1,3)
then,
P^5=
(528,496)
(496,528), as,
3^5+10.3^3+5.3=528 and
5.3^4+10.3^2+1 = 496

X= a^5+C(5,2)a^3b^2+C(5,4)ab^4 = a^5+10a^3b^2+5ab^4
Y= C(5,1)a^4b+C(5,3)a^2b^3+b^5 = 5a^4b+10a^2b^3+b^5
The main things that confuse me are the C(5,2), because I don't know what that means mathematically. The final X= and Y= are also confusing as I'll explain below:

For X: If you have a^5+10a^3b^2+5ab^4 does this mean this:
(a^5) + (10a^3b)^2+(5ab)^4
Where do the 10 and 5 come from? How do they fit in?

For Y:If you have 5a^4b+10a^2b^3+b^5 does it mean this:
(5a^4b)+(10a^2b)^3 + (b^5)
If this is correct for Y then where does the 10 come from?

which then allows you go find the P^5 example:

X= 3^5+10.3^3+5.3=528 and
Y= 5.3^4+10.3^2+1 = 496
Where do the numbers 10.3 and 5.3 come from?

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#### agary12

I was just working over all of my stuff with a friend and he showed something that is extremely simple in comparison to the above method:

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

I'm just wondering, where is the base 2 for everything instead of 3 or 1? How do I explain this?

#### tiny-tim

Homework Helper
3 - 1*3
10 - 2*5
36 - 2*2*3*3
136 - 2*2*17
528 - 2*2*2 *2*2*11
hmm … you're not into this pattern thing, are you?

let's rewrite your figures like this …

3 - 1*3
10 - 2*5
36 - 4*9
136 - 8*17
528 - 16*33 …

now can you see a pattern? If you still don't like that, here's a completely different way …

look at the pairs …

(1 3) (10 26) (36 28) (136 20) (528 496)

try adding or subtracting each pair … what do you get? #### agary12

3 - 1*3
10 - 2*5
36 - 4*9
136 - 8*17
528 - 16*33

In the first column it starts at one, then goes up by one, then up by two then up by four and then up by sixteen. So for the first column in the factor pairs I can see a pattern. The next factor par would be (32 * X). The second numbers are all prime. Now that I look at it again, its the same pattern on the other side as well. 5-3 = (2) 9-5=(4) 17-9=(8) 33-17=(16). So each side of the factor pair increases by 2^2, then 2^4, then 2^6 and so on?
How do you write this mathematically?

#### tiny-tim

Homework Helper
So for the first column in the factor pairs I can see a pattern. The next factor par would be (32 * X). …
Now that I look at it again, its the same pattern on the other side as well. 5-3 = (2) 9-5=(4) 17-9=(8) 33-17=(16). So each side of the factor pair increases by 2^2, then 2^4, then 2^6 and so on?
Good, except it's 22 then 23 then 24 and so on. How do you write this mathematically?
you tell us! #### agary12

Since I figured out how to do it for each component of the factor pair, I just don't know how I would write that for each entry of the matrix.

would it look something like this: 2^(n+1)? For one entry that is. But the problem I have with this is the n variable. Should N represent the increasing exponents or is there a better way to do it? I'm almost certain that each Matrix entry will have 2 terms, but I don't know how to get there.

I believe that the final answer looks like this:

(x y)^n
(y x)
=
(a b)
(b a)

a = [(x+y)^n + (x-y)^n] / 2
b = [(x+y)^n - (x-y)^n] / 2

The problem is that I can't explain it yet.

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#### tiny-tim

Homework Helper
Hint: if you start with something, and add 2 then add 4 then add 8 then add 16 … and so on … and finally add 2n, what have you added altogether? (if you still don't get it, try it in binary)

#### agary12

I don't know what you mean by trying it in binary. You've been adding powers of 2 right? so is it just 2^n?

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