Linear algebra nxn matrix, n=14

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Homework Help Overview

The discussion revolves around computing the determinants of two specific nxn matrices, A and B, using cofactor expansion. The matrices are defined with particular patterns based on the variable n, which is specified as 14 in this context.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants explore the properties of the matrices A and B, questioning whether they are diagonal and how to compute their determinants. There is a focus on understanding the implications of n being 14 and whether it represents a 14x14 matrix. Some participants attempt to derive patterns from smaller values of n to infer the determinant for n=14.

Discussion Status

The discussion is active with various interpretations being explored. Some participants suggest that the determinant may follow a factorial pattern, while others emphasize the need to consider the sign of the determinant based on the matrix structure. There is no explicit consensus on the final determinant value for n=14, and participants are encouraged to justify their reasoning.

Contextual Notes

Participants express uncertainty regarding the correct interpretation of the matrix forms and the calculation of determinants, particularly in relation to the signs and patterns observed in smaller matrices. There is a concern about the implications of having only one attempt left for the homework problem.

concon
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Homework Statement


Use cofactor expansion to compute determinants of nxn matrices
A= (aij)=
[0 0 ... 0 1
0 0 ... 2 0
....
0 (n-1) 0 ... 0
n 0 0 ... 0]

B=(bij)=
[ 0 1 0 ... 0
0 0 2 ... 0
....
0 0 0 ... (n-1)
n 0 0 ... 0]


Homework Equations



det(A) = (aij)(-1^(i+j))det(aij)

The Attempt at a Solution


I though A was diagonal matrix so I tried (n)(n-1)2*1=364
Then I realized to include ... so I thought det(A) = 0
Both were wrong.
Not sure how to solve either det(A) or det(B) when n=14
Does than mean its just n=14 or is it 14x14 matrix?
 
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concon said:
I though A was diagonal matrix so I tried (n)(n-1)2*1=364
You are on the right track, but you forgot one small thing. Try n = 2.
Not sure how to solve either det(A) or det(B) when n=14
Does than mean its just n=14 or is it 14x14 matrix?
How can it be the one without also being the other?
 
haruspex said:
You are on the right track, but you forgot one small thing. Try n = 2.

How can it be the one without also being the other?

1. if n= 2 then det(A) = 0
2. Well if that is true then det(A) while n=14 should be zero, but I already tried that and the website for my homework says that is wrong.
 
concon said:
1. if n= 2 then det(A) = 0
No it isn't. What does A look like with n=2? How do you calculate the det of a 2x2 matrix?
 
haruspex said:
No it isn't. What does A look like with n=2? How do you calculate the det of a 2x2 matrix?

Well if n=2 (meaning a 2x2 matrix)
A=
[ 0 0
0 0]
and det(A) = ad-bc= 0*0-0*0=0 is this correct? Can you please explain how to solve with n=14?
 
concon said:
Well if n=2 (meaning a 2x2 matrix)
A=
[ 0 0
0 0]
and det(A) = ad-bc= 0*0-0*0=0 is this correct? Can you please explain how to solve with n=14?
No, you're misinterpreting the form of the matrix. For every n, the top row ends with a 1; for every n > 1, the second row ends with 2 0; the next, for n > 2, with 3 0 0; etc.
 
haruspex said:
No, you're misinterpreting the form of the matrix. For every n, the top row ends with a 1; for every n > 1, the second row ends with 2 0; the next, for n > 2, with 3 0 0; etc.
So det(A) when n=14 is 14 factorial? I saw that pattern now and it looks like a diagonal matrix so det(A) should be product of the main diagonal entries.

Or, if n=2 then det is is 2. If n=3 then det is 3*2. And if n=4 then det is 4*3*2?
So if n=14, then det is 14!
 
concon said:
So det(A) when n=14 is 14 factorial? I saw that pattern now and it looks like a diagonal matrix so det(A) should be product of the main diagonal entries.

Or, if n=2 then det is is 2. If n=3 then det is 3*2. And if n=4 then det is 4*3*2?
So if n=14, then det is 14!

No, if n=2 det(A)=(-2). If n=3 det(A) is -3*2=(-6). If n=4 det(A) is +24. You have to explain the sign as well as the magnitude.
 
Dick said:
No, if n=2 det(A)=(-2). If n=3 det(A) is -3*2=(-6). If n=4 det(A) is +24. You have to explain the sign as well as the magnitude.

so if n is 14, then det(A) = -1 (14!) ?
 
  • #10
concon said:
so if n is 14, then det(A) = -1 (14!) ?

What do *you* think? Of course, all answers must have some logical basis; what is yours?
 
  • #11
concon said:
so if n is 14, then det(A) = -1 (14!) ?

Why don't you try to use your "relevant equation" to answer that question?
 
  • #12
Ray Vickson said:
What do *you* think? Of course, all answers must have some logical basis; what is yours?
I think that the answer is -1*14!.
Logical basis is the pattern that I can see from small n, that pattern goes 2*3*4*...*n with an alternating sign. Since n=14 occurs at row 14 column 1, sign is negative since -1^(14+1) is -1.
Can you just give me a straight response, is this correct or not? I have only one attempt left for this problem and do not want to waste it. thanks
 

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