Linear algebra nxn matrix, n=14

[concon]

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?

 

[haruspex]

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?

 

[concon]

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.

 

[haruspex]

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?

 

[concon]

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?

 

[haruspex]

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.

 

[concon]

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!

 

[Dick]

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.

 

[concon]

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!) ?

 

[Ray Vickson]

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?

 

[Dick]

so if n is 14, then det(A) = -1 (14!) ?

Why don't you try to use your "relevant equation" to answer that question?

 

[concon]

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