Determinant of a general matrix with variables.

[peripatein]

Hi,

Homework Statement

I was asked to find the determinant of the following two matrices (please see attachment).

Homework Equations

The Attempt at a Solution

I know that the determinant of the matrix on the left is [(-1)^(n-1)]*(n-1), but I have no idea how to formally derive that, nor prove it.
And I have no idea how to find the determinant of the matrix on the right.
Could anyone please guide me through?

 

[aralbrec]

Try row reducing the matrices to upper triangular form before finding the determinant.

 

[peripatein]

I am not sure how to achieve that, and I DO know what an upper triangular form is.

 

[aralbrec]

I am not sure how to achieve that, and I DO know what an upper triangular form is.

I'll do the first row operation for the first matrix:

row2=row2-row3

(determinant doesn't change)

We're trying to get rid of the ones to the left of the main diagonal and that can be done for each row by looking at the row below, except for the last one.

 

[aralbrec]

peripatein, I just worked it out to make sure it was alright. The pattern for the row(n-1)=row(n-1)-row(n) generates -1,1 along the diagonal and zero elsewhere in the row. I set lastrow=lastrow-firstrow to get a 1 in the first column and a -1 in the last. The determinant can be evaluated using the last row and cofactor expansion.

One of the determinants is then zero but the other has a bunch of 1s in the first row. You can get that one into upper triangular form fairly easily with the help of some row exchanges (and the -1 multiplier on the determinant that entails).

 

[peripatein]

Alright, thank you very much!
What about the second matrix, the one on the right, with diagonal 1,2,..,n?

 

[Michael Redei]

Alright, thank you very much!
What about the second matrix, the one on the right, with diagonal 1,2,..,n?

How about subtracting the matrix with all entries equal to n from that second one?

EDIT: Sorry, that was nonsense.

 

[aralbrec]

What about the second matrix, the one on the right, with diagonal 1,2,..,n?

I think the same approach will work. Get rid of the n entries below the main diagonal with the same row subtraction done for the other matrix. The lastrow=lastrow-firstrow will leave a non-zero entry in the first entry around which a cofactor expansion can be done. Then you are left with a matrix that should be fairly easy to get into upper triangular form.

 

[peripatein]

I was unable to attain an upper triangular form in the latter case. May you please assist? We may simplify it and first examine the procedure on a 3x3 matrix.

 

[Michael Redei]

Subtract the last row from each other one.

 

[Dick]

Hey Michael, there is such a thing as giving too much help. I think that's way too much. Just give the hint, ok?

 

[Michael Redei]

I've removed the bulk of my last post.

 

[Dick]

I've removed the bulk of my last post.

Thanks. That's better, right? Let's see where peripatein goes with that.

 

[aralbrec]

Subtract the last row from each other one.

That certainly makes it easier lol.

 

[HallsofIvy]

Hi,

Homework Statement

I was asked to find the determinant of the following two matrices (please see attachment).

Homework Equations

The Attempt at a Solution

I know that the determinant of the matrix on the left is [(-1)^(n-1)]*(n-1), but I have no idea how to formally derive that, nor prove it.
And I have no idea how to find the determinant of the matrix on the right.
Could anyone please guide me through?

Have you at least tried the 2 by 2 and 3 by 3 cases to get an idea of what you should do?

 

[peripatein]

Of course, and for a 3x3 got:
{1-n 0 n-3},{0 2-n n-3},{n n 3}
Which still left me stuck.
How shall I proceed?

 

[Dick]

Of course, and for a 3x3 got:
{1-n 0 n-3},{0 2-n n-3},{n n 3}
Which still left me stuck.
How shall I proceed?

n is the size of the matrix. Here n=3.

 

[peripatein]

This is merely an example, in order to determine the general algorithm.

 

[Dick]

This is merely an example, in order to determine the general algorithm.

Ok, but since n-3=0 your matrix is lower triangular. That's Michael Redei's hint is all about.

 

[peripatein]

Ok, so I got:
n*PI (i-n), where i runs from 1 to n-1.

Should this expression be simplified further/presented differently, or is this indeed how it should be formulated?

 

[Michael Redei]

You can simplify that expression. It will become easier if, instead of multiplying terms (i-n), you multiply (n-i) instead. You'll need some correcting factor though.

 

[peripatein]

[(-1)^(n-1)]n!
Still further?

 

[peripatein]

I meant, does the latter formulation requires further simplification still?

 

[Dick]

I meant, does the latter formulation requires further simplification still?

Looks about as simple as you can get to me.