Linear algebra determinant


by fleeceman10
Tags: algebra, determinant, linear
fleeceman10
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#1
Oct24-10, 03:58 PM
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1. The problem statement, all variables and given/known data
I am finding the determinant by first reducing to row echelon form. Please help I have been agonizing over this lol



2. Relevant equations
I have got to this stage and am now going to do 3rd row - 4th row and replace the 4th row as you do to get to row echelon. At this stage the determinant is 0.8
1 -3 -2 1
0 1 2 -1
0 0 1 -0.8
0 0 1 0

Now I obtain this matrix and since it is now a triangular matrix, the determinant is now -0.8. Yet I have simply added a multiple of one row to another which is meant to leave the determinant unchanged. So what on earth have I done wrong? Thanks for answers.
1 | -3 | -2 | 1
0 | 1 | 2 | -1
0 | 0 | 1 | -0.8
0 | 0 | 0 | -0.8


3. The attempt at a solution
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tiny-tim
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Oct24-10, 04:38 PM
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from your other thread on the same problem
Quote Quote by Pagan Harpoon View Post
Doing that leaves the determinant at +0.8.

Det(The matrix)=det(The matrix without the first row and first column)=det(The matrix without the first and second row and the first and second column).

I arrive at that from expansion along the first column twice.

So you arrive at det(The matrix)=(1)(0)-(1)(-0.8)=0.8

Subtracting the row like you said just zaps that first 1 to 0 making it (0)(0)-(1)(-0.8)=0.8.

Check your calculation again.
fleeceman10
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Oct24-10, 04:44 PM
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Look I'm sorry I don't know what he means. Look at the second matrix, the determinant must be -0.8. I think he means replacing the 3rd row but with row echelon form I have to replace the 4th row

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Oct24-10, 04:52 PM
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Linear algebra determinant


Quote Quote by fleeceman10 View Post
Look I'm sorry I don't know what he means. Look at the second matrix, the determinant must be -0.8. I think he means replacing the 3rd row but with row echelon form I have to replace the 4th row
as he says, your second matrix is wrong, you have to subtract -0.8 from 0, which gives +0.8
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Oct24-10, 04:54 PM
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Quote Quote by tiny-tim View Post
as he says, your second matrix is wrong, you have to subtract -0.8 from 0, which gives +0.8
Why is it not possible to subtract 0 from - 0.8 as is consistent with reducing to row echelon form?
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Oct24-10, 05:10 PM
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Quote Quote by fleeceman10 View Post
Why is it not possible to subtract 0 from - 0.8 as is consistent with reducing to row echelon form?
(there is no need to pm anyone, anyone who answers a thread automatically gets email notification of any new posts)

Your bottom right-hand corner (depending on which row you subtract from the other) will be either

1 -0.8
0 0.8

or

0 -0.8
1 0

both of which have determinant 0.8
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Oct24-10, 05:10 PM
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Quote Quote by HallsofIvy View Post
Expanding this by the fourth row, you have
[tex]-1\left|\begin{array}{ccc}1 & -3 & 1 \\ 0 & 1 & -1 \\ 0 & 0 & -0.8\end{array}\right|[/tex]
which is equal to +0.8, NOT -0.8. The non-zero number, 1, in the fourth row is the "3, 4" position: 3rd column, 4th row and so is multiplied by [tex](-1)^{3+ 4}= (-1)^7= -1[/tex].



The correct answer is (1)(1)(1)(-0.8)= -0.8, not 0.8. Your mistake was in evaluating the previous determinant.
I said that at the first stage, the determinant was 0.8 which you have agreed with so I did not make a mistake. Why has the determinant changed from +0.8 to, as you say, -0.8?
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Oct24-10, 05:17 PM
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Quote Quote by tiny-tim View Post
(there is no need to pm anyone, anyone who answers a thread automatically gets email notification of any new posts)
Ok sorry about that

1 -0.8
0 0.8
What if you do 3rd row - 4th row, replacing 4th row. Then you get
1 -0.8
0 -0.8
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Oct24-10, 05:30 PM
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Quote Quote by fleeceman10 View Post
What if you do 3rd row - 4th row, replacing 4th row.
You're not allowed to do that.

You can replace the 4th row by adding the 3rd row to the 4th row.

You can't muck about with the row you're replacing (you've multiplied it by -1 ) before you start adding to or subtracting other rows from it.
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Oct24-10, 05:39 PM
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Quote Quote by tiny-tim View Post
You're not allowed to do that.

You can replace the 4th row by adding the 3rd row to the 4th row.

You can't muck about with the row you're replacing (you've multiplied it by -1 ) before you start adding to or subtracting other rows from it.
Ah I think I see. Your saying subtracting is basically adding -1 lots of that row. Since I have done this the determinant needs to be multiplied by -1. With solving equations, no notice is given to this which I think has generated my confusion.

Is this correct?
Thanks for the help
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Oct24-10, 05:47 PM
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Quote Quote by fleeceman10 View Post
Ah I think I see. Your saying subtracting is basically adding -1 lots of that row.
Yes, of course.

But you're adding (or subtracting) the wrong row.

Look at your book again.
Since I have done this the determinant needs to be multiplied by -1.
i have no idea what you mean by this.
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Oct24-10, 05:57 PM
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Quote Quote by tiny-tim View Post
Yes, of course.

But you're adding (or subtracting) the wrong row.

Look at your book again.


i have no idea what you mean by this.
When calculating the determinant, each time you take a factor out, the determimant of what is left is multiplied by that constant. Not too sure if that's relevant. So I can muck about with the row that i'm not replacing but can't muck about with the one I am. Why?
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Oct25-10, 02:53 AM
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(just got up )
Quote Quote by fleeceman10 View Post
So I can muck about with the row that i'm not replacing but can't muck about with the one I am. Why?
uhhh?

some techniques work and some don't

there are theorems in books which tell you which techniques work and why
you really do need to go back to your book and study this again, from the beginning

you obviously have somehow got completely the wrong idea about this!


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