Determinant of Triangular Matrix - What Went Wrong?

In summary, the conversation involves a student seeking help with finding the determinant of a matrix by reducing it to row echelon form. They are struggling with understanding why the determinant changes from +0.8 to -0.8 when subtracting rows in the process. The expert helper explains that the determinant is affected by the multiplication of -1 when subtracting a row and that the incorrect row was being manipulated. The student is advised to refer back to their textbook for a better understanding of the techniques involved in calculating determinants.
  • #1
fleeceman10
14
0

Homework Statement


I am finding the determinant by first reducing to row echelon form. Please help I have been agonizing over this lol



Homework 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


The Attempt at a Solution

 
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  • #2
from your other thread on the same problem …
Pagan Harpoon said:
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.
 
  • #3
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
 
  • #4
fleeceman10 said:
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
 
  • #5
tiny-tim said:
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?
 
Last edited:
  • #6
fleeceman10 said:
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
 
  • #7
HallsofIvy said:
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?
 
  • #8
tiny-tim said:
(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
 
  • #9
fleeceman10 said:
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 :redface:) before you start adding to or subtracting other rows from it.
 
  • #10
tiny-tim said:
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 :redface:) 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
 
  • #11
fleeceman10 said:
Ah I think I see. Your saying subtracting is basically adding -1 lots of that row.

Yes, of course. :smile:

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. :confused:
 
  • #12
tiny-tim said:
Yes, of course. :smile:

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

Look at your book again.


i have no idea what you mean by this. :confused:

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?
 
  • #13
(just got up :zzz: …)
fleeceman10 said:
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? :confused:

some techniques work and some don't …

there are theorems in books which tell you which techniques work and why :smile:

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!​
 

1. What is a determinant in linear algebra?

A determinant is a value that is calculated from the elements of a square matrix. It is used to determine if the matrix is invertible, or if a system of linear equations has a unique solution. It is also used to calculate the area or volume of geometric figures in higher dimensions.

2. How is the determinant of a matrix calculated?

The determinant of a matrix is calculated by using a specific formula that involves the elements of the matrix and their positions. For a 2x2 matrix, the formula is ad - bc, where a, b, c, and d are the elements of the matrix. For larger matrices, the calculation involves finding the determinants of smaller submatrices and using them in the formula.

3. What does a determinant of 0 mean?

If the determinant of a matrix is 0, it means that the matrix is not invertible and does not have a unique solution. This is because the determinant is used to determine the invertibility of a matrix. In other words, a determinant of 0 means that the matrix is singular.

4. Can the determinant of a matrix be negative?

Yes, the determinant of a matrix can be negative. The sign of the determinant depends on the number of row operations that are performed on the matrix. If an odd number of row operations are performed, the determinant will be negative. If an even number of row operations are performed, the determinant will be positive.

5. What are the applications of determinants in real life?

Determinants have various applications in real life, especially in fields such as physics, engineering, and economics. They are used to solve systems of linear equations, calculate areas and volumes, and determine the stability of a system. They are also used in computer graphics to rotate and scale images. In economics, determinants are used to analyze the relationships between different variables.

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