Pivotal Condensation for determining Determinant

courtrigrad · Oct 20, 2006

If A =\left(\begin{array}{ccc} 1 & 2 & 3 \\ -1 & 2 & 1 \\ 4 & 1 & -1 \\ \end{array}\right)

and we want to zero the entries in the second row of A (i.e. make it 0 0 1 ).

How do we get A =\left(\begin{array}{ccc} 4 & -4 & 3 \\ 0 & 0 & 1 \\ 3 & 3 & -1 \\ \end{array}\right)?
I have tried using elementary row operations:

A =\left(\begin{array}{ccc} 1 & 2 & 3 \\ -1 & 2 & 1 \\ 4 & 1 & -1 \\ \end{array}\right)

A =\left(\begin{array}{ccc} 1 & 2 & 3 \\ 0 & 4 & 4 \\ 4 & 1 & -1 \\ \end{array}\right)

But then it just gets more complicated.

Any help is appreciated.

Thanks

radou · Oct 20, 2006

The fact that scalar multiplication of a row by a constant multiplies the determinant by that constant might be useful.

courtrigrad · Oct 20, 2006

ah ok. thanks for your help. But you don't know the determinant. Thats what you are trying to find.

courtrigrad · Oct 21, 2006

Does A =\left(\begin{array}{ccc} 4 & -4 & 3 \\ 0 & 0 & 1 \\ 3 & 3 & -1 \\ \end{array}\right)

follow from A =\left(\begin{array}{ccc} -4 & -4 & 0 \\ -1 & 2 & 1 \\ 3 & 3 & 0 \\ \end{array}\right)

Because I used a_{23} as the pivot.

StatusX · Oct 21, 2006

In what sense do you mean "="? They're obviously not the same matrix.

courtrigrad · Oct 21, 2006

The determinants of those matrices are equal.

StatusX · Oct 21, 2006

Be more careful with your notation. Are you trying to show that all the matrices you've called "A" have the same determinant?

HallsofIvy · Oct 21, 2006

Then please don't write "= ".

Yes, it is certainly true that
A =\left(\begin{array}{ccc} 1 & 2 & 3 \\ -1 & 2 & 1 \\ 4 & 1 & -1 \\ \end{array}\right)

gives immediately
\left(\begin{array}{ccc} 1 & 2 & 3 \\ 0 & 4 & 4 \\ 4 & 1 & -1 \\ \end{array}\right)

By just adding the first row to the second row.
Now subtract 4 times the first row from the third:
\left(\begin{array}{ccc} 1 & 2 & 3 \\ 0 & 4 & 4 \\ 0 & -7 & -13 \\ \end{array}\right)
The only "complicated" part is that now we will have to add 7/4 the second row to the third". -13+ (7/4)(4)= -13+ 7= -6

What is the reduced matrix now and what is its determinant?

(I don't know how you would get
\left(\begin{array}{ccc} 4 & -4 & 3 \\ 0 & 0 & 1 \\ 3 & 3 & -1 \\ \end{array}\right)
I didn't try to get that since I can't see why you would want to!)

radou · Oct 21, 2006

Since you're insisting on notation, you should use '... \left| ... \right| ...' in your source. The determinant of a matrix A can't equal the matrix A. But you made a typo, probably.

quasar987 · Oct 21, 2006

\det{A} = \left| \begin{array} {ccc} 4 & -4 & 3 \\ 0 & 0 & 1 \\ 3 & 3 & -1 \\ \end{array} \right|

look at the code courtri

courtrigrad · Oct 21, 2006

Yeah that's what I did. Let me write in better notation:

A =\begin{bmatrix} 1 & 2 & 3 \\ -1 & 2 & 1 \\ 4 & 1 & -1 \\ \end{bmatrix}

how do we get

\det{A} = \left| \begin{array} {ccc} 4 & -4 & 3 \\ 0 & 0 & 1 \\ 3 & 3 & -1 \\ \end{array} \right|
If

A =\begin{bmatrix} 1 & 2 & 3 \\ -1 & 2 & 1 \\ 4 & 1 & -1 \\ \end{bmatrix}

I see how we get

\det{A} = \left| \begin{array} {ccc} 4 & -4 & 0 \\ -1 & 2 & 1 \\ 3 & 3 & 0 \\ \end{array} \right|

The difference between the two is that in the first determinant we zeroed the 3rd column, while in the second one we zeroed the second row (i.e. a_{21} is the pivot)

radou · Oct 21, 2006

courtrigrad said:

Yeah that's what I did. Let me write in better notation:

A =\begin{bmatrix} 1 & 2 & 3 \\ -1 & 2 & 1 \\ 4 & 1 & -1 \\ \end{bmatrix}

how do we get

\det{A} = \left| \begin{array} {ccc} 4 & -4 & 3 \\ 0 & 0 & 1 \\ 3 & 3 & -1 \\ \end{array} \right|

What you wrote implies \begin{bmatrix}1 & 2 & 3 \\ -1 & 2 & 1 \\ 4 & 1 & -1 \\ \end{bmatrix}=\begin{bmatrix}4 & -4 & 3 \\ 0 & 0 & 1 \\ 3 & 3 & -1 \\ \end{bmatrix}. You might rather want to call the matrices on which elementary transformations have been applied as A', A'' etc. Further on, if you wish make the matrix A such that it contains '0, 0, something' in a row, you should look at column operations and how they affect the determinant.

HallsofIvy · Oct 21, 2006

courtrigrad said:

Yeah that's what I did. Let me write in better notation:

A =\begin{bmatrix} 1 & 2 & 3 \\ -1 & 2 & 1 \\ 4 & 1 & -1 \\ \end{bmatrix}

how do we get

\det{A} = \left| \begin{array} {ccc} 4 & -4 & 3 \\ 0 & 0 & 1 \\ 3 & 3 & -1 \\ \end{array} \right|
If

A =\begin{bmatrix} 1 & 2 & 3 \\ -1 & 2 & 1 \\ 4 & 1 & -1 \\ \end{bmatrix}

I see how we get

\det{A} = \left| \begin{array} {ccc} 4 & -4 & 0 \\ -1 & 2 & 1 \\ 3 & 3 & 0 \\ \end{array} \right|

The difference between the two is that in the first determinant we zeroed the 3rd column, while in the second one we zeroed the second row (i.e. a_{21} is the pivot)

WHY "zero the 3rd column"? The point is to get the "echelon" form with zeroes below the main diagonal. Courtrigrad, you said before that you had got to
\left(\begin{array}{ccc} 1 & 2 & 3 \\ 0 & 4 & 4 \\ 0 & -7 & -13 \\ \end{array}\right)

and then had added 7/4 the second row to the third. What did you get?
I pointed out before that "-13+ (7/4)(4)= -13+ 7= -6" and that's the only calculation you need to do. Doesn't that give you the echelon form? Can't you read off the determinant from that?

courtrigrad · Oct 21, 2006

Yes, but that is using the "triangular method" to find the determinant. I am supposed to use pivotal condensation. And this is an example from a book. I already know what the determinant is. Just trying to see how the book came to that conclusion in the example.

radou · Oct 21, 2006

courtrigrad said:

Yes, but that is using the "triangular method" to find the determinant. I am supposed to use pivotal condensation. And this is an example from a book. I already know what the determinant is. Just trying to see how the book came to that conclusion in the example.

'...Important properties of the determinant include the following, which include invariance under elementary row and column operations.

1. Switching two rows or columns changes the sign.

2. Scalars can be factored out from rows and columns.

3. Multiples of rows and columns can be added together without changing the determinant's value.

4. Scalar multiplication of a row by a constant c multiplies the determinant by c.

5. A determinant with a row or column of zeros has value 0.

6. Any determinant with two rows or columns equal has value 0.
...' (quote from Mathworld)

Using these properties should help.

courtrigrad · Oct 21, 2006

I see what they did. They did column operations instead of row operations, because \det A^{T} = \det A

Pivotal Condensation for determining Determinant

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