# Pivotal Condensation for determining Determinant

In summary: So they just turned the rows into columns and did column operations. Then they got A =\begin{bmatrix} 4 & -4 & 3 \\ 0 & 0 & 1 \\ 3 & 3 & -1 \\ \end{bmatrix}Then they did the same thing to get the determinant to be 3. But that's not what I was looking for. I wanted to find the determinant of A =\begin{bmatrix} 1 & 2 & 3 \\ -1 & 2 & 1 \\ 4 & 1 & -1 \\ \end{bmatrix}without doing
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

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

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

Last edited:
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.

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

The determinants of those matrices are equal.

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

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!)

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.

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

look at the code courtri

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)

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.

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?

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.

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.

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

## 1. What is pivotal condensation for determining determinant?

Pivotal condensation is a mathematical method used to calculate the determinant of a square matrix. It involves reducing the matrix into smaller submatrices and using the determinants of these submatrices to calculate the determinant of the original matrix.

## 2. How does pivotal condensation work?

Pivotal condensation works by selecting a row or column in the matrix and using its elements as coefficients to create a smaller submatrix. This submatrix is then multiplied by the determinant of the remaining matrix to calculate the determinant of the original matrix. This process is repeated until the matrix is reduced to a 2x2 matrix, whose determinant can be easily calculated.

## 3. What is the advantage of using pivotal condensation?

Pivotal condensation can simplify the calculation of determinants for large matrices, as it breaks down the problem into smaller, more manageable subproblems. It also allows for the use of properties such as expansion by minors and row operations to further simplify the calculation process.

## 4. Are there any limitations to using pivotal condensation?

One limitation of pivotal condensation is that it can only be used for square matrices. It also requires careful selection of the row or column to use for condensation, as choosing the wrong one may result in a more complex calculation. Additionally, the method can become computationally intensive for very large matrices.

## 5. Can pivotal condensation be used for any type of matrix?

No, pivotal condensation can only be used for square matrices. It is most commonly used for calculating the determinants of 3x3 or 4x4 matrices, but can also be applied to larger square matrices as well.

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