Determinant of a matrix using reduced echelon form

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Homework Help Overview

The discussion revolves around finding the determinant of a matrix through row reduction to echelon form. Participants are examining the implications of achieving a triangular form and the conditions under which the determinant may be zero.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the process of reducing a matrix to echelon form and question whether achieving rows of zeros affects the determinant. There is also exploration of the relationship between linear combinations of rows and the determinant's value.

Discussion Status

The conversation includes multiple interpretations of the determinant's properties, particularly regarding linear dependence and the presence of zero rows. Some participants provide affirmations about the determinant being zero under certain conditions, while others explore the reasoning behind these assertions.

Contextual Notes

There is a focus on the definitions of triangular form and linear combinations, with participants referencing specific properties of determinants without reaching a consensus on all aspects discussed.

TrippingBilly
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Problem statement: Find the determinant of the following matrix by row reduction to echelon form.
|1 3 3 -4 |
|0 1 2 -5 |
|2 5 4 -3 |
|-3 -7 -5 2 |

I reduced this matrix to
|1 3 3 -4 |
|0 1 2 -5 |
|0 -1 -2 5 |
|0 2 4 -10 |

If I reduce this further, the entries in row 3 and 4 become 0. Is this still considered triangular form, and therefore the determinant will be 0?
 
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TrippingBilly said:
Problem statement: Find the determinant of the following matrix by row reduction to echelon form.
|1 3 3 -4 |
|0 1 2 -5 |
|2 5 4 -3 |
|-3 -7 -5 2 |

I reduced this matrix to
|1 3 3 -4 |
|0 1 2 -5 |
|0 -1 -2 5 |
|0 2 4 -10 |

If I reduce this further, the entries in row 3 and 4 become 0. Is this still considered triangular form, and therefore the determinant will be 0?

Correct. It was obvious from the beginning, since the third and second row are proportional (after your first "reduction").
 
Last edited:
Yes and yes.
 
Ok, I didn't reduce it by row-eschelon form, but you should get a determinant of zero.

Why? The second row is a linear combination of the other three.

Upper Triangular form just requires that the entries below the diagonal are all zero. If entries above (or on) the diagonal are zero, that's ok.

ZM
 
If this is true, then will the determinant of any matrix containing a row vector consisting of all zeros will be zero?
 
TrippingBilly said:
If this is true, then will the determinant of any matrix containing a row vector consisting of all zeros will be zero?

Yes, it will.
 
Yes. Think of it this way:

The determinant function can be viewed as a machine that takes n n-dimensional vectors and spits out a number. However, if any of these vectors are linear combinations of any of the other vectors, then the determinant will be zero. A zero vector is a linear combination of every vector.

Another way to see the same thing is just to break the determinant into minors along the row (or column) of zeros. The determinant is then zero.

ZM
 
The exact reason follows directly from the http://www.cs.ut.ee/~toomas_l/linalg/lin1/node14.html" of the determinant.
 
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