Determinant of a matrix using reduced echelon form

AI Thread Summary
The discussion centers on finding the determinant of a matrix through row reduction to echelon form. The matrix was reduced to a form where two rows became proportional, indicating that the determinant is zero. It is confirmed that if any row consists entirely of zeros or if any row is a linear combination of others, the determinant will also be zero. The concept of upper triangular form is clarified, emphasizing that only the entries below the diagonal need to be zero for it to be considered triangular. Overall, the determinant's value reflects the linear independence of the rows or columns of the matrix.
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").
 
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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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