HOw can this determinant be 0 ?

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Discussion Overview

The discussion revolves around the conditions under which the determinant of a specific 3x3 matrix is equal to zero. Participants explore the implications of linear dependence among the rows of the matrix and the geometric interpretation of the determinant in relation to vector arrangements.

Discussion Character

  • Technical explanation
  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • One participant questions how the determinant can be zero given the matrix elements defined as a_{1,j}= j, a_{2,j}= j+3, and a_{3,j}= 6+j.
  • Another participant asserts that the rows of the matrix are not linearly independent, providing an example of a linear combination that results in one row being expressible in terms of others.
  • A participant explains the formula for calculating the determinant of a 3x3 matrix and applies it to the specific case, showing that the determinant evaluates to zero.
  • Some participants discuss the geometric interpretation of the determinant, suggesting that the rows or columns of the matrix can be viewed as vectors that lie in the same plane, leading to a volume of zero for the associated parallelepiped.
  • There is a clarification regarding the parallelism of the vectors, with one participant correcting a misunderstanding about the nature of the vectors in the matrix.

Areas of Agreement / Disagreement

Participants generally agree that the determinant is zero due to linear dependence among the rows, but there is some disagreement regarding the specific nature of the vectors and their relationships.

Contextual Notes

Some assumptions about linear independence and vector arrangements are discussed, but the implications of these assumptions are not fully resolved.

zetafunction
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HOw can this determinant be 0 ??

let be a 3x3 matrix with elements

[tex]a_{1,j}= j[/tex]

[tex]a_{2,j}= j+3[/tex]

[tex]a_{3,j}= 6+j[/tex]

with i=1,2,3 and j=1,2,3 ...

but DetA=0 apparently there is no zeros or other condition that points that determinant should be 0, is there any explanation ??
 
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Because the rows aren't linearly independent:

2*[4 5 6] - 1*[1 2 3] = [7 8 9]
 


The determinant of a 3x3 matrix

a b c
d e f
g h i

is: aei+bfg+cdh-afh-bdi-ceg


In your case

a b c
d e f =
g h i

1 2 3
4 5 6
7 8 9



Therefore the determinant is:
1x5x9 + 2x6x7 + 3x4x8 - 1x6x8 - 2x4x9 - 3x5x7
= 45 + 84 + 96 - 48 -72 - 105
= 225 - 225
= 0
 


If you look at the 3x3 matrix as an array of 3 vectors (row or column), all of them are parallel.
 


Gear300 said:
If you look at the 3x3 matrix as an array of 3 vectors (row or column), all of them are parallel.

?? [1,2,3] and [4,5,6] are parallel?
 


Gear300 said:
If you look at the 3x3 matrix as an array of 3 vectors (row or column), all of them are parallel.
No, they aren't. But they all lie in the same plane. Another way of looking at this is that that the determinant of a matrix having three vectors, u, v, w, say, as rows (or columns) is a "triple product": [itex]u\cdot(v\times w)[/itex]. And that can be interpreted as the volume of the parallelopiped having the three vectors as edges. That volume is 0 if and only if the three vectors are all in the same plane so that the parallelopiped has no "height".
 
Last edited by a moderator:


g_edgar said:
?? [1,2,3] and [4,5,6] are parallel?

HallsofIvy said:
No, they aren't. But they all lie in the same plane. Another way of looking at this that that the determinant of a matrix having three vectors, u, v, w, say, as rows (or columns) is a "triple product": [itex]u\cdot(v\times w)[/itex]. And that can be interpreted as the volume of the parallelopiped having the three vectors as edges. That volume is 0 if and only if the three vectors are all in the same plane so that the parallelopiped has no "height".

Sorry about that...mixed up the words...they lie on the same plane as HallsofIvy stated.
 
Last edited:

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