HOw can this determinant be 0 ?

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The determinant of the given 3x3 matrix is zero because the rows are not linearly independent, meaning they lie in the same plane. This results in a volume of zero for the parallelepiped formed by the vectors represented by the rows or columns of the matrix. The calculation of the determinant confirms this, as it simplifies to zero. The discussion clarifies that while the vectors are not parallel, they are coplanar, leading to the determinant being zero. Understanding the geometric interpretation of the determinant is crucial in this context.
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HOw can this determinant be 0 ??

let be a 3x3 matrix with elements

a_{1,j}= j

a_{2,j}= j+3

a_{3,j}= 6+j

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": u\cdot(v\times w). 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".
 
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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": u\cdot(v\times w). 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.
 
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