# HOw can this determinant be 0 ?

1. Sep 9, 2009

### zetafunction

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 ??

2. Sep 9, 2009

### Moo Of Doom

Re: HOw can this determinant be 0 ??

Because the rows aren't linearly independent:

2*[4 5 6] - 1*[1 2 3] = [7 8 9]

3. Oct 7, 2009

### zgozvrm

Re: HOw can this determinant be 0 ??

The determinant of a 3x3 matrix

a b c
d e f
g h i

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

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

4. Oct 7, 2009

### Gear300

Re: HOw can this determinant be 0 ??

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

5. Oct 8, 2009

### g_edgar

Re: HOw can this determinant be 0 ??

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

6. Oct 8, 2009

### HallsofIvy

Re: HOw can this determinant be 0 ??

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".

Last edited by a moderator: Oct 9, 2009
7. Oct 8, 2009

### Gear300

Re: HOw can this determinant be 0 ??

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

Last edited: Oct 8, 2009