Question

I need a proof to show that the determinant of (product of a 3*2 matrix and a 2*3 matrix )is zero

Answer 1

I guess we can take generic case

Answer 2

Can be it be zero I am in doubt..

Answer 3

Is it the general case or particular??

Answer 4

why does it have to be zero?
that need not be true!

Answer 5

yes it s zero,,,
u may take any cases...
jst take a 3*2 matrix. multiply it with any other 2*3 matrix. then find the determinant of that. it will b zero.

Answer 6

I don't think it's true. Please recheck the statement.

Answer 7

ok i did three different examples and it seems u r right!

Answer 8

Product of 3*2 and 2*3 will give you another matrix of 3*3 dimension..
Now you want to find its determinant how it can be zero..
You can take any example of 3*3 matrix directly and how it can be zero??

Answer 9

it has to be matrices with real elements, right?

Answer 10

I just got amazed, yeah it is true..

Answer 11

Check for linear in dependencies when you choose the matrices.

Answer 12

same here @waterineyes !

Answer 13

@champs
hey u jst take an example n look wether is it correct or nt... then argue :P

Answer 14

What is the trick or logic behind it..
We multiply two general matrices and determinant got is what zero..

I think it is the case when we multiply 2*3 and 3*2 matrices and not 3*3 and 3*3 matrices..
What you think @helder_edwin

Answer 15

i dont knw much.... can anyone please mathematically prove this thing ??
i jst hav to present it in an xhibition.

Answer 16

@waterineyes i'm not sure!

Answer 17

i have to go.
i'll try to log in later

Answer 18

If you have a 3 by 2 matrix, the three rows can't be independent of one another.
So a 3 by 2 looks like

( r1
r2
a r1 +b r2)

where r1 and r2 are 2-d row vectors and a and b are numbers.

Similarly, in a 2 by 3, the columns can't be independent. So, the
2 by 3 looks like

(c1 c2 c3=(d c1+ e c2))

where c1 and c2 are 2-d column vectors and d and e are numbers.

Multiply them out to get the 3 by 3, and you'll see that the third column of the 3 by 3 is a linear combination of the first two colums, so the determinant has to be zero.

Answer 19

@josephjoshy: That's linear dependencies of a vector space.

Answer 20

@josephjoshy sorry i couldn't log in yesterday. these days i've been busy.
But here is what i got
\[ \large \begin{pmatrix}
r_1\\ r_2 \\ r_3
\end{pmatrix}\cdot
\begin{pmatrix}
c_1 & c_2 & c_3
\end{pmatrix}=
\begin{pmatrix}
r_1^tc_1 & r_1^tc_2 & r_1^tc_3\\
r_2^tc_1 & r_2^tc_2 & r_2^tc_3\\
r_3^tc_1 & r_3^tc_2 & r_3^tc_3
\end{pmatrix} \]
now the first matrix cannot have linearly independent rows so at least
\[ \large r_3=\alpha r_1+\beta r_2 \]
if you replace this in the third row of the product matrix you get
\[ \large \begin{pmatrix}
(\alpha r_1+\beta r_2)^tc_1 & (\alpha r_1+\beta r_2)^tc_2 &
(\alpha r_1+\beta r_2)^tc_3
\end{pmatrix}= \]
\[ \large =\begin{pmatrix}
\alpha r_1^tc_1+\beta r_2^tc_1 & \alpha r_1^tc_2+\beta r_2^tc_2
& \alpha r_1^tc_3+\beta r_2^tc_3
\end{pmatrix} \]
and this is a linear combination of rows 1 and 2 of the product matrix.
Hence its determinant must be zero.