I need a proof to show that the determinant of (product of a 3*2 matrix and a 2*3 matrix )is zero
I guess we can take generic case
Can be it be zero I am in doubt..
Is it the general case or particular??
why does it have to be zero? that need not be true!
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.
I don't think it's true. Please recheck the statement.
ok i did three different examples and it seems u r right!
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??
it has to be matrices with real elements, right?
I just got amazed, yeah it is true..
Check for linear in dependencies when you choose the matrices.
same here @waterineyes !
@champs hey u jst take an example n look wether is it correct or nt... then argue :P
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
i dont knw much.... can anyone please mathematically prove this thing ?? i jst hav to present it in an xhibition.
@waterineyes i'm not sure!
i have to go. i'll try to log in later
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.
@josephjoshy: That's linear dependencies of a vector space.
@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.
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