there's lots of such matrices that have this property.
–
Michael Mendrin
·
1 year, 1 month ago

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@Michael Mendrin
–
So it's actually simple problem if we are asked to find one each with non-zero determinants: say that it's impossible.
–
Gian Sanjaya
·
1 year, 1 month ago

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@Gian Sanjaya
–
lol, then you will get impossible marks in the result !!!
–
Syed Baqir
·
1 year, 1 month ago

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@Michael Mendrin
–
The product of the determinants of the 2 matrix has the same value as the determinant of their product, so either must have 0 determinant.
–
Gian Sanjaya
·
1 year, 1 month ago

@Syed Baqir
–
I guess you need to consider 2 matrices. And then, you multiply them (I told that for 2x2 matrices!!!), next you calculate the determinant of the result. I hope that you can do it yourself using this.
–
Gian Sanjaya
·
1 year, 1 month ago

@Syed Baqir
–
Yeah, I just said about if the determinant is 0. In fact, if exactly one 0 exist, a 2x2 matrix wouldn't have zero determinant.
–
Gian Sanjaya
·
1 year, 1 month ago

As I said, finding two matrices with non-zero determinants will take a bit longer. Normally it shouldn't be possible, but I've heard of algebras where this is possible, i.e., the elements aren't restricted to reals. You didn't specify if the elements had to be reals or even complex.
–
Michael Mendrin
·
1 year, 1 month ago

@Michael Mendrin
–
But sir can you tell us how you got the answer I mean the method ?
–
Syed Baqir
·
1 year, 1 month ago

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@Syed Baqir
–
Trial and error. But it's a LOT harder problem if the matrices have non-zero determinants! Don't ask me tonight, I'm going to sleep pretty soon!
–
Michael Mendrin
·
1 year, 1 month ago

## Comments

Sort by:

TopNewestWell, this one works

\(A=B=\begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}\)

and this one too

\(A=\begin{pmatrix} 0 & 1 \\ 0 & -1 \end{pmatrix}\quad and\quad B=\begin{pmatrix} 1 & 1 \\ 0 & 0 \end{pmatrix}\)

there's lots of such matrices that have this property. – Michael Mendrin · 1 year, 1 month ago

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– Gian Sanjaya · 1 year, 1 month ago

So it's actually simple problem if we are asked to find one each with non-zero determinants: say that it's impossible.Log in to reply

– Syed Baqir · 1 year, 1 month ago

lol, then you will get impossible marks in the result !!!Log in to reply

– Gian Sanjaya · 1 year, 1 month ago

The product of the determinants of the 2 matrix has the same value as the determinant of their product, so either must have 0 determinant.Log in to reply

– Syed Baqir · 1 year, 1 month ago

Can you show us the method ?Log in to reply

– Gian Sanjaya · 1 year, 1 month ago

I guess you need to consider 2 matrices. And then, you multiply them (I told that for 2x2 matrices!!!), next you calculate the determinant of the result. I hope that you can do it yourself using this.Log in to reply

– Syed Baqir · 1 year, 1 month ago

But the elements must not contain any 0Log in to reply

– Gian Sanjaya · 1 year, 1 month ago

Yeah, I just said about if the determinant is 0. In fact, if exactly one 0 exist, a 2x2 matrix wouldn't have zero determinant.Log in to reply

– Syed Baqir · 1 year, 1 month ago

But our matrix must be non- zeroLog in to reply

– Michael Mendrin · 1 year, 1 month ago

You mean non-zero determinant? That will take me a little longer.Log in to reply

– Syed Baqir · 1 year, 1 month ago

I mean matrix in which none of the elements are 0Log in to reply

\(A=\begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix}\quad and\quad B=\begin{pmatrix} -1 & 1 \\ 1 & -1 \end{pmatrix}\)

As I said, finding two matrices with non-zero determinants will take a bit longer. Normally it shouldn't be possible, but I've heard of algebras where this is possible, i.e., the elements aren't restricted to reals. You didn't specify if the elements had to be reals or even complex. – Michael Mendrin · 1 year, 1 month ago

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B = \( \begin{pmatrix} 1 & -1 \\ -1 & 1 \end{pmatrix} \) – Syed Baqir · 1 year, 1 month ago

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– Michael Mendrin · 1 year, 1 month ago

Well obviously it can work both ways.Log in to reply

– Syed Baqir · 1 year, 1 month ago

But sir can you tell us how you got the answer I mean the method ?Log in to reply

– Michael Mendrin · 1 year, 1 month ago

Trial and error. But it's a LOT harder problem if the matrices have non-zero determinants! Don't ask me tonight, I'm going to sleep pretty soon!Log in to reply

Any way nice solution I was using brute force method .

If you have time, dont forget to enlighten this note with brilliant solution !! – Syed Baqir · 1 year, 1 month ago

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– Michael Mendrin · 1 year, 1 month ago

I will sleep on a brilliant solution tonight! Good night!Log in to reply

– Syed Baqir · 1 year, 1 month ago

Good Night !!Log in to reply