@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.

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.

@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!

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## Comments

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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.

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

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

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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.

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Can you show us the method ?

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But the elements must not contain any 0

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But our matrix must be non- zero

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You mean non-zero determinant? That will take me a little longer.

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

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.

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

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Any way nice solution I was using brute force method .

If you have time, dont forget to enlighten this note with brilliant solution !!

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