MatriX

Find two non zero matrix such that :

AB = BA = 0

#Algebra #7 #9 #4 #99999

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`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
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`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
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Comments

Well, 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 - 5 years, 9 months ago

But our matrix must be non- zero

Syed Baqir - 5 years, 9 months ago

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

Michael Mendrin - 5 years, 9 months ago

@Michael Mendrin – I mean matrix in which none of the elements are 0

Syed Baqir - 5 years, 9 months ago

@Syed Baqir – Here you go

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

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 - 5 years, 9 months ago

@Michael Mendrin – You are correct but I think B must be like this:

B = $\begin{pmatrix} 1 & -1 \\ -1 & 1 \end{pmatrix}$

Syed Baqir - 5 years, 9 months ago

@Syed Baqir – Well obviously it can work both ways.

Michael Mendrin - 5 years, 9 months ago

@Michael Mendrin – But sir can you tell us how you got the answer I mean the method ?

Syed Baqir - 5 years, 9 months ago

@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 - 5 years, 9 months ago

@Michael Mendrin – hehe, But I will ask you tomorrow :P,

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 - 5 years, 9 months ago

@Syed Baqir – I will sleep on a brilliant solution tonight! Good night!

Michael Mendrin - 5 years, 9 months ago

@Michael Mendrin – Good Night !!

Syed Baqir - 5 years, 9 months 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.

Gian Sanjaya - 5 years, 9 months ago

But the elements must not contain any 0

Syed Baqir - 5 years, 9 months 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 - 5 years, 9 months ago

Can you show us the method ?

Syed Baqir - 5 years, 9 months 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 - 5 years, 9 months ago

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 - 5 years, 9 months ago

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

Syed Baqir - 5 years, 9 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$