This discussion board is a place to discuss our Daily Challenges and the math and science
related to those challenges. Explanations are more than just a solution — they should
explain the steps and thinking strategies that you used to obtain the solution. Comments
should further the discussion of math and science.

When posting on Brilliant:

Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .

Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.

Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.

Stay on topic — we're all here to learn more about math and science, not to hear about your favorite get-rich-quick scheme or current world events.

Markdown

Appears as

*italics* or _italics_

italics

**bold** or __bold__

bold

- bulleted - list

bulleted

list

1. numbered 2. list

numbered

list

Note: you must add a full line of space before and after lists for them to show up correctly

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!

@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.

Easy Math Editor

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in`\(`

...`\)`

or`\[`

...`\]`

to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## 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.

Log in to reply

But our matrix must be non- zero

Log in to reply

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

Log in to reply

Log 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.

Log in to reply

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

Log in to reply

Log in to reply

Log in to reply

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 !!

Log in to reply

Log in to reply

Log in to reply

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

But the elements must not contain any 0

Log in to reply

Log in to reply

Can you show us the method ?

Log in to reply

Log in to reply

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

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

Log in to reply