1. ## Solve determinant

$
\left(
\begin
{array}
{ccccc}
1&n&n&...&n\\
n&2&n&...&n\\
n&n&3&...&n\\
...&...&...&...&...\\
n&n&n&...&n
\end
{array}
\right)
$

The solution is $(-1)^{(n-1)}*n!$

Would someone be so kind and explain me how to do it? Thanks in advance!

2. Subtract the last row from every other row, this leaves us

$\left(\begin{array}{ccccc}1-n&0&0&...&0\\0&2-n&0&...&0\\0&0&3-n&...&0\\...&...&...&...&...\\n&n&n&...&n\end{arra y}\right)$

3. Originally Posted by SimonM
Subtract the last row from every other row, this leaves us

$\left(\begin{array}{ccccc}1-n&0&0&...&0\\0&2-n&0&...&0\\0&0&3-n&...&0\\...&...&...&...&...\\n&n&n&...&n\end{arra y}\right)$
then subtract the last column from every other column and you're done!

4. ## re

thanks guys. But what will the result be in your cases?

5. Originally Posted by babanner
thanks guys. But what will the result be in your cases?
Have you done what was suggested? What do you get?

6. ## re

well, i have posted in the first post what the final result should be. However in the above posts the final result seems to be different. In your cases i think the result will be: $(n-1)*(n-2)* ... *n$

7. ## Re

Thanks guys. I figured it out

I didn't notice that $n*(n-1)*(n-2)*(n-3) ... *1$ is the same as $1*2*3*4*...*(n-1)*n$ which both equal $n!$