# Finding the Determinant

• October 22nd 2013, 02:46 PM
vidomagru
Finding the Determinant
Prove that the determinant of the matrix

$\begin{bmatrix} 1&n&n&...&n\\ n&2&n&...&n\\ ...&...&...&...&...\\ n&n&n&...&n \end{bmatrix}$

is equal to $(-1)^{n-1}n!$.

Do I approach this with induction? where do I start?
• October 22nd 2013, 03:38 PM
emakarov
Re: Finding the Determinant
Subtract the last row from every other row.
• October 22nd 2013, 04:57 PM
HallsofIvy
Re: Finding the Determinant
emakarov is suggesting a row operation. Do you know how row operations on a matrix change the determinant?

If you swap two rows, you multiply the determinant by -1.

If you multiply a row by a constant, you multiply the determinant by that constant.

If you add a multiple of a row to another row (such as adding 1 times the last two to each other row) does not change the determinant.
• October 22nd 2013, 08:29 PM
vidomagru
Re: Finding the Determinant
Quote:

Originally Posted by emakarov
Subtract the last row from every other row.

THANK YOU! I would not have caught that trick. So here is what I did:

$\begin{bmatrix} 1&n&n&...&n&n\\ n&2&n&...&n&n\\ n&n&3&...&n&n\\ ...&...&...&...&...&...\\ n&n&n&...&n&n \end{bmatrix}$

$= \begin{bmatrix} 1-n&0&0&...&0&0\\ 0&2-n&0&...&0&0\\ 0&0&3-n&...&0&0\\ ...&...&...&...&...&...\\ 0&0&0&...&(n-1)-n&0 \\ 0&0&0&...&0&0 \end{bmatrix}$

$= \begin{bmatrix} -1(n-1)&0&0&...&0&0\\ 0&-1(n-2)&0&...&0&0\\ 0&0&-1(n-3)&...&0&0\\ ...&...&...&...&...&...\\ 0&0&0&...&-1&0 \\ 0&0&0&...&0&0 \end{bmatrix}$

Which of course is a diagonal matrix so the determinant is the product of the diagonal terms, however the last one is 0, which throws this off. What did I do wrong?

I can see that the first $n-1$ terms give you $(-1)^{n-1}(n-1)!$ but where does the last $n$ come from?
• October 23rd 2013, 12:05 AM
emakarov
Re: Finding the Determinant
Quote:

Originally Posted by emakarov
Subtract the last row from every other row.

The hint was to subtract the last row from every row except the last one. Subtracting a row v from a different row w, (as HallsofIvy wrote, formally this is replacing w with the sum w + (-1)v) does not change the determinant. Subtracting a row from itself definitely changes the determinant in general: it turns it to zero!
• October 23rd 2013, 04:04 AM
vidomagru
Re: Finding the Determinant
Quote:

Originally Posted by emakarov
The hint was to subtract the last row from every row except the last one. Subtracting a row v from a different row w, (as HallsofIvy wrote, formally this is replacing w with the sum w + (-1)v) does not change the determinant. Subtracting a row from itself definitely changes the determinant in general: it turns it to zero!

So if you don't subtract the last row from itself then you would have n|diagonalmatrix| -n|diagonalmatrix| +n|diagonalmatrix| ... Which does not seem to give n!.
• October 23rd 2013, 06:56 AM
emakarov
Re: Finding the Determinant
I'll borrow your wonderful LaTeX code.

Expanding the determinant along the last column (Laplace's formula), we get

$\begin{vmatrix} -1(n-1)&0&0&\dots&0&0\\ 0&-1(n-2)&0&\dots&0&0\\ 0&0&-1(n-3)&\dots&0&0\\ \dots&\dots&\dots&\dots&\dots&\dots\\ 0&0&0&\dots&-1&0 \\ n&n&n&\dots&n&n \end{vmatrix}={}$

$n\begin{vmatrix} -1(n-1)&0&0&\dots&0\\ 0&-1(n-2)&0&\dots&0\\ 0&0&-1(n-3)&\dots&0\\ \dots&\dots&\dots&\dots&\dots\\ 0&0&0&\dots&-1\end{vmatrix}={}$

$n(-1)^{n-1}(n-1)!=(-1)^{n-1}n!$.