1. ## Finding the Determinant

Prove that the determinant of the matrix

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

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

Do I approach this with induction? where do I start?

2. ## Re: Finding the Determinant

Subtract the last row from every other row.

3. ## 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.

4. ## Re: Finding the Determinant

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:

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

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

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

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 $\displaystyle n-1$ terms give you $\displaystyle (-1)^{n-1}(n-1)!$ but where does the last $\displaystyle n$ come from?

5. ## Re: Finding the Determinant

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!

6. ## Re: Finding the Determinant

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!.

7. ## Re: Finding the Determinant

I'll borrow your wonderful LaTeX code.

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

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

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

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