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?
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?
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.
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?
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!
I'll borrow your wonderful LaTeX code.
Expanding the determinant along the last column (Laplace's formula), we get
$\displaystyle \begin{vmatrix} -1(n-1)&0&0&\dots&0&0\\ 0&-1(n-2)&0&\dots&0&0\\ 0&0&-1(n-3)&\dots&0&0\\ \dots&\dots&\dots&\dots&\dots&\dots\\ 0&0&0&\dots&-1&0 \\ n&n&n&\dots&n&n \end{vmatrix}={}$
$\displaystyle n\begin{vmatrix} -1(n-1)&0&0&\dots&0\\ 0&-1(n-2)&0&\dots&0\\ 0&0&-1(n-3)&\dots&0\\ \dots&\dots&\dots&\dots&\dots\\ 0&0&0&\dots&-1\end{vmatrix}={}$
$\displaystyle n(-1)^{n-1}(n-1)!=(-1)^{n-1}n!$.