# Determinant problem/Solution

A solution: The approach used here is to use Gaussian Elimination. Divide the top row by ${\displaystyle m}$:

${\displaystyle {\Delta \over m}={\begin{vmatrix}1&-{1 \over m}&-{1 \over m}&\dots &-{1 \over m}\\-1&m&-1&\dots &-1\\-1&-1&m&\dots &-1\\\vdots &\vdots &\vdots &\ddots &\vdots \\-1&-1&-1&\dots &m\end{vmatrix}}}$

Anticipate addition of rows along the second column:

${\displaystyle m-{1 \over m}={m^{2} \over m}-{1 \over m}={m^{2}-1 \over m}={(m+1)(m-1) \over m}}$

This should become the leading element of the second row.

Anticipate addition of rows along the third and latter columns:

${\displaystyle -{1 \over m}-1={-1-m \over m}=-{(1+m) \over m}}$

This should become the trailing elements of the second row.

Add the first row to the second row and the rows below it:

${\displaystyle {\Delta \over m}={\begin{vmatrix}1&-{1 \over m}&-{1 \over m}&\dots &-{1 \over m}\\0&{(m+1)(m-1) \over m}&-{(m+1) \over m}&\dots &-{(m+1) \over m}\\0&-{(m+1) \over m}&{(m+1)(m-1) \over m}&\dots &-{(m+1) \over m}\\\vdots &\vdots &\vdots &\ddots &\vdots \\0&-{(m+1) \over m}&-{(m+1) \over m}&\dots &{(m+1)(m-1) \over m}\end{vmatrix}}}$

Divide the second row by ${\displaystyle m-1}$:

${\displaystyle {\Delta \over m(m-1)}={\begin{vmatrix}1&-{1 \over m}&-{1 \over m}&\dots &-{1 \over m}\\0&{(m+1) \over m}&-{(m+1) \over m(m-1)}&\dots &-{(m+1) \over m(m-1)}\\0&-{(m+1) \over m}&{(m+1)(m-1) \over m}&\dots &-{(m+1) \over m}\\\vdots &\vdots &\vdots &\ddots &\vdots \\0&-{(m+1) \over m}&-{(m+1) \over m}&\dots &{(m+1)(m-1) \over m}\end{vmatrix}}}$

Anticipate addition along third column:

${\displaystyle {(m+1)(m-1) \over m}-{(m+1) \over m(m-1)}}$
${\displaystyle \qquad ={(m+1)(m-1)^{2}-(m+1) \over m(m-1)}}$
${\displaystyle \qquad ={(m+1)[(m-1)^{2}-1] \over m(m-1)}}$
${\displaystyle \qquad ={(m+1)[{\cancel {(m-1+1)}}(m-1-1)] \over {\cancel {m}}(m-1)}}$
${\displaystyle \qquad ={(m+1)(m-2) \over (m-1)}}$

This should become the new leading element of the third row.

Anticipate addition along fourth and latter columns:

${\displaystyle -{(m+1) \over m}-{(m+1) \over m(m-1)}}$
${\displaystyle \qquad =-{\Bigg [}{(m+1)(m-1)+(m+1) \over m(m-1)}{\Bigg ]}}$
${\displaystyle \qquad =-{(m+1){\cancel {[m-1+1]}} \over {\cancel {m}}(m-1)}}$
${\displaystyle \qquad =-{(m+1) \over (m-1)}}$

This should become the trailing elements of the third row.

Add the second row to the rows below it:

${\displaystyle {\Delta \over m(m-1)}={\begin{vmatrix}1&-{1 \over m}&-{1 \over m}&\dots &\dots &-{1 \over m}\\0&{(m+1) \over m}&-{(m+1) \over m(m-1)}&\dots &\dots &-{(m+1) \over m(m-1)}\\0&0&{(m+1)(m-2) \over (m-1)}&-{(m+1) \over (m-1)}&\dots &-{(m+1) \over (m-1)}\\0&0&-{(m+1) \over (m-1)}&{(m+1)(m-2) \over (m-1)}&\dots &-{(m+1) \over (m-1)}\\\vdots &\vdots &\vdots &\vdots &\ddots &\vdots \\0&0&-{(m+1) \over (m-1)}&-{(m+1) \over (m-1)}&\dots &{(m+1)(m-2) \over (m-1)}\end{vmatrix}}}$

Divide the third row by ${\displaystyle m-2}$:

${\displaystyle {\Delta \over m(m-1)(m-2)}={\begin{vmatrix}1&-{1 \over m}&-{1 \over m}&\dots &\dots &-{1 \over m}\\0&{(m+1) \over m}&-{(m+1) \over m(m-1)}&\dots &\dots &-{(m+1) \over m(m-1)}\\0&0&{(m+1) \over (m-1)}&-{(m+1) \over (m-1)(m-2)}&\dots &-{(m+1) \over (m-1)(m-2)}\\0&0&-{(m+1) \over (m-1)}&{(m+1)(m-2) \over (m-1)}&\dots &-{(m+1) \over (m-1)}\\\vdots &\vdots &\vdots &\vdots &\ddots &\vdots \\0&0&-{(m+1) \over (m-1)}&-{(m+1) \over (m-1)}&\dots &{(m+1)(m-2) \over (m-1)}\end{vmatrix}}}$

Anticipate addition along the fourth column:

${\displaystyle {(m+1)(m-2) \over (m-1)}-{(m+1) \over (m-1)(m-2)}}$
${\displaystyle \qquad ={(m+1)(m-2)^{2}-(m+1) \over (m-1)(m-2)}}$
${\displaystyle \qquad ={(m+1)[(m-2)^{2}-1] \over (m-1)(m-2)}}$
${\displaystyle \qquad ={(m+1)[{\cancel {(m-1)}}(m-3)] \over {\cancel {(m-1)}}(m-2)}}$
${\displaystyle \qquad ={(m+1)(m-3) \over (m-2)}}$

This should become the leading element of the fourth row.

Anticipate addition along the fifth and latter columns:

${\displaystyle -{(m+1) \over (m-1)}-{(m+1) \over (m-1)(m-2)}}$
${\displaystyle \qquad =-{[(m+1)(m-2)+(m+1)] \over (m-1)(m-2)}}$
${\displaystyle \qquad =-{(m+1)[m-2+1] \over (m-1)(m-2)}}$
${\displaystyle \qquad =-{(m+1){\cancel {(m-1)}} \over {\cancel {(m-1)}}(m-2)}}$
${\displaystyle \qquad =-{(m+1) \over (m-2)}}$

This should become the trailing elements of the fourth row.

Add the third row to the rows below it:

${\displaystyle {\Delta \over m(m-1)(m-2)}={\begin{vmatrix}1&-{1 \over m}&-{1 \over m}&\dots &\dots &\dots &-{1 \over m}\\0&{(m+1) \over m}&-{(1+m) \over m(m-1)}&\dots &\dots &\dots &-{(1+m) \over m(m-1)}\\0&0&{(m+1) \over (m-1)}&-{(m+1) \over (m-1)(m-2)}&\dots &\dots &-{(m+1) \over (m-1)(m-2)}\\0&0&0&{(m+1)(m-3) \over (m-2)}&-{(m+1) \over (m-2)}&\dots &-{(m+1) \over (m-2)}\\0&0&0&-{(m+1) \over (m-2)}&{(m+1)(m-3) \over (m-2)}&\dots &-{(m+1) \over (m-2)}\\\vdots &\vdots &\vdots &\vdots &\vdots &\ddots &\vdots \\0&0&0&-{(m+1) \over (m-2)}&-{(m+1) \over (m-2)}&\dots &{(m+1)(m-3) \over (m-2)}\end{vmatrix}}}$

Divide the fourth row by ${\displaystyle m-3}$:

${\displaystyle {\Delta \over m(m-1)(m-2)(m-3)}={\begin{vmatrix}1&-{1 \over m}&-{1 \over m}&\dots &\dots &\dots &-{1 \over m}\\0&{(m+1) \over m}&-{(m+1) \over m(m-1)}&\dots &\dots &\dots &-{(m+1) \over m(m-1)}\\0&0&{(m+1) \over (m-1)}&-{(m+1) \over (m-1)(m-2)}&\dots &\dots &-{(m+1) \over (m-1)(m-2)}\\0&0&0&{(m+1) \over (m-2)}&-{(m+1) \over (m-2)(m-3)}&\dots &-{(m+1) \over (m-2)(m-3)}\\0&0&0&-{(m+1) \over (m-2)}&{(m+1)(m-3) \over (m-2)}&\dots &-{(m+1) \over (m-2)}\\\vdots &\vdots &\vdots &\vdots &\vdots &\ddots &\vdots \\0&0&0&-{(m+1) \over (m-2)}&-{(m+1) \over (m-2)}&\dots &{(m+1)(m-3) \over (m-2)}\end{vmatrix}}}$

Add fourth row to rows below it:

${\displaystyle {\Delta \over m(m-1)(m-2)(m-3)}={\begin{vmatrix}1&-{1 \over m}&-{1 \over m}&\dots &\dots &\dots &\dots &-{1 \over m}\\0&{(m+1) \over m}&-{(m+1) \over m(m-1)}&\dots &\dots &\dots &\dots &-{(m+1) \over m(m-1)}\\0&0&{(m+1) \over (m-1)}&-{(m+1) \over (m-1)(m-2)}&\dots &\dots &\dots &-{(m+1) \over (m-1)(m-2)}\\0&0&0&{(m+1) \over (m-2)}&-{(m+1) \over (m-2)(m-3)}&\dots &\dots &-{(m+1) \over (m-2)(m-3)}\\0&0&0&0&{(m+1)(m-4) \over (m-3)}&-{(m+1) \over (m-3)}&\dots &-{(m+1) \over (m-3)}\\0&0&0&0&-{(m+1) \over (m-3)}&{(m+1)(m-4) \over (m-3)}&\dots &-{(m+1) \over (m-3)}\\\vdots &\vdots &\vdots &\vdots &\vdots &\vdots &\ddots &\vdots \\0&0&0&0&-{(m+1) \over (m-3)}&-{(m+1) \over (m-3)}&\dots &{(m+1)(m-4) \over (m-3)}\end{vmatrix}}}$

For purposes of calculating a determinant, the non-diagonal elements of an upper-triangular matrix do not matter. By induction the equation ends up looking thus:

${\displaystyle {\Delta \over m(m-1)(m-2)(m-3)...(2)(1)}={\begin{vmatrix}{(m+1) \over (m+1)}&-{1 \over m}&-{1 \over m}&\dots &\dots &\dots &\dots &\dots &-{1 \over m}\\0&{(m+1) \over m}&-{(m+1) \over m(m-1)}&\dots &\dots &\dots &\dots &\dots &-{(m+1) \over m(m-1)}\\0&0&{(m+1) \over (m-1)}&-{(m+1) \over (m-1)(m-2)}&\dots &\dots &\dots &\dots &-{(m+1) \over (m-1)(m-2)}\\0&0&0&{(m+1) \over (m-2)}&-{(m+1) \over (m-2)(m-3)}&\dots &\dots &\dots &-{(m+1) \over (m-2)(m-3)}\\0&0&0&0&{(m+1) \over (m-3)}&-{(m+1) \over (m-3)(m-4)}&\dots &\dots &-{(m+1) \over (m-3)(m-4)}\\0&0&0&0&0&{(m+1) \over (m-4)}&-{(m+1) \over (m-4)(m-5)}&\dots &-{(m+1) \over (m-4)(m-5)}\\0&0&0&0&0&0&{(m+1) \over (m-5)}&\dots &-{(m+1) \over (m-5)(m-6)}\\\vdots &\vdots &\vdots &\vdots &\vdots &\vdots &\vdots &\ddots &\vdots \\0&0&0&0&0&0&\dots &0&{(m+1) \over 2}\end{vmatrix}}}$

The determinant of the upper triangular matrix equals the product of its diagonal elements:

${\displaystyle {\Delta \over m(m-1)(m-2)...(1)}={(m+1)^{m} \over (m+1)(m)(m-1)(m-2)...(3)(2)}}$
${\displaystyle \Delta ={(m+1)^{m} \over (m+1)}=(m+1)^{m-1}}$.