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Question Number 168813 by safojontoshtemirov last updated on 18/Apr/22

Answered by aleks041103 last updated on 18/Apr/22

$$Wecanaddand/orsubtractdifferent$$$$rowsfromeachotherwithoutchanging$$$$thedeterminant.$$$$\Rightarrow d_n=\left|\begin{array}{cccccc}1& 1& 1& \dots & 1& 1\\ 1& 2& 2& \dots & 2& 2\\ 1& 2& 3& \dots & 3& 3\\ ⋮& ⋮& ⋮& \ddots & ⋮& ⋮\\ 1& 2& 3& \dots & n-1& n-1\\ 1& 2& 3& \dots & n-1& n\end{array}\right|$$$$subtractingthe(n-1)throwfromthe$$$$n-throwweget:$$$$d_n=\left|\begin{array}{cccccc}1& 1& 1& \dots & 1& 1\\ 1& 2& 2& \dots & 2& 2\\ 1& 2& 3& \dots & 3& 3\\ ⋮& ⋮& ⋮& \ddots & ⋮& ⋮\\ 1& 2& 3& \dots & n-1& n-1\\ 0& 0& 0& \dots & 0& 1\end{array}\right|=$$$$={(-1)}^{(n-1)+(n-1)}\left|\begin{array}{ccccc}1& 1& 1& \dots & 1\\ 1& 2& 2& \dots & 2\\ 1& 2& 3& \dots & 3\\ ⋮& ⋮& ⋮& \ddots & ⋮\\ 1& 2& 3& \dots & n-1\end{array}\right|=$$$$={(-1)}^{2(n-1)}{d}_{n-1}=d_{n-1}$$$$\Rightarrow d_n=d_{n-1}\Rightarrow d_n=d_1=\mid 1\mid$$$$\Rightarrow d_n=1$$

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