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Question Number 205164 by depressiveshrek last updated on 11/Mar/24

$$\mathrm{Find}\mathrm{the}\mathrm{determinant}:$$$$\left|\begin{array}{ccccc}1-x& 2& 3& \dots & n\\ 1& 2-x& 3& \dots & n\\ 1& 2& 3-x& \dots & n\\ ⋮& ⋮& ⋮& \ddots & ⋮\\ 1& 2& 3& \dots & n-x\end{array}\right|$$

Answered by aleks041103 last updated on 12/Mar/24

$$Bysubtractingthefirstrowfromallother$$$$\left|\begin{array}{ccccc}1-x& 2& 3& \dots & n\\ 1& 2-x& 3& \dots & n\\ 1& 2& 3-x& \dots & n\\ ⋮& ⋮& ⋮& \ddots & ⋮\\ 1& 2& 3& \dots & n-x\end{array}\right|=$$$$=\left|\begin{array}{ccccc}1-x& 2& 3& \dots & n\\ x& -x& 0& \dots & 0\\ x& 0& -x& \dots & 0\\ ⋮& ⋮& ⋮& \ddots & ⋮\\ x& 0& 0& \dots & -x\end{array}\right|$$$$\left|\begin{array}{ccccc}{a}_0& b_1& b_2& \dots & b_n\\ c_1& a_1& 0& \dots & 0\\ c_2& 0& a_2& \dots & 0\\ ⋮& ⋮& ⋮& \ddots & ⋮\\ c_n& 0& 0& \dots & a_n\end{array}\right|=$$$$=a_0{a}_1...a_n-c_1{b}_1{a}_2...a_n-c_2{b}_2{a}_1{a}_3...a_n-...=$$$$=a_0{a}_1...a_n-\underset{k=1}{\sum ^n}{b}_k{c}_k\prod _{s\ne k}{a}_s$$$$\Rightarrow \left|\begin{array}{ccccc}1-x& 2& 3& \dots & n\\ x& -x& 0& \dots & 0\\ x& 0& -x& \dots & 0\\ ⋮& ⋮& ⋮& \ddots & ⋮\\ x& 0& 0& \dots & -x\end{array}\right|=$$$$=(1-x){(-x)}^{n-1}-\underset{k=1}{\sum ^{n-1}}(k+1)x{(-x)}^{n-2}=$$$$=(1-x+\underset{k=2}{\sum ^n}k){(-1)}^{n-1}{x}^{n-1}=$$$$={(-1)}^n(x-\frac{n(n+1)}{2})x^{n-1}$$

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