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Question Number 205164 by depressiveshrek last updated on 11/Mar/24
Find the determinant:   determinant (((1−x),2,3,…,n),(1,(2−x),3,…,n),(1,2,(3−x),…,n),(⋮,⋮,⋮,⋱,⋮),(1,2,3,…,(n−x)))
$$\mathrm{Find}\:\mathrm{the}\:\mathrm{determinant}: \\ $$$$\begin{vmatrix}{\mathrm{1}−{x}}&{\mathrm{2}}&{\mathrm{3}}&{\ldots}&{{n}}\\{\mathrm{1}}&{\mathrm{2}−{x}}&{\mathrm{3}}&{\ldots}&{{n}}\\{\mathrm{1}}&{\mathrm{2}}&{\mathrm{3}−{x}}&{\ldots}&{{n}}\\{\vdots}&{\vdots}&{\vdots}&{\ddots}&{\vdots}\\{\mathrm{1}}&{\mathrm{2}}&{\mathrm{3}}&{\ldots}&{{n}−{x}}\end{vmatrix} \\ $$
Answered by aleks041103 last updated on 12/Mar/24
By subtracting the first row from all other   determinant (((1−x),2,3,…,n),(1,(2−x),3,…,n),(1,2,(3−x),…,n),(⋮,⋮,⋮,⋱,⋮),(1,2,3,…,(n−x)))=  = determinant (((1−x),2,3,…,n),(x,(−x),0,…,0),(x,0,(−x),…,0),(⋮,⋮,⋮,⋱,⋮),(x,0,0,…,(−x)))   determinant ((a_0 ,b_1 ,b_2 ,…,b_n ),(c_1 ,a_1 ,0,…,0),(c_2 ,0,a_2 ,…,0),(⋮,⋮,⋮,⋱,⋮),(c_n ,0,0,…,a_n ))=  =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 −Σ_(k=1) ^n b_k c_k Π_(s≠k) a_s   ⇒ determinant (((1−x),2,3,…,n),(x,(−x),0,…,0),(x,0,(−x),…,0),(⋮,⋮,⋮,⋱,⋮),(x,0,0,…,(−x)))=  =(1−x)(−x)^(n−1) −Σ_(k=1) ^(n−1) (k+1)x(−x)^(n−2) =  =(1−x+Σ_(k=2) ^n k)(−1)^(n−1) x^(n−1) =  =(−1)^n (x−((n(n+1))/2))x^(n−1)
$${By}\:{subtracting}\:{the}\:{first}\:{row}\:{from}\:{all}\:{other} \\ $$$$\begin{vmatrix}{\mathrm{1}−{x}}&{\mathrm{2}}&{\mathrm{3}}&{\ldots}&{{n}}\\{\mathrm{1}}&{\mathrm{2}−{x}}&{\mathrm{3}}&{\ldots}&{{n}}\\{\mathrm{1}}&{\mathrm{2}}&{\mathrm{3}−{x}}&{\ldots}&{{n}}\\{\vdots}&{\vdots}&{\vdots}&{\ddots}&{\vdots}\\{\mathrm{1}}&{\mathrm{2}}&{\mathrm{3}}&{\ldots}&{{n}−{x}}\end{vmatrix}= \\ $$$$=\begin{vmatrix}{\mathrm{1}−{x}}&{\mathrm{2}}&{\mathrm{3}}&{\ldots}&{{n}}\\{{x}}&{−{x}}&{\mathrm{0}}&{\ldots}&{\mathrm{0}}\\{{x}}&{\mathrm{0}}&{−{x}}&{\ldots}&{\mathrm{0}}\\{\vdots}&{\vdots}&{\vdots}&{\ddots}&{\vdots}\\{{x}}&{\mathrm{0}}&{\mathrm{0}}&{\ldots}&{−{x}}\end{vmatrix} \\ $$$$\begin{vmatrix}{{a}_{\mathrm{0}} }&{{b}_{\mathrm{1}} }&{{b}_{\mathrm{2}} }&{\ldots}&{{b}_{{n}} }\\{{c}_{\mathrm{1}} }&{{a}_{\mathrm{1}} }&{\mathrm{0}}&{\ldots}&{\mathrm{0}}\\{{c}_{\mathrm{2}} }&{\mathrm{0}}&{{a}_{\mathrm{2}} }&{\ldots}&{\mathrm{0}}\\{\vdots}&{\vdots}&{\vdots}&{\ddots}&{\vdots}\\{{c}_{{n}} }&{\mathrm{0}}&{\mathrm{0}}&{\ldots}&{{a}_{{n}} }\end{vmatrix}= \\ $$$$={a}_{\mathrm{0}} {a}_{\mathrm{1}} …{a}_{{n}} −{c}_{\mathrm{1}} {b}_{\mathrm{1}} {a}_{\mathrm{2}} …{a}_{{n}} −{c}_{\mathrm{2}} {b}_{\mathrm{2}} {a}_{\mathrm{1}} {a}_{\mathrm{3}} …{a}_{{n}} −…= \\ $$$$={a}_{\mathrm{0}} {a}_{\mathrm{1}} …{a}_{{n}} −\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{b}_{{k}} {c}_{{k}} \underset{{s}\neq{k}} {\prod}{a}_{{s}} \\ $$$$\Rightarrow\begin{vmatrix}{\mathrm{1}−{x}}&{\mathrm{2}}&{\mathrm{3}}&{\ldots}&{{n}}\\{{x}}&{−{x}}&{\mathrm{0}}&{\ldots}&{\mathrm{0}}\\{{x}}&{\mathrm{0}}&{−{x}}&{\ldots}&{\mathrm{0}}\\{\vdots}&{\vdots}&{\vdots}&{\ddots}&{\vdots}\\{{x}}&{\mathrm{0}}&{\mathrm{0}}&{\ldots}&{−{x}}\end{vmatrix}= \\ $$$$=\left(\mathrm{1}−{x}\right)\left(−{x}\right)^{{n}−\mathrm{1}} −\underset{{k}=\mathrm{1}} {\overset{{n}−\mathrm{1}} {\sum}}\left({k}+\mathrm{1}\right){x}\left(−{x}\right)^{{n}−\mathrm{2}} = \\ $$$$=\left(\mathrm{1}−{x}+\underset{{k}=\mathrm{2}} {\overset{{n}} {\sum}}{k}\right)\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} {x}^{{n}−\mathrm{1}} = \\ $$$$=\left(−\mathrm{1}\right)^{{n}} \left({x}−\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{2}}\right){x}^{{n}−\mathrm{1}} \\ $$

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