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Find-det-A-where-A-is-an-n-n-matrix-of-the-form-A-1-1-1-1-1-1-1-1-1-1-1-1-1-1-1-1-1-1-1-1-

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Question Number 3965 by Yozzii last updated on 25/Dec/15
Find det(A) where A is an n×n matrix of the form A= ((λ,1,1,…,…,…,…,1),(1,λ,1,…,…,…,…,1),(1,1,λ,…,…,…,…,1),(⋮,⋮,⋮,⋱,…,…,…,1),(⋮,⋮,⋮,⋮,⋱,…,…,1),(⋮,⋮,⋮,⋮,⋮,⋱,…,1),(⋮,⋮,⋮,⋮,⋮,⋮,⋱,1),(1,1,1,1,1,1,1,λ) ) λ=constant for leading diagonal elements 1s for all other elements.
Finddet(A)whereAisann×nmatrixoftheformA=(λ11…………11λ1…………111λ…………1⋮⋮⋮⋱………1⋮⋮⋮⋮⋱……1⋮⋮⋮⋮⋮⋱…1⋮⋮⋮⋮⋮⋮⋱11111111λ)λ=constantforleadingdiagonalelements1sforallotherelements.
Commented by 123456 last updated on 25/Dec/15
(λ+1)^2 −1=λ^2 +2λ+1−1=λ(λ+2) λ+1−1=λ (λ+2)^2 −1=λ^2 +4λ−4−1=λ^2 −4λ−5 ⋮
(λ+1)2−1=λ2+2λ+1−1=λ(λ+2)λ+1−1=λ(λ+2)2−1=λ2+4λ−4−1=λ2−4λ−5⋮
Answered by 123456 last updated on 25/Dec/15
A= determinant ((λ,1,1,…,1),(1,λ,1,…,1),(1,1,λ,…,1),(⋮,⋮,⋮,⋱,1),(1,1,1,1,λ)) A=(1/λ^(n−1) ) determinant ((λ,1,1,…,1),(λ,λ^2 ,λ,…,λ),(λ,λ,λ^2 ,…,λ),(⋮,⋮,⋮,⋱,λ),(λ,λ,λ,λ,λ^2 )) A=(1/λ^(n−1) ) determinant ((λ,1,1,…,1),(0,(λ^2 −1),(λ−1),…,(λ−1)),(0,(λ−1),(λ^2 −1),…,(λ−1)),(⋮,⋮,⋮,⋱,(λ−1)),(0,(λ−1),(λ−1),(λ−1),(λ^2 −1))) A=(((λ−1)/λ))^(n−1) determinant ((λ,1,1,…,1),(0,(λ+1),1,…,1),(0,1,(λ+1),…,1),(⋮,⋮,⋮,⋱,1),(0,1,1,1,(λ+1))) A=(((λ−1)/λ))^(n−1) (1/((λ+1)^(n−2) )) determinant ((λ,1,1,…,1),(0,(λ+1),1,…,1),(0,(λ+1),((λ+1)^2 ),…,(λ+1)),(⋮,⋮,⋮,⋱,(λ+1)),(0,(λ+1),(λ+1),(λ+1),((λ+1)^2 ))) A=(((λ−1)/λ))^(n−1) (1/((λ+1)^(n−2) )) determinant ((λ,1,1,…,1),(0,(λ+1),1,…,1),(0,0,(λ(λ+2)),…,λ),(⋮,⋮,⋮,⋱,λ),(0,0,λ,λ,(λ(λ+2)))) A=(((λ−1)/λ))^(n−1) (λ/((λ+1)^(n−2) )) determinant ((λ,1,1,…,1),(0,(λ+1),1,…,1),(0,0,(λ+2),…,1),(⋮,⋮,⋮,⋱,1),(0,0,1,1,(λ+2))) continue
A=|λ11…11λ1…111λ…1⋮⋮⋮⋱11111λ|A=1λn−1|λ11…1λλ2λ…λλλλ2…λ⋮⋮⋮⋱λλλλλλ2|A=1λn−1|λ11…10λ2−1λ−1…λ−10λ−1λ2−1…λ−1⋮⋮⋮⋱λ−10λ−1λ−1λ−1λ2−1|A=(λ−1λ)n−1|λ11…10λ+11…101λ+1…1⋮⋮⋮⋱10111λ+1|A=(λ−1λ)n−11(λ+1)n−2|λ11…10λ+11…10λ+1(λ+1)2…λ+1⋮⋮⋮⋱λ+10λ+1λ+1λ+1(λ+1)2|A=(λ−1λ)n−11(λ+1)n−2|λ11…10λ+11…100λ(λ+2)…λ⋮⋮⋮⋱λ00λλλ(λ+2)|A=(λ−1λ)n−1λ(λ+1)n−2|λ11…10λ+11…100λ+2…1⋮⋮⋮⋱10011λ+2|continue
Commented by prakash jain last updated on 25/Dec/15
From the line in above A=(((λ−1)/λ))^(n−1) determinant ((λ,1,1,…,1),(0,(λ+1),1,…,1),(0,1,(λ+1),…,1),(⋮,⋮,⋮,⋱,1),(0,1,1,1,(λ+1))) Let us say D(λ,n) is the determinant. D(λ,n)=(((λ−1)^(n−1) )/λ^(n−2) )D(λ+1,n−1) D(λ,2)=λ^2 −1=(λ−1)(λ+1) D(λ,3)=(((λ−1)^2 )/λ)D(λ+1,2)=(((λ−1)^2 )/λ)[(λ+1)^2 −1] =(λ−1)^2 (λ+2) D(λ,4)=(((λ−1)^3 )/λ^2 )λ^2 (λ+3)=(λ−1)^3 (λ+3) D(λ,n)=(λ−1)^(n−1) (λ+n−1)
FromthelineinaboveA=(λ−1λ)n−1|λ11…10λ+11…101λ+1…1⋮⋮⋮⋱10111λ+1|LetussayD(λ,n)isthedeterminant.D(λ,n)=(λ−1)n−1λn−2D(λ+1,n−1)D(λ,2)=λ2−1=(λ−1)(λ+1)D(λ,3)=(λ−1)2λD(λ+1,2)=(λ−1)2λ[(λ+1)2−1]=(λ−1)2(λ+2)D(λ,4)=(λ−1)3λ2λ2(λ+3)=(λ−1)3(λ+3)D(λ,n)=(λ−1)n−1(λ+n−1)
Commented by Yozzii last updated on 25/Dec/15
It should be D(λ,n)=(λ−1)^(n−1) (λ+n−1) ?
ItshouldbeD(λ,n)=(λ−1)n−1(λ+n−1)?
Commented by prakash jain last updated on 25/Dec/15
yes. corrected.
yes.corrected.
Answered by Yozzii last updated on 26/Dec/15
Perhaps some sort of recursive formula for ∣A_n ∣ could be found? n=2: A_2 = ((λ,1),(1,λ) )⇒∣A_2 ∣=λ^2 −1 n=3: A_3 = ((λ,1,1),(1,λ,1),(1,1,λ) ) ⇒∣A_3 ∣=λ determinant ((λ,1),(1,λ))− determinant ((1,1),(1,λ))+ determinant ((1,λ),(1,1)) Now, − determinant ((1,1),(1,λ))= determinant ((1,λ),(1,1)) and determinant ((λ,1),(1,λ))=∣A_2 ∣. ∴ ∣A_3 ∣=λ∣A_2 ∣−2 determinant ((1,1),(1,λ)) ∣A_3 ∣=λ∣A_2 ∣−2(λ−1) ∣A_3 ∣=λ(λ^2 −1)−2(λ−1)=(λ−1)(λ^2 +λ−2) n=4:A_4 = ((λ,1,1,1),(1,λ,1,1),(1,1,λ,1),(1,1,1,λ) ) ∣A_4 ∣=λ determinant ((λ,1,1),(1,λ,1),(1,1,λ))− determinant ((1,1,1),(1,λ,1),(1,1,λ))+ determinant ((1,λ,1),(1,1,1),(1,1,λ))− determinant ((1,λ,1),(1,1,λ),(1,1,1)) Also, determinant ((1,λ,1),(1,1,1),(1,1,λ))=− determinant ((1,1,1),(1,λ,1),(1,1,λ))=− determinant ((1,λ,1),(1,1,λ),(1,1,1)) and A_3 = determinant ((λ,1,1),(1,λ,1),(1,1,λ)). determinant ((1,1,1),(1,λ,1),(1,1,λ))=1×(λ−1)(λ+1)−2(λ−1) =(λ−1)(λ+1−1−1) =(λ−1)^2 ∴∣A_4 ∣=λ∣A_3 ∣−3 determinant ((1,1,1),(1,λ,1),(1,1,λ)) ∣A_4 ∣=λ∣A_3 ∣−3(λ−1)^2 n=5: A_5 = ((λ,1,1,1,1),(1,λ,1,1,1),(1,1,λ,1,1),(1,1,1,λ,1),(1,1,1,1,λ) ) ∣A_5 ∣=λ∣A_4 ∣− determinant ((1,1,1,1),(1,λ,1,1),(1,1,λ,1),(1,1,1,λ))+ determinant ((1,λ,1,1),(1,1,1,1),(1,1,λ,1),(1,1,1,λ))− determinant ((1,λ,1,1),(1,1,λ,1),(1,1,1,1),(1,1,1,λ))+ determinant ((1,λ,1,1),(1,1,λ,1),(1,1,1,λ),(1,1,1,1)) It can be shown that ∣A_5 ∣=λ∣A_4 ∣−4 determinant ((1,1,1,1),(1,λ,1,1),(1,1,λ,1),(1,1,1,λ)). determinant ((1,1,1,1),(1,λ,1,1),(1,1,λ,1),(1,1,1,λ))=1×∣A_3 ∣−3 determinant ((1,1,1),(1,λ,1),(1,1,λ)) =(λ−1)[(λ^2 −2λ+1)] =(λ−1)^3 ∴∣A_5 ∣=λ∣A_4 ∣−4(λ−1)^3 . I conject that ∣A_(n+1) ∣=λ∣A_n ∣−n(λ−1)^(n−1) (n≥2) A_2 =λ^2 −1. This is like u_(n+1) =λu_n −n(λ−1)^(n−1) (n≥2). From Mr. Jain′s contribution, ∣A_n ∣=(λ−1)^(n−1) (λ+n−1)=D(λ,n) By my recurrence relation, ∣A_(n+1) ∣=λ(λ−1)^(n−1) (λ+n−1)−n(λ−1)^(n−1) ∣A_(n+1) ∣=(λ−1)^(n−1) (λ^2 +(n−1)λ−n) ∣A_(n+1) ∣=(λ−1)^(n−1) (λ(λ−1)+n(λ−1)) ∣A_(n+1) ∣=(λ−1)^n (λ+n)=D(λ,n+1)
Perhapssomesortofrecursiveformulafor∣An∣couldbefound?n=2:A2=(λ11λ)⇒∣A2∣=λ2−1n=3:A3=(λ111λ111λ)⇒∣A3∣=λ|λ11λ|−|111λ|+|1λ11|Now,−|111λ|=|1λ11|and|λ11λ|=∣A2∣.∴∣A3∣=λ∣A2∣−2|111λ|∣A3∣=λ∣A2∣−2(λ−1)∣A3∣=λ(λ2−1)−2(λ−1)=(λ−1)(λ2+λ−2)n=4:A4=(λ1111λ1111λ1111λ)∣A4∣=λ|λ111λ111λ|−|1111λ111λ|+|1λ111111λ|−|1λ111λ111|Also,|1λ111111λ|=−|1111λ111λ|=−|1λ111λ111|andA3=|λ111λ111λ|.|1111λ111λ|=1×(λ−1)(λ+1)−2(λ−1)=(λ−1)(λ+1−1−1)=(λ−1)2∴∣A4∣=λ∣A3∣−3|1111λ111λ|∣A4∣=λ∣A3∣−3(λ−1)2n=5:A5=(λ11111λ11111λ11111λ11111λ)∣A5∣=λ∣A4∣−|11111λ1111λ1111λ|+|1λ11111111λ1111λ|−|1λ1111λ11111111λ|+|1λ1111λ1111λ1111|Itcanbeshownthat∣A5∣=λ∣A4∣−4|11111λ1111λ1111λ|.|11111λ1111λ1111λ|=1×∣A3∣−3|1111λ111λ|=(λ−1)[(λ2−2λ+1)]=(λ−1)3∴∣A5∣=λ∣A4∣−4(λ−1)3.Iconjectthat∣An+1∣=λ∣An∣−n(λ−1)n−1(n⩾2)A2=λ2−1.Thisislikeun+1=λun−n(λ−1)n−1(n⩾2).FromMr.Jain′scontribution,∣An∣=(λ−1)n−1(λ+n−1)=D(λ,n)Bymyrecurrencerelation,∣An+1∣=λ(λ−1)n−1(λ+n−1)−n(λ−1)n−1∣An+1∣=(λ−1)n−1(λ2+(n−1)λ−n)∣An+1∣=(λ−1)n−1(λ(λ−1)+n(λ−1))∣An+1∣=(λ−1)n(λ+n)=D(λ,n+1)
Commented by Yozzii last updated on 26/Dec/15
I had hoped of solving the recursive formula for ∣A_n ∣...Generating function?
Ihadhopedofsolvingtherecursiveformulafor∣An∣…Generatingfunction?

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