Matrix Determinant 01
\( Let\,\,f\left( x \right) =\left( p_1-x \right) \left( p_2-x \right) ...\left( p_n-x \right) \,\,and\,\,let \\ \varDelta _n=\,\,\left| \begin{matrix} p_1& a& a& a& ...& a& a\\ b& p_2& a& a& ...& a& a\\ b& b& p_3& a& ...& a& a\\ b& b& b& p_4& ...& a& a\\ \vdots& \vdots& \vdots& \vdots& & \vdots& \vdots\\ b& b& b& b& ...& p_{n-1}& a\\ b& b& b& b& ...& b& p_n\\ \end{matrix} \right| \\ If\,\,a\ne b\,\,then \\ \varDelta _n=\frac{Xf\left( a \right) -Yf\left( b \right)}{b-a} \\ What\,\,is\,\,X+Y\,\,? \\
\)
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1 solution
Can you explain more detail the process from Δ n = ( p n − a ) Δ n − 1 + a f n − 1 ( b ) to Δ n = b − a b f n ( a ) − a f n ( b ) ? Thank you very much
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Δ 1 = p 1 = b − a b f 1 ( a ) − a f 1 ( b ) , so the result holds for n = 1 . If Δ n − 1 = b − a b f n − 1 ( a ) − a f n − 1 ( b ) then Δ n = ( p n − a ) [ b − a b f n − 1 ( a ) − a f n − 1 ( b ) ] + a f n − 1 ( b ) = b − a b f n ( a ) − a f n ( b ) and so the result follows by induction.
An elementary row operation (subtract the last but one row from the last) gives Δ n = ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ p 1 b b b ⋮ b b a p 2 b b ⋮ b b a a p 3 b ⋮ b b a a a p 4 ⋮ b b . . . . . . . . . . . . . . . . . . a a a a ⋮ p n − 1 b a a a a ⋮ a p n ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ = ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ p 1 b b b ⋮ b 0 a p 2 b b ⋮ b 0 a a p 3 b ⋮ b 0 a a a p 4 ⋮ b 0 . . . . . . . . . . . . . . . . . . a a a a ⋮ p n − 1 b − p n − 1 a a a a ⋮ a p n − a ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ and hence Δ n = ( p n − a ) Δ n − 1 + ( p n − 1 − b ) ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ p 1 b b b ⋮ b b a p 2 b b ⋮ b b a a p 3 b ⋮ b b a a a p 4 ⋮ b b . . . . . . . . . . . . . . . . . . a a a a ⋮ p n − 2 b a a a a ⋮ a a ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ where the last determinant is of an ( n − 1 ) × ( n − 1 ) matrix. Further row operations (subtracting the last row from all the others) give ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ p 1 b b b ⋮ b b a p 2 b b ⋮ b b a a p 3 b ⋮ b b a a a p 4 ⋮ b b . . . . . . . . . . . . . . . . . . a a a a ⋮ p n − 2 b a a a a ⋮ a a ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ = ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ p 1 − b 0 0 0 ⋮ 0 b a − b p 2 − b 0 0 ⋮ 0 b a − b a − b p 3 − b 0 ⋮ 0 b a − b a − b a − b p 4 − b ⋮ 0 b . . . . . . . . . . . . . . . . . . a − b a − b a − b a − b ⋮ p n − 2 − b b 0 0 0 0 ⋮ 0 a ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ = a f n − 2 ( b ) so we have the recurrence relation Δ n = ( p n − a ) Δ n − 1 + a f n − 1 ( b ) It is now an easy induction to show that Δ n = b − a b f n ( a ) − a f n ( b ) for all n ≥ 1 , making the answer a + b .