Find the determinant: determinant ((5,3,0,0,…,0,0),(2,5,3,0,…,0,0),(0,2,5,3,…,0,0),(⋮,⋮,⋮,⋮,⋱,⋮,⋮),(0,0,0,0,…,5,3),(0,0,0,0,…,2,5))


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Question Number 205156 by depressiveshrek last updated on 11/Mar/24

$Find the determinant :$ $\| \begin{matrix} 5 & 3 & 0 & 0 & \dots & 0 & 0 \\ 2 & 5 & 3 & 0 & \dots & 0 & 0 \\ 0 & 2 & 5 & 3 & \dots & 0 & 0 \\ ⋮ & ⋮ & ⋮ & ⋮ & ⋱ & ⋮ & ⋮ \\ 0 & 0 & 0 & 0 & \dots & 5 & 3 \\ 0 & 0 & 0 & 0 & \dots & 2 & 5 \end{matrix} \|$
	Answered by pi314 last updated on 11/Mar/24
	$Δ_n = determinant (((5 3 0 0......0 0)),((2 5 3 0......0 0)),((0 2 5 3......0 0)),((..................5 3)),((0 0 0 0....... 2 5))) Δ_n = 5.Δ_(n−1) −2.3Δ_(n−2) Δ_n =5Δ_(n−1) −6Δ_(n−2) Δ_2 = determinant (((5 3)),((2 5)))=19 Δ_1 =5 Δ_3 = determinant (((5 3 0)),((2 5 3)),((0 2 5)),())=5∣19∣−30=65 X^2 −5X+6=0⇒(X−3)(X−2)=0;X∈{2,3} Δ_n =a(2)^n +b(3^n ) 4a+9b=19;2a+3b=5 3b+10=19⇒b=3;a=−2 Δ_n =−2^(n+1) +3^(n+1) ;Δ_3 =−(2)^4 +(3)^4 =81−16=65 Δ_4 = determinant (((5 3 0 0)),((2 5 3 0)),((0 2 5 3)),((0 0 2 5)))=5.65−6.19=211=−(2)^5 +3^5 ..True Δ_n =−(2)^(n+1) +(3)^(n+1)$
	$Δ_{n} = \| \begin{matrix} 5 3 0 0 . . . . . . 0 0 \\ 2 5 3 0 . . . . . . 0 0 \\ 0 2 5 3 . . . . . . 0 0 \\ . . . . . . . . . . . . . . . . . . 5 3 \\ 0 0 0 0 . . . . . . . 2 5 \end{matrix} \|$ $Δ_{n} = 5 . Δ_{n - 1} - 2.3 Δ_{n - 2}$ $Δ_{n} = 5 Δ_{n - 1} - 6 Δ_{n - 2}$ $Δ_{2} = \| \begin{matrix} 5 3 \\ 2 5 \end{matrix} \| = 19$ $Δ_{1} = 5$ $Δ_{3} = \| \begin{matrix} 5 3 0 \\ 2 5 3 \\ 0 2 5 \end{matrix} \| = 5 ∣ 19 ∣ - 30 = 65$ $X^{2} - 5 X + 6 = 0 \Rightarrow (X - 3) (X - 2) = 0; X \in {2, 3}$ $Δ_{n} = a {(2)}^{n} + b (3^{n})$ $4 a + 9 b = 19; 2 a + 3 b = 5$ $3 b + 10 = 19 \Rightarrow b = 3; a = - 2$ $Δ_{n} = - 2^{n + 1} + 3^{n + 1}; Δ_{3} = - {(2)}^{4} + {(3)}^{4} = 81 - 16 = 65$ $Δ_{4} = \| \begin{matrix} 5 3 0 0 \\ 2 5 3 0 \\ 0 2 5 3 \\ 0 0 2 5 \end{matrix} \| = 5.65 - 6.19 = 211 = - {(2)}^{5} + 3^{5} . . T r u e$ $Δ_{n} = - {(2)}^{n + 1} + {(3)}^{n + 1}$
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