First slide

Introduction to Determinants

Question

Suppose $a, b$ are two real numbers and $f (n) = α^{n} + β^{n}$ . Let
$Δ = |\begin{array}{ccc} 3 & 1 + f (1) & 1 + f (2) \\ 1 + f (1) & 1 + f (2) & 1 + f (3) \\ 1 + f (2) & 1 + f (3) & 1 + f (4) \end{array}|$
If $Δ = k (α - 1)^{2} (β - 1)^{2} (α - β)^{2},$ then $k$ is equal to

Moderate

A

1
B

$4 α β$
C

9
D

$α^{2} β^{2}$

Solution

$\begin{array}{l} Δ = |\begin{array}{ccc} 1 + 1 + 1 & 1 + α + β & 1 + α^{2} + β^{2} \\ 1 + α + β & 1 + α^{2} + β^{2} & 1 + α^{3} + β^{3} \\ 1 + α^{2} + β^{2} & 1 + α^{3} + β^{3} & 1 + α^{4} + β^{4} \end{array}| \\ = |\begin{array}{ccc} 1 & 1 & 1 \\ 1 & α & β \\ 1 & α^{2} & β^{2} \end{array}| |\begin{array}{ccc} 1 & 1 & 1 \\ 1 & α & β \\ 1 & α^{2} & β^{2} \end{array}| = Δ_{1}^{2} \end{array}$
where $Δ_{1} = |\begin{array}{ccc} 1 & 1 & 1 \\ 1 & α & β \\ 1 & α^{2} & β^{2} \end{array}|$
Applying $C_{3} \to C_{3} - C_{2}$ and $C_{2} \to C_{2} - C_{1}$ we get
$\begin{aligned} Δ_{1} = |\begin{array}{ccc} 1 & 0 & 0 \\ 1 & α - 1 & β - α \\ 1 & α^{2} - 1 & β^{2} - α^{2} \end{array}| \\ = (α - 1) (β - α) |\begin{array}{cc} 1 & 1 \\ α + 1 & β + α \end{array}| \end{aligned}$
$= (α - 1) (β - 1) (β - α)$
Thus $Δ = (α - 1)^{2} (β - 1)^{2} (α - β)^{2}$
$∴ k = 1$

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