First slide

Introduction to Determinants

Question

If $x$ is a positive integer, and $∆ (x) =$ $|\begin{array}{ccc} x! & (x + 1)! & (x + 2)! \\ (x + 1)! & (x + 2)! & (x + 3)! \\ (x + 2)! & (x + 3)! & (x + 4)! \end{array}|$ , then $∆ (x)$ is equal to

Moderate

A

$2 x! (x + 1)!$
B

$2 x! (x + 1)! (x + 2)!$
C

$2 x! (x + 3)!$
D

$2 (x + 1)! (x + 2)! (x + 3)!$

Solution

Taking $x!$ common from $R_{1}, (x + 1)!$ from $R_{2}$ and $(x + 2)!$ from $R_{3}$ to obtain
$Δ (x) = x! (x + 1)! (x + 2)! Δ_{1} (x)$ where
$Δ_{1} (x) = |\begin{array}{lll} 1 x + 1 (x + 1) (x + 2) \\ 1 x + 2 (x + 2) (x + 3) \\ 1 x + 3 (x + 3) (x + 4) \end{array}|$
Applying $R_{3} \to R_{3} - R_{2}, R_{2} \to R_{2} - R_{1},$ we get
$Δ_{1} (x) = |\begin{array}{ccc} 1 & x + 1 & (x + 1) (x + 2) \\ 0 & 1 & 2 (x + 2) \\ 0 & 1 & 2 (x + 3) \end{array}| = 2$

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