First slide

Introduction to Determinants

Question

If $Δ (x) = |\begin{array}{ccc} 1 & 1 & 1 \\ {(e^{x} + e^{- x})}^{2} & {(π^{x} + π^{- x})}^{2} & 2 \\ {(e^{x} - e^{- x})}^{2} & {(π^{x} - π^{- x})}^{2} & - 2 \end{array}|$ then $∆ (x)$ equal to

Moderate

A

$x^{2}$
B

$x^{2} - 1$
C

$e^{x^{2}} - π^{x^{2}}$
D

0

Solution

Using $R_{2} \to R_{2} - R_{3}$ and ${(a^{x} + a^{- x})}^{2} - {(a^{x} - a^{- x})}^{2}$ $= 4 a^{x} a^{- x} = 4$ we get
$Δ (x) = |\begin{array}{ccc} 1 & 1 & 1 \\ 4 & 4 & 4 \\ {(e^{x} - e^{- x})}^{2} & {(π^{x} - π^{- x})}^{2} & - 2 \end{array}| = 0$
[since R₁ and R₂ are proportional]

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