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The value of ...

Question

The value of determinant $Δ = ∣ ∣ ∣ ∣ ∣ ∣ ∣ \begin{matrix} sin (n π) & tan (2 n π) & cos ((2 n + 1) \frac{π}{2}) 1 & 0 & - 1 ln (7) & cos (\frac{π}{4}) & - 2 \end{matrix} ∣ ∣ ∣ ∣ ∣ ∣ ∣$ is

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Solution

For, $n \in I$
$sin (n π) = 0$
$tan (2 n π) = 0 and cos ((2 n + 1) \frac{π}{2}) = 0$
So, $Δ$ will become
$Δ = ∣ ∣ ∣ ∣ ∣ ∣ \begin{matrix} 0 & 0 & 0 1 & 0 & - 1 ln (7) & cos (\frac{π}{4}) & - 2 \end{matrix} ∣ ∣ ∣ ∣ ∣ ∣$
As all the elements of $R_{1}$ is $0.$
Hence, the value of $Δ = 0$

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