With out expanding show that the value of the following determinant is zero:
$∣ ∣ ∣ ∣ \begin{matrix} 49 & 1 & 6 39 & 7 & 4 26 & 2 & 3 \end{matrix} ∣ ∣ ∣ ∣$

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Solution

$Δ = ∣ ∣ ∣ ∣ \begin{matrix} 49 & 1 & 6 39 & 7 & 4 26 & 2 & 3 \end{matrix} ∣ ∣ ∣ ∣$
$C_{1} \to C_{1} - 8 C_{3}$
$= ∣ ∣ ∣ ∣ \begin{matrix} 1 & 1 & 6 7 & 7 & 4 2 & 2 & 3 \end{matrix} ∣ ∣ ∣ ∣$
Since $C_{1}$ and $C_{2}$ are identical
$Δ = 0$

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