Evaluate the following determinants without expansion as far as possible.
$∣ ∣ ∣ ∣ \begin{matrix} a - b & b - c & c - a b - c & c - a & a - b c - a & a - b & b - c \end{matrix} ∣ ∣ ∣ ∣ = 0$

True

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False

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Solution

The correct option is A True
$∣ ∣ ∣ ∣ \begin{matrix} a - b & b - c & c - a b - c & c - a & a - b c - a & a - b & b - c \end{matrix} ∣ ∣ ∣ ∣$

Applying $R_{1} \to R_{1} + (R_{2} + R_{3})$

$∣ ∣ ∣ ∣ \begin{matrix} a - b + (b - c + c - a) & b - c + (c - a + a - b) & c - a + (a - b + b - c) b - c & c - a & a - b c - a & a - b & b - c \end{matrix} ∣ ∣ ∣ ∣$

$\Rightarrow ∣ ∣ ∣ ∣ \begin{matrix} 0 & 0 & 0 b - c & c - a & a - b c - a & a - b & b - c \end{matrix} ∣ ∣ ∣ ∣ = 0$

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Similar questions

Using the property of determinants and without expanding, prove that:

∣ ∣ ∣ ∣ \begin{matrix} a - b & b - c & c - a b - c & c - a & a - b c - a & a - b & b - c \end{matrix} ∣ ∣ ∣ ∣

= 0

∣ ∣ ∣ ∣ \begin{matrix} a - b & b - c & c - a b - c & c - a & a - b c - a & a - b & b - c \end{matrix} ∣ ∣ ∣ ∣ = 0

∣ ∣ ∣ ∣ \begin{matrix} b + c & c + a & a + b a + b & b + c & c + a c + a & a + b & b + c \end{matrix} ∣ ∣ ∣ ∣ = 2 ∣ ∣ ∣ ∣ \begin{matrix} a & b & c c & a & b b & c & a \end{matrix} ∣ ∣ ∣ ∣

∣ ∣ ∣ ∣ \begin{matrix} a - b & b - c & c - a b - c & c - a & a - b c - a & a - b & b - c \end{matrix} ∣ ∣ ∣ ∣

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