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Property 4

Prove that:a-...

Question

Prove that :

$|\begin{matrix} a - b - c & 2 a & 2 a \\ 2 b & b - c - a & 2 b \\ 2 c & 2 c & c - a - b \end{matrix}| = {(a + b + c)}^{3}$

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Solution

$Let LHS = ∆ = |a - b - c 2 a 2 a 2 b b - c - a 2 b 2 c 2 c c - a - b| \Rightarrow ∆ = |a + b + c a + b + c a + b + c 2 b b - c - a 2 b 2 c 2 c c - a - b| [Applying R_{1} \to R_{1} + R_{2} + R_{3}] = (a + b + c) |1 1 1 2 b b - c - a 2 b 2 c 2 c c - a - b| = (a + b + c) |0 1 1 b + c + a b - c - a 2 b 0 2 c c - a - b| [Applying C_{1} \to C_{1} - C_{2}] = (a + b + c) \{(a + b + c) \times |1 1 2 c c - a - b|\} [Expanding along C_{1}] = {(a + b + c)}^{3} = RHS$

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∣ ∣ ∣ ∣ \begin{matrix} a - b - c & 2 a & 2 a 2 b & b - c - a & 2 b 2 c & 2 c & c - a - b \end{matrix} ∣ ∣ ∣ ∣ = {(a + b + c)}^{3}

Q. Prove the following :

∣ ∣ ∣ ∣ \begin{matrix} a - b - c & 2 a & 2 a 2 b & b - c - a & 2 b 2 c & 2 c & c - a - b \end{matrix} ∣ ∣ ∣ ∣ = {(a + b + c)}^{3} .

∣ ∣ ∣ ∣ \begin{matrix} a - b - c & 2 a & 2 a 2 b & b - c - a & 2 b 2 c & 2 c & c - a - b \end{matrix} ∣ ∣ ∣ ∣ = (a + b + c)^{3}

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Q. By using properties of determination, show that

∣ ∣ ∣ ∣ \begin{matrix} a - b - c & 2 a & 2 a 2 b & b - c - a & 2 b 2 c & 2 c & c - a - b \end{matrix} ∣ ∣ ∣ ∣ = {(a + b + c)}^{3}

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

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