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Determinant

Find the dete...

Question

Find the determinant of given matrix $⎡ ⎢ ⎣ \begin{matrix} a - b - c & 2 a & 2 a 2 b & b - c - a & 2 b 2 c & 2 c & c - a - b \end{matrix} ⎤ ⎥ ⎦$

A

2 (a + b + c)^{3}

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B

(a - b - c)^{3}

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C

2 (a - b - c)^{3}

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D

(a + b + c)^{3}

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Solution

The correct option is D $(a + b + c)^{3}$
$∣ ∣ ∣ ∣ \begin{matrix} a - b - c & 2 a & 2 a 2 b & b - c - a & 2 b 2 c & 2 c & c - a - b \end{matrix} ∣ ∣ ∣ ∣$
$= a - b - c ∣ ∣ ∣ \begin{matrix} b - c - a & 2 b 2 c & c - a - b \end{matrix} ∣ ∣ ∣ - 2 a ∣ ∣ ∣ \begin{matrix} 2 b & 2 b 2 c & c - a - b \end{matrix} ∣ ∣ ∣ + 2 a ∣ ∣ ∣ \begin{matrix} 2 b & b - c - a 2 c & 2 c \end{matrix} ∣ ∣ ∣$
$= a - b - c [(b^{2} - c^{2} - a^{2}) - (4 c b)] - 2 a [(2 b c - 2 b a - 2 b^{2} - 4 c b)] + 2 a [(4 c b - 2 c b - 2 c^{2} - 2 a c)]$
$= (a + b + c)^{3}$ ..... (simplified)

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Similar questions

Q. 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}

∣ ∣ ∣ ∣ \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}

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} .

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