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If $$\left| {\begin{matrix} {a - b - c} & {2a} & {2a} \\ {2b} & {b - c - a} & {2b} \\ {2c} & {2c} & {c - a - b} \\ \end{matrix} } \right|$$ = (a + b + c) (x + a + b + c)2, x $$ \ne $$ 0, then x is equal to :

JEE · Math · previous-year question

  1. A.–2(a + b + c)correct
  2. B.2(a + b + c)
  3. C.abc
  4. D.–(a + b + c)

Answer

A. –2(a + b + c)

Explanation

$$\left| {\begin{matrix} {a - b - c} & {2a} & {2a} \\ {2b} & {b - c - a} & {2b} \\ {2c} & {2c} & {c - a - b} \\ \end{matrix} } \right|$$ R1 $$ \to $$ R1 + R2 + R3 $$ = \left| {\begin{matrix} {a + b + c} & {a + b + c} & {a + b + c} \\ {2b} & {b - c - a} & {2b} \\ {2c} & {2c} & {c - a - b} \\ \end{matrix} } \right|$$ $$ = \left( {a + b + c} \right)\left| {\begin{matrix} 1 & 0 & 0 \\ {2b} & { - \left( {a + b + c} \right)} & 0 \\ {2c} & {2c} & {c - a - b} \\ \end{matrix} } \right|$$ $$=$$ (a + b + c) (a + b + c)2 $$ \Rightarrow $$ x $$=$$ $$-$$ 2(a + b + c)

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