JEE MAIN - Mathematics (2019 - 11th January Evening Slot - No. 12)
If $$\left| {\matrix{
{a - b - c} & {2a} & {2a} \cr
{2b} & {b - c - a} & {2b} \cr
{2c} & {2c} & {c - a - b} \cr
} } \right|$$
= (a + b + c) (x + a + b + c)2, x $$ \ne $$ 0,
then x is equal to :
= (a + b + c) (x + a + b + c)2, x $$ \ne $$ 0,
then x is equal to :
–2(a + b + c)
2(a + b + c)
abc
–(a + b + c)
Explanation
$$\left| {\matrix{
{a - b - c} & {2a} & {2a} \cr
{2b} & {b - c - a} & {2b} \cr
{2c} & {2c} & {c - a - b} \cr
} } \right|$$
R1 $$ \to $$ R1 + R2 + R3
$$ = \left| {\matrix{ {a + b + c} & {a + b + c} & {a + b + c} \cr {2b} & {b - c - a} & {2b} \cr {2c} & {2c} & {c - a - b} \cr } } \right|$$
$$ = \left( {a + b + c} \right)\left| {\matrix{ 1 & 0 & 0 \cr {2b} & { - \left( {a + b + c} \right)} & 0 \cr {2c} & {2c} & {c - a - b} \cr } } \right|$$
$$=$$ (a + b + c) (a + b + c)2
$$ \Rightarrow $$ x $$=$$ $$-$$ 2(a + b + c)
R1 $$ \to $$ R1 + R2 + R3
$$ = \left| {\matrix{ {a + b + c} & {a + b + c} & {a + b + c} \cr {2b} & {b - c - a} & {2b} \cr {2c} & {2c} & {c - a - b} \cr } } \right|$$
$$ = \left( {a + b + c} \right)\left| {\matrix{ 1 & 0 & 0 \cr {2b} & { - \left( {a + b + c} \right)} & 0 \cr {2c} & {2c} & {c - a - b} \cr } } \right|$$
$$=$$ (a + b + c) (a + b + c)2
$$ \Rightarrow $$ x $$=$$ $$-$$ 2(a + b + c)
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