JEE MAIN - Mathematics (2018 (Offline) - No. 5)
If $$\left| {\matrix{
{x - 4} & {2x} & {2x} \cr
{2x} & {x - 4} & {2x} \cr
{2x} & {2x} & {x - 4} \cr
} } \right| = \left( {A + Bx} \right){\left( {x - A} \right)^2}$$
then the ordered pair (A, B) is equal to :
then the ordered pair (A, B) is equal to :
(4, 5)
(-4, -5)
(-4, 3)
(-4, 5)
Explanation
$$\left| {\matrix{
{x - 4} & {2x} & {2x} \cr
{2x} & {x - 4} & {2x} \cr
{2x} & {2x} & {x - 4} \cr
} } \right|$$
Applying c1 $$ \to $$ c1 + c2 + c3
$$ = \,\,\,\,\left| {\matrix{ {5x - 4} & {2x} & {2x} \cr {5x - 4} & {x - 4} & {2x} \cr {5x - 4} & {2x} & {x - 4} \cr } } \right|$$
Taking common (5x $$-$$ 4) from c1
$$ = \,\,\,\,\left( {5x - 4} \right)\left| {\matrix{ 1 & {2x} & {2x} \cr 1 & {x - 4} & {2x} \cr 1 & {2x} & {x - 4} \cr } } \right|$$
Apply R2 $$ \to $$R2 $$-$$ R1 and R3 $$ \to $$R3 $$-$$ R1
$$ = \,\,\,\,\left( {5x - 4} \right)\left| {\matrix{ 1 & {2x} & {2x} \cr 0 & { - \left( {x + 4} \right)} & 0 \cr 0 & 0 & { - \left( {x + 4} \right)} \cr } } \right|$$
$$ = \,\,\,\,\left( {5x - 4} \right){\left( {x + 4} \right)^2}$$
So, (A + Bx) (x $$-$$ A)2 = (5x $$-$$ 4) (x + 4)2
By comparing both sides we get, A = $$-$$ 4 and B = 5
Applying c1 $$ \to $$ c1 + c2 + c3
$$ = \,\,\,\,\left| {\matrix{ {5x - 4} & {2x} & {2x} \cr {5x - 4} & {x - 4} & {2x} \cr {5x - 4} & {2x} & {x - 4} \cr } } \right|$$
Taking common (5x $$-$$ 4) from c1
$$ = \,\,\,\,\left( {5x - 4} \right)\left| {\matrix{ 1 & {2x} & {2x} \cr 1 & {x - 4} & {2x} \cr 1 & {2x} & {x - 4} \cr } } \right|$$
Apply R2 $$ \to $$R2 $$-$$ R1 and R3 $$ \to $$R3 $$-$$ R1
$$ = \,\,\,\,\left( {5x - 4} \right)\left| {\matrix{ 1 & {2x} & {2x} \cr 0 & { - \left( {x + 4} \right)} & 0 \cr 0 & 0 & { - \left( {x + 4} \right)} \cr } } \right|$$
$$ = \,\,\,\,\left( {5x - 4} \right){\left( {x + 4} \right)^2}$$
So, (A + Bx) (x $$-$$ A)2 = (5x $$-$$ 4) (x + 4)2
By comparing both sides we get, A = $$-$$ 4 and B = 5
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