JEE MAIN - Mathematics (2021 - 26th February Morning Shift - No. 10)
The value of $$\left| {\matrix{
{(a + 1)(a + 2)} & {a + 2} & 1 \cr
{(a + 2)(a + 3)} & {a + 3} & 1 \cr
{(a + 3)(a + 4)} & {a + 4} & 1 \cr
} } \right|$$ is :
$$-$$2
0
(a + 2)(a + 3)(a + 4)
(a + 1)(a + 2)(a + 3)
Explanation
Given, $$\Delta $$ = $$\left| {\matrix{
{(a + 1)(a + 2)} & {a + 2} & 1 \cr
{(a + 2)(a + 3)} & {a + 3} & 1 \cr
{(a + 3)(a + 4)} & {a + 4} & 1 \cr
} } \right|$$
R2 $$ \to $$ R2 $$-$$ R1 and R3 $$ \to $$ R3 $$-$$ R1
$$\Delta$$ = $$\left| {\matrix{ {(a + 1)(a + 2)} & {a + 2} & 1 \cr {(a + 2)(a + 3 - a - 1)} & 1 & 0 \cr {{a^2} + 7a + 12 - {a^2} - 3a - 2} & 2 & 0 \cr } } \right|$$
$$ = \left| {\matrix{ {{a^2} + 3a + 2} & {a + 2} & 1 \cr {2(a + 2)} & 1 & 0 \cr {4a + 10} & 2 & 0 \cr } } \right|$$
$$ = 4(a + 2) - 4a - 10$$
$$ = 4a + 8 - 4a - 10 = - 2$$
R2 $$ \to $$ R2 $$-$$ R1 and R3 $$ \to $$ R3 $$-$$ R1
$$\Delta$$ = $$\left| {\matrix{ {(a + 1)(a + 2)} & {a + 2} & 1 \cr {(a + 2)(a + 3 - a - 1)} & 1 & 0 \cr {{a^2} + 7a + 12 - {a^2} - 3a - 2} & 2 & 0 \cr } } \right|$$
$$ = \left| {\matrix{ {{a^2} + 3a + 2} & {a + 2} & 1 \cr {2(a + 2)} & 1 & 0 \cr {4a + 10} & 2 & 0 \cr } } \right|$$
$$ = 4(a + 2) - 4a - 10$$
$$ = 4a + 8 - 4a - 10 = - 2$$
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