JEE MAIN - Mathematics (2021 - 20th July Morning Shift - No. 21)
Let a, b, c, d in arithmetic progression with common difference $$\lambda$$. If $$\left| {\matrix{
{x + a - c} & {x + b} & {x + a} \cr
{x - 1} & {x + c} & {x + b} \cr
{x - b + d} & {x + d} & {x + c} \cr
} } \right| = 2$$, then value of $$\lambda$$2 is equal to ________________.
Answer
1
Explanation
$$\left| {\matrix{
{x + a - c} & {x + b} & {x + a} \cr
{x - 1} & {x + c} & {x + b} \cr
{x - b + d} & {x + d} & {x + c} \cr
} } \right| = 2$$
$${C_2} \to {C_2} - {C_3}$$
$$ \Rightarrow \left| {\matrix{ {x - 2\lambda } & \lambda & {x + a} \cr {x - 1} & \lambda & {x + b} \cr {x + 2\lambda } & \lambda & {x + c} \cr } } \right| = 2$$
$${R_2} \to {R_2} - {R_1},{R_3} \to {R_3} - {R_1}$$
$$ \Rightarrow \left| {\matrix{ {x - 2\lambda } & 1 & {x + a} \cr {2\lambda - 1} & 0 & \lambda \cr {4\lambda } & 0 & {2\lambda } \cr } } \right| = 2$$
$$ \Rightarrow 1(4{\lambda ^2} - 4{\lambda ^2} + 2\lambda ) = 2$$
$$ \Rightarrow {\lambda ^2} = 1$$
$${C_2} \to {C_2} - {C_3}$$
$$ \Rightarrow \left| {\matrix{ {x - 2\lambda } & \lambda & {x + a} \cr {x - 1} & \lambda & {x + b} \cr {x + 2\lambda } & \lambda & {x + c} \cr } } \right| = 2$$
$${R_2} \to {R_2} - {R_1},{R_3} \to {R_3} - {R_1}$$
$$ \Rightarrow \left| {\matrix{ {x - 2\lambda } & 1 & {x + a} \cr {2\lambda - 1} & 0 & \lambda \cr {4\lambda } & 0 & {2\lambda } \cr } } \right| = 2$$
$$ \Rightarrow 1(4{\lambda ^2} - 4{\lambda ^2} + 2\lambda ) = 2$$
$$ \Rightarrow {\lambda ^2} = 1$$
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