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 (integer)
1
Solution
$$\left| {\matrix{
{x + a - c} & {x + b} & {x + a} \cr
{x - 1} & {x + c} & {x + b} \cr
{x - b + d} & {x + d} & {x + c} \cr
} } \right| = 2$$<br><br>${C_2} \to {C_2} - {C_3}$<br><br>$$ \Rightarrow \left| {\matrix{
{x - 2\lambda } & \lambda & {x + a} \cr
{x - 1} & \lambda & {x + b} \cr
{x + 2\lambda } & \lambda & {x + c} \cr
} } \right| = 2$$<br><br>${R_2} \to {R_2} - {R_1},{R_3} \to {R_3} - {R_1}$<br><br>$$ \Rightarrow \left| {\matrix{
{x - 2\lambda } & 1 & {x + a} \cr
{2\lambda - 1} & 0 & \lambda \cr
{4\lambda } & 0 & {2\lambda } \cr
} } \right| = 2$$<br><br>$\Rightarrow 1(4{\lambda ^2} - 4{\lambda ^2} + 2\lambda ) = 2$<br><br>$\Rightarrow {\lambda ^2} = 1$
About this question
Subject: Mathematics · Chapter: Matrices and Determinants · Topic: Types of Matrices and Operations
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