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  • Let a, b,c,d be in arithmetic progression with common difference . If <img alt="\left|\begin{array}{lll} x+a-c & x+b & x+a \\ x-1 & x+c & x+b \\ x-b+d & x+d & x+c \end{array}\ri

Let a, b,c,d be in arithmetic progression with common difference \lambda. If \left|\begin{array}{lll} x+a-c & x+b & x+a \\ x-1 & x+c & x+b \\ x-b+d & x+d & x+c \end{array}\right|=2 then value of \lambda ^{2} is equal to __________
 

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best_answer

As a,b,c,d are in AP with common difference = \lambda

So,

\begin{aligned} &b-a=c-b=d-b=2 \\ &c-a=d-b=22 \end{aligned}

So determinant is

\left|\begin{array}{lll} x-2 \lambda & x+b & x+a \\ x-1 & x+c & x+b \\ x+2 \lambda & x+d & x+c \end{array}\right|

R_{2} \rightarrow R_{2}-R_{1}, R_{3} \rightarrow R_{3}-R_{2}

=\left|\begin{array}{ccc} x-2 \lambda & x+b & x+a \\ 2 \lambda-1 & \lambda & \lambda \\ 2 \lambda+1 & \lambda & \lambda \end{array}\right|

\begin{aligned} &c_{3} \rightarrow c_{3}-c_{2} \\ &=\left|\begin{array}{ccc} x-2 \lambda & x+b & -\lambda \\ 2 \lambda-1 & \lambda & 0 \\ 2 \lambda+1 & \lambda & 0 \end{array}\right| \end{aligned}

Expanding along column 3

\begin{aligned} &-\lambda((2 \lambda-1) \lambda-\lambda(2 \lambda+1))=2 \text { (given) } \\ &\Rightarrow-\lambda^{2}(-2)=2 \\ &\Rightarrow \lambda^{2}=1 \end{aligned}

Hence, the correct answer is 1

Posted by

Kuldeep Maurya

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