# Let a, b ,c, d be in

Question:

Let $\mathrm{a}, \mathrm{b}, \mathrm{c}, \mathrm{d}$ 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

Solution:

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

$\mathrm{C}_{2} \rightarrow \mathrm{C}_{2}-\mathrm{C}_{3}$

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

$\mathrm{R}_{2} \rightarrow \mathrm{R}_{2}-\mathrm{R}_{1}, \quad \mathrm{R}_{3} \rightarrow \mathrm{R}_{3}-\mathrm{R}_{1}$

$\Rightarrow \lambda\left|\begin{array}{ccc}x-2 \lambda & 1 & x+a \\ 2 \lambda-1 & 0 & \lambda \\ 4 \lambda & 0 & 2 \lambda\end{array}\right|=2$

$\Rightarrow 1\left(4 \lambda^{2}-4 \lambda^{2}+2 \lambda\right)=2$

$\Rightarrow \lambda^{2}=1$

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