Question

Let $a, b, c, d$ be in arithmetic progression with common difference $\lambda$. If $\begin{vmatrix}x+a-c & x+b & x+a \\ x-1 & x+c & x+b \\ x-b+d & x+d & x+c\end{vmatrix}=2$ then value of $\lambda^{2}$ is equal to _______.

Answer 1

$\begin{vmatrix}x+a-c & x+b & x+a \\ x-1 & x+c & x+b \\ x-b+d & x+d & x+c\end{vmatrix}=2$
$C_{2} \rightarrow C_{2}-C_{3}$
$\Rightarrow\begin{vmatrix}x-2 \lambda & \lambda & x+a \\ x-1 & \lambda & x+b \\ x+2 \lambda & \lambda & x+c\end{vmatrix}=2$
$R_{2} \rightarrow R_{2}-R_{1}, R_{3} \rightarrow R_{3}-R_{1}$
$\Rightarrow \lambda\begin{vmatrix}x-2 \lambda & 1 & x+a \\ 2 \lambda-1 & 0 & \lambda \\ 4 \lambda & 0 & 2 \lambda\end{vmatrix}=2$
$\Rightarrow 1\left(4 \lambda^{2}-4 \lambda^{2}+2 \lambda\right)=2$
$\Rightarrow \lambda^{2}=1$