Question

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

Answer 1

Answer: 1

$\left|\begin{array}{ccc}x+a-c& x+b& x+a\\ x-1& x+c& x+b\\ x-b+d& x+d& x+c\end{array}\right|=2$
$C_2\to C_2-C_3$
$\Rightarrow \left|\begin{array}{ccc}x-2\lambda & \lambda & x+a\\ x-1& \lambda & x+b\\ x+2\lambda & \lambda & x+c\end{array}\right|=2$
$R_2\to R_2-R_1,R_3\to R_3-R_1$
$\Rightarrow \lambda \left|\begin{array}{ccc}x-2\lambda & 1& x+a\\ 2\lambda -1& 0& \lambda \\ 4\lambda & 0& 2\lambda \end{array}\right|=2$
$\Rightarrow 1\left(4{\lambda }^2-4{\lambda }^2+2\lambda \right)=2$
$\Rightarrow {\lambda }^2=1$