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

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