Mathematics - Determinant Question with Solution | TestHub

Since, $a, b, c \& d$ are in $\mathrm{A}.\mathrm{P}.$, therefore

$b-a=c-b=d-c=\lambda$

We have,

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

$C_2\to C_2-C_3$

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

$R_2\to R_2-R_1, R_3\to R_3-R_1$

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

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

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

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