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If $a,b,c$ are in A.P with common difference d and $\begin{vmatrix}x+1&x+a&x+b\\x+a&x+b&x+c\\x-b+1&x-1&x-a+c\end{vmatrix}$ has absolute value 2 then d is

$\begin{array}{1 1}(A)\;a&(B)\;-2\\(C)\;\pm 1&(D)\;\text{None of these}\end{array}$

1 Answer

Let $a=1+d$
$\Delta =\begin{vmatrix}x+1&x+1+d&x+1+2d\\x+1+d&x+1+2d&x+1+3d\\x-2d&x-1&x+2d\end{vmatrix}$
Applying $C_2-C_1=C_2,C_3-C_2=C_3$
$\begin{vmatrix} x+1&d&d\\x+1+d&d&d\\x-2d&2d-1&2d+1\end{vmatrix}$
$\Rightarrow \begin{vmatrix}x+1&d&d\\d&0&0\\-2d-1&d-1&d+1\end{vmatrix}$
$\Rightarrow -d^2[d+1-(d-1)]=-2d^2$
$\therefore|\Delta |=2d^2$
$\Rightarrow 2$(Given)
$\Rightarrow d=\pm 1$
Hence (C) is the correct answer.
answered Apr 23, 2014 by sreemathi.v

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