Question

x, y, z are in AP with common difference d.

$\left|\begin{array}{ccc}x+y& y& 1\\ y+z& z& 1\\ z+x& x& 1\end{array}\right|$ =

• A. 2 $d^2$
• B. -3$d^2$
• C. - $d^2$
• D. d

Answer 1

The correct option is B -3$d^2$

You can use the property of the determinant that if the terms of one row is given as a sum of two other terms then the determinant can be written as the sum of two determinants as given below,

$\left|\begin{array}{ccc}x+y& y& 1\\ y+z& z& 1\\ z+x& x& 1\end{array}\right|$ = $\left|\begin{array}{ccc}x& y& 1\\ y& z& 1\\ z& x& 1\end{array}\right|$ + $\left|\begin{array}{ccc}y& y& 1\\ z& z& 1\\ x& x& 1\end{array}\right|$

$\left|\begin{array}{ccc}x& y& 1\\ y& z& 1\\ z& x& 1\end{array}\right|$ + 0

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

= $\left|\begin{array}{ccc}x& y& 1\\ y-x& z-y& 0\\ z-x& x-y& 0\end{array}\right|$

= $\left|\begin{array}{ccc}x& y& 1\\ d& d& 0\\ 2d& -d& 0\end{array}\right|$

= + 1$(-d^2-2d^2)$
= - $3d^2$