Question

The determinant $\left|\begin{array}{ccc}x& y& x+y\\ y& x+y& x\\ x+y& x& y\end{array}\right|$ is divisible by

• A. $x-y$
• B. $x^2-y^2+xy$
• C. $x^2+xy+y^2$
• D. $x^2-xy+y^2$

Answer 1

The correct option is D

$x^2-xy+y^2$

Explanation for the correct answer:

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

Replace $R_1$ by $R_1+R_2+R_3$ we have,

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

$=\left|\begin{array}{ccc}2x+2y& 2x+2y& 2x+2y\\ y& x+y& x\\ x+y& x& y\end{array}\right|$

Taking common factor $2\left(x+y\right)$ from $R_1$, we have

$∆=2\left(x+y\right)\left|\begin{array}{ccc}1& 1& 1\\ y& x+y& x\\ x+y& x& y\end{array}\right|$

Noe replace $C_2$by $C_2-C_1$and $C_3$ by $C_3-C_1$ we have,

$∆=2\left(x+y\right)\left|\begin{array}{ccc}1& 0& 0\\ y& x& x-y\\ x+y& -y& -x\end{array}\right|$

$=2\left(x+y\right)\left(1\left(-x^2-\left(-y\right)\left(x-y\right)\right)+0+0\right)$

$=2\left(x+y\right)\left(-x^2+xy-y^2\right)$

$=-2\left(x+y\right)\left(x^2-xy+y^2\right)$

Hence $∆$ is divisible by $x^2-xy+y^2$

Therefore, Option D is the correct option.