Question

The number of positive integral solutions of the equation $\Delta =12$
where $\Delta =\left|\begin{array}{ccc}y+z& z& y\\ z& z+x& x\\ y& x& x+y\end{array}\right|=0$ is

• A. $3$
• B. $15$
• C. $4$
• D. $2^3{3}^2$

Answer 1

The correct option is A $3$

Applying $C_1\to C_1+C_2+C_3$
$\Delta =2\left|\begin{array}{ccc}y+z& z& y\\ z+x& z+x& x\\ x+y& x& x+y\end{array}\right|=0$
Applying $C_2\to C_2-C_1,C_3\to C_3-C_1$
$\Delta =2\left|\begin{array}{ccc}y+z& -y& -z\\ z+x& 0& -z\\ x+y& -y& 0\end{array}\right|=2\left|\begin{array}{ccc}0& -y& -z\\ x& 0& -z\\ x& -y& 0\end{array}\right|=2xyz\left|\begin{array}{ccc}0& 1& 1\\ 1& 0& 1\\ 1& 1& 0\end{array}\right|$
Again applying $C_1\to C_1+C_2+C_3$
$\Delta =2xyz\left|\begin{array}{ccc}2& 1& 1\\ 2& 0& 1\\ 2& 1& 0\end{array}\right|$
Taking $2$ common and applying $C_2\to C_2-C_1,C_3\to C_3-C_1$
$\Delta =4xyz\left|\begin{array}{ccc}1& 0& 0\\ 1& -1& 0\\ 1& 0& -1\end{array}\right|=4xyz$
As $4xyz=12\Rightarrow xyz=3$
The number $3$ can be assigned to any of $x,y,z$. Therefore the number of positive integral solution of $\Delta =12$ is $3$