Question

The number of ordered triplets $(x,y,z),x,y,z\in I$, satisfying the equation $\left|D\right|=8$, where
$D=\left[\begin{array}{ccc}y+z& z& y\\ z& z+x& x\\ y& x& x+y\end{array}\right]$ is

• A. $3$
• B. $8$
• C. $16$
• D. $12$

Answer 1

The correct option is A $3$

Applying $R_1\to R_1-R_2-R_3\Rightarrow \left|\begin{array}{ccc}0& -2x& -2x\\ z& z+x& x\\ y& x& x+y\end{array}\right|$

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

Now, $c_2\to c_2-c_3=(-2x)\left|\begin{array}{ccc}0& 0& 1\\ z& z& x\\ y& -y& x+y\end{array}\right|$

Expanding along $R_1$

$=(-2x)(-2yz)=4xyz$

Given $|D|=8$

$\Rightarrow |4xyz|=8$

$\Rightarrow |xyz|=2$

Only possibilities are $(1,1,2),(1,2,1),(2,1,1)$

$\Rightarrow 3$

$\Rightarrow$ Option (A).