Question

The number of positive integral solutions of the equation $\left|\begin{array}{ccc}{x}^3+1& x^{2}y& x^{2}z\\ x{y}^2& y^3+1& y^{2}z\\ x{z}^2& y{z}^2& z^3+1\end{array}\right|=11$ is

• A. 0
• B. 3
• C. 6
• D. 12

Answer 1

Answer: (B)

Multiply $R_1$ by $x;R_2$ by $y$ and $R_3$ by $z$ and divide the determinant by $xyz$
$\frac{1}{xyz}\left|\begin{array}{ccc}{x}^4+x& x^{3}y& x^{3}z\\ x{y}^3& y^4+y& y^{3}z\\ x{z}^3& y{z}^3& z^4+z\end{array}\right|=11$
$=\frac{xyz}{xyz}\left|\begin{array}{ccc}{x}^3+1& x^3& x^3\\ y^3& y^3+1& y^3\\ z^3& z^3& z^3+1\end{array}\right|=11$
use $R_1\to R_1+R_2+R_3$
$D=\left(x^3+y^3+z^3+1\right)\left|\begin{array}{ccc}1& 1& 1\\ y^3& y^3+1& y^3\\ z^3& z^3& z^3+1\end{array}\right|=11$
hence $x^3+y^3+z^3=10$ (as the det. has the value 1)
$(2,1,1),(1,2,1),(1,1,2)\Rightarrow (B)$