Question

If $a,b,c,x$ are positive integers, then $\left|\begin{array}{ccc}{a}^2+x& ab& ac\\ ab& b^2+x& bc\\ ac& bc& c^2+x\end{array}\right|$ is divisible by

• A. $x^2$
• B. $x^3$
• C. $x^4$
• D. $a^2+b^2+c^2$

Answer 1

The correct option is A $x^2$

$\Delta =\left|\begin{array}{ccc}{a}^2+x& ab& ac\\ ab& b^2+x& bc\\ ac& bc& c^2+x\end{array}\right|$

$=\frac{1}{abc}\left|\begin{array}{ccc}{a}^3+ax& a^{2}b& a^{2}c\\ a{b}^2& b^3+bx& b^{2}c\\ a{c}^2& b{c}^2& c^3+cx\end{array}\right|$

$=\left|\begin{array}{ccc}{a}^2+x& a^2& a^2\\ b^2& b^2+x& b^2\\ c^2& c^2& c^2+x\end{array}\right|$

Applying $R_1\to R_1+R_2+R_3,$
$\Delta =\left|\begin{array}{ccc}{a}^2+b^2+c^2+x& a^2+b^2+c^2+x& a^2+b^2+c^2+x\\ b^2& b^2+x& b^2\\ c^2& c^2& c^2+x\end{array}\right|$

$=(a^2+b^2+c^2+x)\left|\begin{array}{ccc}1& 1& 1\\ b^2& b^2+x& b^2\\ c^2& c^2& c^2+x\end{array}\right|$

Applying $C_2\to C_2-C_1$ and $C_3\to C_3-C_1$
$\Delta =(a^2+b^2+c^2+x)\left|\begin{array}{ccc}1& 0& 0\\ b^2& x& 0\\ c^2& 0& x\end{array}\right|$

$=(a^2+b^2+c^2+x)x^2$
$\therefore$ determinant is divisible by $x^2$