Question

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

• A. $abc$
• B. $x^2$
• C. $x^3$
• D. None of these

Answer 1

Answer: (B)

$\Delta =\frac{1}{a}\left|\begin{array}{ccc}{a}^3+ax& ab& ac\\ a^{2}b& b^2+x& bc\\ a^{2}c& bc& c^2+x\end{array}\right|$
Applying $C_1\to C_1+b{C}_2+c{C}_3$ and taking $a^2+b^2+c^2+x$ common,
$\Delta =\frac{1}{a}\left(a^2+b^2+c^2+x\right)\left|\begin{array}{ccc}a& ab& ac\\ b& b^2+x& bc\\ c& bc& c^2+x\end{array}\right|$
Applying $C_2\to C_2-b{C}_1$ and $C_3\to C_3-c{C}_1$ we get
$\Delta =\frac{1}{a}\left(a^2+b^2+c^2+x\right)\left|\begin{array}{ccc}a& 0& 0\\ b& x& 0\\ c& 0& x\end{array}\right|$
$=\frac{1}{a}\left(a^2+b^2+c^2+x\right)\left(a{x}^2\right)=x^2\left(a^2+b^2+c^2+x\right)$
Thus, $\Delta$ is divisible by $x$ and $x^2$.