Question

Determinant $ \Delta= \begin{vmatrix}a^{2}+x&ab&ac\\ ab&b^{2}+x&bc\\ ac&bc&c^{2}+x\end{vmatrix}$ is divisible by

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

Answer 1

Answer: (B)

$ \Delta=\frac{1}{a} \begin{vmatrix}a^{3}+ax&ab&ac\\ a^{2}b&b^{2}+x&bc\\ a^{2}c&bc&c^{2}+x\end{vmatrix}$
Applying $C_{1} \rightarrow C_{1}+b C_{2}+c C_{3}$ and taking $a^{2}+b^{2}+c^{2}+x$ common,
$ \Delta=\frac{1}{a} \left(a^{2}+b^{2}+c^{2}+x\right)\begin{vmatrix}a&ab&ac\\ b&b^{2}+x&bc\\ c&bc&c^{2}+x\end{vmatrix}$
Applying $C_{2} \rightarrow C_{2}-b C_{1}$ and $C_{3} \rightarrow C_{3}-c C_{1}$ we get
$ \Delta=\frac{1}{a} \left(a^{2}+b^{2}+c^{2}+x\right)\begin{vmatrix}a&0&0\\ b&x&0\\ c&0&x\end{vmatrix}$
$=\frac{1}{a}\left(a^{2}+b^{2}+c^{2}+x\right)\left(a x^{2}\right)=x^{2}\left(a^{2}+b^{2}+c^{2}+x\right)$
Thus, $\Delta$ is divisible by $x$ and $x^{2}$.