Question

$\left| \begin{matrix} a^{2} + x^{2} & ab & ca \\ ab & b^{2} + x^{2} & bc \\ ca & bc & c^{2} + x^{2} \end{matrix} \right|$ is divisor of.

• A. $a^{2}$
• B. $b^{2}$
• C. $c^{2}$
• D. $x^{2}$

Answer 1

Answer: (D)

$x^{2}$

Answer 2

$\left| \begin{matrix} a^{2} + x^{2} & ab & ca \\ ab & b^{2} + x^{2} & bc \\ ca & bc & c^{2} + x^{2} \end{matrix} \right|$

Multiply $C_{1},C_{2},C_{3}$ by $a,b,c$ respectively and hence divide by abc

∴ $\Delta = \frac{1}{abc}\left| \begin{matrix} a(a^{2} + x^{2}) & ab^{2} & c^{2}a \\ a^{2}b & b(b^{2} + x^{2}) & bc^{2} \\ ca^{2} & b^{2}c & c(c^{2} + x^{2}) \end{matrix} \right|$

Now take out a, b and c common from $R_{1},R_{2}$ and $R_{3}$,

∴ $\Delta = \left| \begin{matrix} a^{2} + x^{2} & b^{2} & c^{2} \\ a^{2} & b^{2} + x^{2} & c^{2} \\ a^{2} & b^{2} & c^{2} + x^{2} \end{matrix} \right|$

Now applying $C_{1} \rightarrow C_{1} + C_{2} + C_{3}$

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

⇒ $\Delta = x^{4}(a^{2} + b^{2} + c^{2} + x^{2})$

Hence, it is divisible by $x^{2}$.