Question

If $a,b,c$ are positive integers, then the determinant $\Delta = \left| \begin{matrix} a^{2} + x & ab & ac \\ ab & b^{2} + x & bc \\ ac & bc & c^{2} + x \end{matrix} \right|$ is divisible by.

• A. $x^{3}$
• B. $x^{2}$
• C. $(a^{2} + b^{2} + c^{2})$
• D. None of these

Answer 1

Answer: (B)

$x^{2}$

Answer 2

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

= $\mathbf{(abc)(ac + ab + bc) = a + b + c}$

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

**{**Applying $R_{1} \rightarrow R_{1} + R_{2} + R_{3}\}$

$\mathbf{= (}\mathbf{a}^{\mathbf{2}}\mathbf{+}\mathbf{b}^{\mathbf{2}}\mathbf{+}\mathbf{c}^{\mathbf{2}}\mathbf{+ x)}\left| \begin{matrix} \mathbf{1} & \mathbf{0} & \mathbf{0} \\ \mathbf{b}^{\mathbf{2}} & \mathbf{x} & \mathbf{0} \\ \mathbf{c}^{\mathbf{2}} & \mathbf{0} & \mathbf{x} \end{matrix} \right|$ $\{$ Applying $\left. \ \begin{array}{r} \mathbf{\&}\mathbf{C}_{\mathbf{2}}\mathbf{\rightarrow}\mathbf{C}_{\mathbf{2}}\mathbf{-}\mathbf{C}_{\mathbf{1}} \\ \mathbf{\&}\mathbf{C}_{\mathbf{3}}\mathbf{\rightarrow}\mathbf{C}_{\mathbf{3}}\mathbf{-}\mathbf{C}_{\mathbf{1}} \end{array} \right\}$

= $\mathbf{x}^{\mathbf{2}}\mathbf{(}\mathbf{a}^{\mathbf{2}}\mathbf{+}\mathbf{b}^{\mathbf{2}}\mathbf{+}\mathbf{c}^{\mathbf{2}}\mathbf{+ x)}$.

Hence $\Delta$ is divisible by $x^{2}$ as well as by x.