Q75 / 180 · COMEDK 2020 · Mathematics

The value of $\left|\begin{array}{lll}x & p & q \\ p & x & q \\ p & q & x\end{array}\right|$ is

  • Correct answer
    A
    $(x-p)(x-q)(x+p+q)$
  • B
    $x(x-p)(x-q)$
  • C
    $p q(x-p)(x-q)$
  • D
    $(p-q)(x-q)(x-p)$

Solution

We have $$ \left[\begin{array}{lll} x & p & q \\ p & x & q \\ p & q & x \end{array}\right] $$ Applying $C_{1} \rightarrow C_{1}+C_{2}+C_{33}$, we get $$ \left|\begin{array}{lll} x+p+q & p & q \\ x+p+q & x & q \\ x+p+q & q & x \end{array}\right| $$ Taking common $(x+p+q)$ from $C_{1}$, we get $$ (x+p+q)\left|\begin{array}{lll} 1 & p & q \\ 1 & x & q \\ 1 & q & k \end{array}\right| $$ Applying $R_{1} \rightarrow R_{1}-R_{2}, R_{2} \rightarrow R_{2}-R_{3}$, we get $$ \begin{aligned} &(x+p+q)\left|\begin{array}{ccc} 0 & p-x & 0 \\ 0 & x-q & q-x \\ 1 & q & x \end{array}\right| \\ &=(x+p+q) \cdot 1[(p-x)(q-x)-0] \\ &=\{(x-p)(x-q)(x+p+q) \end{aligned} $$

Hence, the correct answer is option: 'A'.

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