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Question: Answered & Verified by Expert
The value of $\left|\begin{array}{lll}x & p & q \\ p & x & q \\ p & q & x\end{array}\right|$ is
MathematicsDeterminantsCOMEDKCOMEDK 2020
Options:
  • 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:
2503 Upvotes Verified Answer
The correct answer is: $(x-p)(x-q)(x+p+q)$
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
$$
\begin{aligned}
&(x+p+q)\left|\begin{array}{ccc}
0 & p-x & 0 \\
0 & x-q & q-x \\
1 & q & x
\end{array}\right| \\
&=(x+p+q) \cdot 1[(p-x)(q-x)-0] \\
&=\{(x-p)(x-q)(x+p+q)
\end{aligned}
$$

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