Join the Most Relevant JEE Main 2025 Test Series & get 99+ percentile! Join Now
Search any question & find its solution
Question: Answered & Verified by Expert
$\left|\begin{array}{ccc}a-b-c & 2 a & 2 a \\ 2 b & b-c-a & 2 b \\ 2 c & 2 c & c-a-b\end{array}\right|$ is equal to
MathematicsDeterminantsAP EAMCETAP EAMCET 2008
Options:
  • A $0$
  • B $a+b+c$
  • C $(a+b+c)^2$
  • D $(a+b+c)^3$
Solution:
1649 Upvotes Verified Answer
The correct answer is: $(a+b+c)^3$
Let $\Delta=\left[\begin{array}{ccc}a-b-c & 2 a & 2 a \\ 2 b & b-c-a & 2 b \\ 2 c & 2 c & c-a-b\end{array}\right]$
Applying $\quad R_1 \rightarrow R_1+R_2+R_3 \quad$ and taking common $(a+b+c)$ from $R_1$
$$
=(a+b+c)\left|\begin{array}{ccc}
1 & 1 & 1 \\
2 b & b-c-a & 2 b \\
2 c & 2 c & c-a-b
\end{array}\right|
$$
Applying $C_2 \rightarrow C_2-C_1$ and $C_3 \rightarrow C_3-C_1$,
$$
\begin{aligned}
& =(a+b+c)\left|\begin{array}{ccc}
1 & 0 & 0 \\
2 b & -b-c-a & 0 \\
2 c & 0 & -a-b-c
\end{array}\right| \\
& =(a+b+c)[(-b-c-a)(-a-b-c)] \\
& =(a+b+c)^3
\end{aligned}
$$

Looking for more such questions to practice?

Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.