$\left|\begin{array}{ccc}a-b-c & 2 a & 2 a \\ 2 b & b-c-a & 2 b \\ 2 c & 2 c & c-a-b\end{array}\right|$ is…
$\left|\begin{array}{ccc}a-b-c & 2 a & 2 a \\ 2 b & b-c-a & 2 b \\ 2 c & 2 c & c-a-b\end{array}\right|$ is equal to
- $0$
- $a+b+c$
- $(a+b+c)^2$
- $(a+b+c)^3$
Solution
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}
$
Asked in: AP EAMCET 2008
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