Q45 / 160 · AP EAMCET 2024 (21 May Shift 2) · Mathematics
$\left|\begin{array}{ccc}
a+b+2 c & a & b \\
c & b+c+2 a & b \\
c & a & c+a+2 b
\end{array}\right|=$
- A$(a+b+c)^3$
- Correct answerB$2(a+b+c)^3$
- C$3(a+b+c)^3$
- D$(a+b+c)$
Solution
$\Delta=\left|\begin{array}{ccc}a+b+2 c & a & b \\ c & b+c+2 a & b \\ c & a & c+a+2 b\end{array}\right|$
Applying, $C_1 \rightarrow C_1+C_2+C_3$, we get $\Rightarrow \Delta=2(a+b+c)\left|\begin{array}{ccc} 1 & a & b \\ 1 & 2 a+b+c & b \\ 1 & a & a+2 b+c \end{array}\right|$
Applying, $R_2 \rightarrow R_2-R_1 ; R_3 \rightarrow R_3-R_1$
$\begin{aligned} & \Rightarrow \Delta=2(a+b+c)\left|\begin{array}{ccc}1 & a & b \\ 0 & a+b+c & 0 \\ 0 & 0 & a+b+c\end{array}\right| \\ & \Rightarrow \Delta=2(a+b+c)^3 .\end{aligned}$
Applying, $C_1 \rightarrow C_1+C_2+C_3$, we get $\Rightarrow \Delta=2(a+b+c)\left|\begin{array}{ccc} 1 & a & b \\ 1 & 2 a+b+c & b \\ 1 & a & a+2 b+c \end{array}\right|$
Applying, $R_2 \rightarrow R_2-R_1 ; R_3 \rightarrow R_3-R_1$
$\begin{aligned} & \Rightarrow \Delta=2(a+b+c)\left|\begin{array}{ccc}1 & a & b \\ 0 & a+b+c & 0 \\ 0 & 0 & a+b+c\end{array}\right| \\ & \Rightarrow \Delta=2(a+b+c)^3 .\end{aligned}$
Hence, the correct answer is option: 'B'.