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Question
Prove the following using properties of determinants :
\[\begin{vmatrix}a + b + 2c & a & b \\ c & b + c + 2a & b \\ c & a & c + a + 2b\end{vmatrix} = 2\left( a + b + c \right) {}^3\]
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Solution
Let \[\bigtriangleup = \begin{vmatrix}a + b + 2c & a & b \\ c & b + c + 2a & b \\ c & a & c + a + 2b\end{vmatrix}\]
Applying \[C_1 \to C_1 + C_2 + C_3\] , we get:
\[\bigtriangleup = \begin{vmatrix}2a + 2b + 2c & a & b \\ 2a + 2b + 2c & b + c + 2a & b \\ 2a + 2b + 2c & a & c + a + 2b\end{vmatrix}\]
\[\Rightarrow \bigtriangleup = 2\left( a + b + c \right)\begin{vmatrix}1 & a & b \\ 1 & b + c + 2a & b \\ 1 & a & c + a + 2b\end{vmatrix}\]
Now, applying
\[R_2 \to R_2 - R_1 \text { and } R_3 \to R_3 - R_1\] , we get:
\[\Rightarrow \bigtriangleup = 2\left( a + b + c \right)\begin{vmatrix}1 & a & b \\ 0 & b + c + a & 0 \\ 0 & 0 & c + a + b\end{vmatrix}\]
\[\Rightarrow \bigtriangleup = 2 \left( a + b + c \right)^3\]
∴ \[\begin{vmatrix}a + b + 2c & a & b \\ c & b + c + 2a & b \\ c & a & c + a + 2b\end{vmatrix} = 2\left( a + b + c \right) {}^3\]
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