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Prove that : ∣ ∣ ∣ ∣ a − B − C 2 a 2 a 2 B B − C − a 2 B 2 C 2 C C − a − B ∣ ∣ ∣ ∣ = ( a + B + C ) 3

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Question

Prove that :

\[\begin{vmatrix}a - b - c & 2a & 2a \\ 2b & b - c - a & 2b \\ 2c & 2c & c - a - b\end{vmatrix} = \left( a + b + c \right)^3\]

 

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Solution

\[\text{ Let LHS } = ∆ = \begin{vmatrix} a - b - c & 2a & 2a\\ 2b & b - c - a & 2b\\2c & 2c & c - a - b \end{vmatrix}\] 
\[ \Rightarrow ∆ = \begin{vmatrix} a + b + c & a + b + c & a + b + c \\ 2b & b - c - a & 2b\\2c & 2c & c - a - b \end{vmatrix} \left[\text{ Applying }R_1 \to R_1 + R_2 + R_3 \right]\] 
\[ = \left( a + b + c \right)\begin{vmatrix} 1 & 1 & 1 \\ 2b & b - c - a & 2b\\ 2c & 2c & c - a - b \end{vmatrix}\]
\[ = \left( a + b + c \right) \begin{vmatrix} 0 & 1 & 1 \\ b + c + a & b - c - a & 2b\\ 0 & 2c & c - a - b \end{vmatrix} \left[\text{ Applying }C_1 \to C_1 - C_2 \right]\] 
\[ = \left( a + b + c \right)\left\{ \left( a + b + c \right) \times \begin{vmatrix} 1 & 1 \\ 2c & c - a - b \end{vmatrix} \right\} \left[\text{ Expanding along }C_1 \right]\] 
\[ = \left( a + b + c \right)^3 \] 
\[ = RHS\]

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Chapter 5: Determinants - Exercise 6.2 [Page 59]

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R.D. Sharma Mathematics Volume 1 and 2 [English] Class 12
Chapter 5 Determinants
Exercise 6.2 | Q 15 | Page 59

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