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Prove that : ∣ ∣ ∣ ∣ a B − C C − B a − C B C − a A − B B − a C ∣ ∣ ∣ ∣ = ( a + B − C ) ( B + C − a ) ( C + a − B ) - Mathematics

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Question

Prove that :

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

 

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Solution

\[\text{ Let LHS }= \Delta = \begin{vmatrix} a & b - c & c - b\\a - c & b & c - a\\a - b & b - a & c \end{vmatrix}\] 
\[\Delta = \begin{vmatrix} a & 0 & c - b + a\\a - c & b + c - a & 0\\a - b & b + c - a & c + a - b \end{vmatrix} \left[\text{ Applying }C_2 \to C_2 + C_3\text{ and }C_3 \to C_1 + C_3 \right]\] 
\[ = \left( b + c - a \right)\left( c + a - b \right)\begin{vmatrix} a & 0 & 1\\a - c & 1 & 0\\a - b & 1 & 1 \end{vmatrix} \left[\text{ Taking out common factor from }C_2\text{ and }C_3 \right]\] 
\[ = \left( b + c - a \right)\left( c + a - b \right)\left\{ \left( a \times \begin{vmatrix} 1 & 0\\1 & 1 \end{vmatrix} \right) + \left( 1 \times \begin{vmatrix} a - c & 1\\a - b & 1 \end{vmatrix} \right) \right\} \left[\text{ Expanding along }R_1 \right]\] 
\[ = \left( a + b - c \right)\left( b + c - a \right)\left( c + a - b \right)\] 
\[ = RHS\] 

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Chapter 6: Determinants - Exercise 6.2 [Page 59]

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RD Sharma Mathematics [English] Class 12
Chapter 6 Determinants
Exercise 6.2 | Q 27 | Page 59

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