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

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प्रश्न

Prove that :

\[\begin{vmatrix}a^2 & a^2 - \left( b - c \right)^2 & bc \\ b^2 & b^2 - \left( c - a \right)^2 & ca \\ c^2 & c^2 - \left( a - b \right)^2 & ab\end{vmatrix} = \left( a - b \right) \left( b - c \right) \left( c - a \right) \left( a + b + c \right) \left( a^2 + b^2 + c^2 \right)\]
बेरीज
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उत्तर

\[\text{ Let LHS }= \Delta = \begin{vmatrix} a^2 & a^2 - \left( b - c \right)^2 & bc\\ b^2 & b^2 - \left( c - a \right)^2 & ca\\ c^2 & c^2 - \left( a - b \right)^2 & ab \end{vmatrix}\]
\[ \Rightarrow ∆ = \begin{vmatrix} a^2 & - \left( b - c \right)^2 & bc\\ b^2 & - \left( c - a \right)^2 & ca\\ c^2 & - \left( a - b \right)^2 & ab \end{vmatrix} \left[\text{ Applying }C_2 \to C_2 - C_1 \right]\] 
\[ = \left( - 1 \right)\begin{vmatrix} a^2 & \left( b - c \right)^2 & bc\\ b^2 & \left( c - a \right)^2 & ca\\ c^2 & \left( a - b \right)^2 & ab \end{vmatrix}\] 
\[ = - \begin{vmatrix} a^2 & b^2 + c^2 & bc\\ b^2 & c^2 + a^2 & ca\\ c^2 & a^2 + b^2 & ab \end{vmatrix} \left[\text{ Applying }C_2 \to C_2 - 2 C_3 \right]\] 
\[ = - \begin{vmatrix} a^2 + b^2 + c^2 & b^2 + c^2 & bc\\ b^2 + c^2 + a^2 & c^2 + a^2 & ca\\ c^2 + a^2 + b^2 & a^2 + b^2 & ab \end{vmatrix} \left[\text{ Applying }C_1 \to C_1 + C_2 \right]\] 
\[ = - \left( a^2 + b^2 + c^2 \right)\begin{vmatrix} 1 & b^2 + c^2 & bc\\1 & c^2 + a^2 & ca\\1 & a^2 + b^2 & ab \end{vmatrix}\]
\[ = - \left( a^2 + b^2 + c^2 \right)\begin{vmatrix} 1 & b^2 + c^2 & bc\\0 & \left( c^2 + a^2 \right) - \left( b^2 + c^2 \right) & ca - bc\\0 & \left( a^2 + b^2 \right) - \left( b^2 + c^2 \right) & ab - bc \end{vmatrix} \left[\text{ Applying }R_2 \to R_2 - R_1 \text{ and }R_3 \to R_3 - R_1 \right]\] 
\[ = \left( \left( a^2 + b^2 + c^2 \right) \right)\begin{vmatrix} 1 & b^2 + c^2 & bc\\0 & a^2 - b^2 & c\left( a - b \right)\\0 & a^2 - c^2 & b \left( a - c \right) \end{vmatrix}\]
\[ = - \left( a^2 + b^2 + c^2 \right)\left( a - b^{} \right)\left( a - c \right)\begin{vmatrix} 1 & b^2 + c^2 & bc\\0 & a + b^{} & c\\0 & a^{} + c^{} & b \end{vmatrix} \left[\text{ Taking }\left( a - b \right)\text{ common from }R_2\text{ and }\left( a - c \right)\text{ common from }R_3 \right]\] 
\[ = \left( a^2 + b^2 + c^2 \right)\left( a - b^{} \right)\left( c - a \right) \times \left\{ 1 \times \begin{vmatrix} a + b^{} & c\\ a^{} + c^{} & b \end{vmatrix} \right\} \left[ \because \left( c - a \right) = - \left( a - c \right) \right] \left[\text{ Expanding along }C_1 \right]\] 
\[ = \left( a^2 + b^2 + c^2 \right)\left( a - b^{} \right) \left( c - a \right) \left( ab + b^2 - ac - c^2 \right) \] 
\[= \left( a^2 + b^2 + c^2 \right)\left( a - b \right)\left( c - a \right)\left\{ a\left( b - c \right) + \left( b + c \right)\left( b - c \right) \right\}\] 
\[ = \left( a - b \right)\left( c - a \right)\left( b - c \right)\left( a + b + c \right)\left( a^2 + b^2 + c^2 \right)\]
 = RHS
Hence proved

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पाठ 5: Determinants - Exercise 6.2 [पृष्ठ ५९]

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आर.डी. शर्मा Mathematics Volume 1 and 2 [English] Class 12
पाठ 5 Determinants
Exercise 6.2 | Q 22 | पृष्ठ ५९

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