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Prove that : ∣ ∣ ∣ ∣ B + C a − B a C + a B − C B a + B C − a C ∣ ∣ ∣ ∣ = 3 a B C − a 3 − B − C - Mathematics

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प्रश्न

Prove that :

\[\begin{vmatrix}b + c & a - b & a \\ c + a & b - c & b \\ a + b & c - a & c\end{vmatrix} = 3abc - a^3 - b - c^3\]

 

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उत्तर

\[\text{Let LHS }= \Delta = \begin{vmatrix} b + c & a - b & a\\c + a & b - c & b \\a + b & c - a & c \end{vmatrix}\] 

\[\Delta = \left( b + c \right) \begin{vmatrix} b - c & b \\c - a & c \end{vmatrix} - \left( a - b \right) \begin{vmatrix} c + a & b \\ a + b & c \end{vmatrix} + a \begin{vmatrix} c + a & b - c \\a + b & c - a \end{vmatrix} \left[\text{ Expanding }\right] \] 
\[ = \left( b + c \right)\left\{ bc - c^2 - bc + ab \right\} - \left( a - b \right)\left\{ c^2 + ac - ab - b^2 \right\} + a\left\{ c^2 - a^2 - ab + ac - b^2 + bc \right\}\] 
\[ = b c^2 - c^3 + abc - a c^2 - a^2 c + a^2 b + a b^2 + b c^2 + abc - a b^2 - b^3 + a c^2 - a^3 - a^2 c - a b^2 + abc\] 
\[ \Rightarrow \Delta = 3abc - a^3 - b^3 - c^3 \left[\text{ Simplyfying }\right]\] 
\[ = RHS\]

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अध्याय 6: Determinants - Exercise 6.2 [पृष्ठ ५८]

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आरडी शर्मा Mathematics [English] Class 12
अध्याय 6 Determinants
Exercise 6.2 | Q 12 | पृष्ठ ५८

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