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The Value of the Determinant ∣ ∣ ∣ ∣ a − B B + C a B − C C + a B C − a A + B C ∣ ∣ ∣ ∣ (A) a 3 + B 3 + C 3 (B) 3bc (C) a 3 + B 3 + C 3 − 3 a B C (D) None of These

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Question

The value of the determinant  

\[\begin{vmatrix}a - b & b + c & a \\ b - c & c + a & b \\ c - a & a + b & c\end{vmatrix}\]



Options

  • \[a^3 + b^3 + c^3\]

  • 3bc

  • \[a^3 + b^3 + c^3 - 3abc\]

  • none of these

MCQ
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Solution

\[\begin{vmatrix}a - b & b + c & a \\ b - c & c + a & b \\ c - a & a + b & c\end{vmatrix}\]
\[ = \begin{vmatrix}- b & b + c + a & a \\ - c & c + a + b & b \\ - a & a + b + c & c\end{vmatrix} \left[\text{ Applying }C_1 \to C_1 - C_3\text{ and }C_2 \to C_2 + C_3 \right]\]
\[ = \left( - 1 \right)\left( a + b + c \right)\begin{vmatrix}b & 1 & a \\ c & 1 & b \\ a & 1 & c\end{vmatrix} \left[\text{ Taking }\left( - 1 \right)\text{ common from }C_1\text{ and }\left( a + b + c \right)\text{ common from }C_2 \right]\]
\[ = \left( - 1 \right)\left( a + b + c \right)\begin{vmatrix}b & 1 & a \\ c - b & 0 & b - a \\ a - b & 0 & c - a\end{vmatrix} \left[\text{ Applying }R_2 \to R_2 - R_1\text{ and }R_3 \to R_3 - R_1 \right]\]
\[ = \left( - 1 \right)\left( a + b + c \right)\left[ - \left( c - b \right)\left( c - a \right) + \left( b - a \right)\left( a - b \right) \right]\]
\[ = \left( - 1 \right)\left( a + b + c \right)\left[ - c^2 + ac + bc - ab + ba - b^2 - a^2 + ab \right]\]
\[ = \left( - 1 \right)\left( a + b + c \right)\left( - a^2 - b^2 - c^2 + ab + bc + ac \right)\]
\[ = \left( a + b + c \right)\left( a^2 + b^2 + c^2 - ab - bc - ac \right)\]
\[ = a^3 + a b^2 + a c^2 - a^2 b - abc - a^2 c + b a^2 + b^3 + b c^2 - a b^2 - b^2 c - abc + c a^2 + c b^2 + c^3 - acb - b c^2 - a c^2 \]
\[ = a^3 + b^3 + c^3 - 3abc\]

 

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Chapter 5: Determinants - Exercise 6.7 [Page 95]

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R.D. Sharma Mathematics Volume 1 and 2 [English] Class 12
Chapter 5 Determinants
Exercise 6.7 | Q 24 | Page 95

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