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The Determinant ∣ ∣ ∣ ∣ ∣ B 2 − a B B − C B C − a C a B − a 2 a − B B 2 − a B B C − C a C − a A B − a 2 ∣ ∣ ∣ ∣ ∣ (A) a B C (B) ( C − a ) ( a − B ) (D) None of These

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Question

The determinant  \[\begin{vmatrix}b^2 - ab & b - c & bc - ac \\ ab - a^2 & a - b & b^2 - ab \\ bc - ca & c - a & ab - a^2\end{vmatrix}\]


 

Options

  • \[abc\left( b - c \right)\left( c - a \right)\left( a - b \right)\]

  • \[\left( b - c \right)\left( c - a \right)\left( a - b \right)\]

  • \[\left( a + b + c \right)\left( b - c \right)\left( c - a \right)\left( a - b \right)\]

  • none of these

MCQ
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Solution

\[\begin{vmatrix}b^2 - ab & b - c & bc - ac \\ ab - a^2 & a - b & b^2 - ab \\ bc - ca & c - a & ab - a^2\end{vmatrix}\]
\[ = \begin{vmatrix}b\left( b - a \right) & b - c & c\left( b - a \right) \\ a\left( b - a \right) & a - b & b\left( b - a \right) \\ c\left( b - a \right) & c - a & a\left( b - a \right)\end{vmatrix}\]
\[ = \left( b - a \right)^2 \begin{vmatrix}b & b - c & c \\ a & a - b & b \\ c & c - a & a\end{vmatrix} \left[\text{ Taking }\left( b - a \right)\text{ common from }C_1\text{ and }C_3 \right]\]
\[ = \left( b - a \right)^2 \begin{vmatrix}0 & b - c & c \\ 0 & a - b & b \\ 0 & c - a & a\end{vmatrix} \left[\text{ Applying }C_1 \to C_1 - C_2 - C_3 \right]\]
\[ = 0\]

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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 26 | Page 95

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