Question

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

 

पर्याय

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

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

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

• none of these

Answer 1

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