The value of the determinant left|begin{array | vedclass

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(C) Let $D = \left|\begin{array}{lll}b^2-a b & b-c & b c-a c \ a b-a^2 & a-b & b^2-a b \ b c-a c & c-a & a b-a^2\end{array}ight|$.Taking $(b-a)$ c

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$0$

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(C) Let $D = \left|\begin{array}{lll}b^2-a b & b-c & b c-a c \ a b-a^2 & a-b & b^2-a b \ b c-a c & c-a & a b-a^2\end{array}ight|$.Taking $(b-a)$ common from $C_1$ and $C_3$:$D = (b-a)(b-a) \left|\begin{array}{lll}b & b-c & c \ a & a-b & b \ c & c-a & a\end{array}ight|$.Applying the operation $C_2 ightarrow C_2 + C_1 - C_3$:$C_2 + C_1 - C_3 = (b-c) + b - c = 2b - 2c$ (This does not simplify to zero directly).Let us re-evaluate the operation: $C_2 ightarrow C_2 + C_1 - C_3$ is not correct. Let us perform $C_2 ightarrow C_2 + C_1 - C_3$ on the matrix $\left|\begin{array}{lll}b & b-c & c \ a & a-b & b \ c & c-a & a\end{array}ight|$.$C_2 + C_1 = (b-c+b, a-b+a, c-a+c) = (2b-c, 2a-b, 2c-a)$.Actually,notice that $C_1 - C_3 = (b-c, a-b, c-a)$,which is exactly $C_2$.Thus,$C_2 - (C_1 - C_3) = 0$.Since column $C_2$ becomes a zero column,the value of the determinant is $0$.

The value of the determinant $\left|\begin{array}{lll}b^2-a b & b-c & b c-a c \\ a b-a^2 & a-b & b^2-a b \\ b c-a c & c-a & a b-a^2\end{array}\right|$ is

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