Question

If  \[∆_1 = \begin{vmatrix}1 & 1 & 1 \\ a & b & c \\ a^2 & b^2 & c^2\end{vmatrix}, ∆_2 = \begin{vmatrix}1 & bc & a \\ 1 & ca & b \\ 1 & ab & c\end{vmatrix},\text{ then }\]}

Options

• \[∆_1 + ∆_2 = 0\]

• \[∆_1 + 2 ∆_2 = 0\]

• \[∆_1 = ∆_2\]

• none of these

Answer 1

(a) \[∆_1 + ∆_2 = 0\] 
\[\Delta_{2 =} \begin{vmatrix} 1 & bc & a\\1 & ca & b\\1 & ab & c \end{vmatrix}\]
\[ = \frac{1}{abc}\begin{vmatrix} a & abc & a^2 \\b & bca & b^2 \\c & cab & c^2 \end{vmatrix} [ R_1 , R_2 , R_3\text{ are multiplied by a, b and c respectively, therefore we divide by abc}]\]
\[ = \frac{abc}{abc} \begin{vmatrix} a & 1 & a^2 \\b & 1 & b^2 \\c & 1 & c^2 \end{vmatrix} \left[\text{ Taking abc common from }C_2 \right]\]
\[ = - \begin{vmatrix} 1 & a & a^2 \\1 & b & b^2 \\1 & c & c^2 \end{vmatrix} C_1 \leftrightarrow C_2 \]
We know that the value of a determinant remains unchanged if its rows and columns are interchanged . So, 
\[ ∆_2 = - \begin{vmatrix} 1 & 1 & 1\\a & b & c\\ a^2 & b^2 & c^2 \end{vmatrix} \]
\[ = - \Delta_1 \]
\[ \Rightarrow \Delta_1 + \Delta_2 = 0\]