Question

Evaluate :

\[\begin{vmatrix}a & b + c & a^2 \\ b & c + a & b^2 \\ c & a + b & c^2\end{vmatrix}\]

Answer 1

\[∆ = \begin{vmatrix}a & b + c & a^2 \\ b & c + a & b^2 \\ c & a + b & c^2\end{vmatrix} \]
\[\text{ When }a = b,\text{ the first two rows become identical }. \text{ Hence, } \text{a - b is a factor }. \]
\[\text{ Similarly, when b = c the second and third rows become identical }. \text{ So, b - c is also a factor }. \]
\[\text{ Also, when }c = a, \text{the third and first rows become identical} .\text{ Hence, c - a is also a factor }. \]
\[\text{ The product of diagonal elements }, a(c + a) c^2 \text{ is }4 .\text{ So, the other factor should be a linear in a, b and c . It should also remain unaltered when any two letters are changed }. \text{ Let this factor be } \lambda (a + b + c) . \]
\[\text{Here, } \lambda \text{ is a constant . To find this, we have }\]
\[a = 0, b = 1, c = 2\]
\[\begin{vmatrix}0 & 3 & 2 \\ 1 & 2 & 1 \\ 2 & 1 & 4\end{vmatrix} = \lambda(a - b)(b - c)(c - a)(a + b + c)\]
\[\begin{vmatrix}0 & 3 & 2 \\ 1 & 2 & 1 \\ 2 & 1 & 4\end{vmatrix} = \lambda(0 - 1)(1 - 2)(2 - 1)(0 + 1 + 2)\]
\[ \Rightarrow - 6 = 6\lambda\]
\[ \Rightarrow \lambda = - 1\]
\[\text{ Thus, } \]
\[ \begin{vmatrix}a & b + c & a^2 \\ b & c + a & b^2 \\ c & a + b & c^2\end{vmatrix} = - ((a + b + c))(a - b)(b - c)(c - a)\]