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(xii)\( \left|\begin{array}{ccc}a & b & c \\ a^{2} & b^{2} & c^{2} \\ b+c & c+a & a+b\end{array}\right|=(b-c)(c-a)(a-b)(a+b+c) \) |
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Answer» \(\begin{vmatrix}a&b&c\\a^2&b^2&c^2\\b+&c+a&a+b\end{vmatrix}\) Applying C2→C2→C1, C3→C3 - C1 \(=\begin{vmatrix}a&b-a&c-a\\a^2&(b-a)(b+a)&(c-a)(c+a)\\b+c&-(b-a)&-(c-a)\end{vmatrix}\) Applying C2→C2/b-a, C3→C3/c-a \(=(b-a)(c-a)\begin{vmatrix}a&1&1\\a^2&(b+a)&(c+a)\\b+c&-1&-1\end{vmatrix}\) Applying C3→ \(\frac{C_3-C_2}{c-b}\) \(=(b-a)(c-a)(c-b)\begin{vmatrix}a&1&0\\a^2&(b+a)&1\\b+c&-1&0\end{vmatrix}\) on expanding determinant along column C3 = (a - b) (b -c) (c - a) (-1)\(\begin{vmatrix}a& 1\\b+c&-1\end{vmatrix}\) = (a - b) (b - c) (c - a) (-1) (-a - b - c) = (a - b) (b - c) (c - a) (a + b +c) |
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