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\( \left|\begin{array}{ccc}a+b+c & -c & -b \\ -c & a+b+c & -a \\ -b & -a & a+b+c\end{array}\right|=2(a+b)(b+c)(c+a) \) |
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Answer» \(\begin{vmatrix}a+b+c&-c&-b\\-c&a+b+c&-a\\-b&-a&a+b+c\end{vmatrix}\) Applying C1→ C1 + C2 & C2 → C2 + C3 \(=\begin{vmatrix}a+b&-(b+c)&-b\\a+b&b+c&-a\\-(a+b)&b+c&a+b+c\end{vmatrix}\) Applying C1 → \(\frac{C_1}{a+b}\) & C2 → \(\frac{C_2}{b+c}\) = (a + b)(b + c)\(\begin{vmatrix}1&-1&-b\\1&1&-a\\-1&1&a+b+c\end{vmatrix}\) Applying R1 → R2 - R1, & R3 → R3 + R1 = (a + b)(b + c)\(\begin{vmatrix}1&-1&-b\\0&2&-a+b\\0&0&a+c\end{vmatrix}\) = 2(a + b) (b + c) (c + a) (Expand determinant along C1) |
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