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Question

If $$\begin{vmatrix} a & a+1 & a-1 \\ -b & b+1 & b-1 \\ c & c-1 & c+1 \end{vmatrix}+\begin{vmatrix} a+1 & b+1 & c-1 \\ a-1 & b-1 & c+1 \\ { \left( -1 \right)  }^{ n+2 }a & { \left( -1 \right)  }^{ n+2 }b & { \left( -1 \right)  }^{ n }c \end{vmatrix}=0$$. Then the value of $$n$$ is


A
zero
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B
any even integer
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C
any odd integer
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D
any integer
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Solution

The correct option is C any integer
$$\begin{vmatrix}a+1&b+1&c-1\\a-1&b-1&c+1\\(-1)^{n+2}a&(-1)^{n+2}b&(-1)^{n}c\end{vmatrix}=\begin{vmatrix}2&2&-2\\a-1&b-1&c+1\\(-1)^{n}a&(-1)^{n}b&(-1)^{n}c\end{vmatrix}=2(-1)^{n}\begin{vmatrix}1&1&-1\\a-1&b-1&c+1\\a&b&c\end{vmatrix}$$                                                                                                                                                       $$=2(-1)^{n}\begin{vmatrix}1&1&-1\\a&b&c\\a&b&c\end{vmatrix}=0$$
$$\implies n $$ is any integer

Mathematics

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