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Question

Let a, b, c be such that b$$(a+c)\neq 0$$. If $$\begin{vmatrix} a & a+1 & a-1\\ -b & b+1 & b-1 \\ c & c-1 & c+1\end{vmatrix}+\begin{vmatrix} a+1 & b+1 & c-1 \\ a-1 & b-1 & c+1 \\ (-1)^{n+2}a & (-1)^{n+1}b & (-1)^nc\end{vmatrix}=0$$
Then the value of n is?


A
0
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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 D Any odd integer
$$\begin{vmatrix} a & a+1 & a-1 \\ -b & b+1 & b-1\\ c & c-1 & c+1\end{vmatrix}+\begin{vmatrix} (-1)^{n+2}a & a+1 & a-1 \\ (-1)^{n+1}b & b+1 & b-1\\ (-1)^nc & c-1 & c+1\end{vmatrix}$$
$$=\begin{vmatrix} a+(-1)^{n+2}a & a+1 & a-1\\ -b+(-1)^{n+1}b & b+1 & b-1\\ c+(-1)^nc & c-1 & c+1\end{vmatrix}$$
$$=0$$ if n is an odd integer.

Mathematics

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