Let a, b, c be such that  b(a+c) \neq 0.If

then the value of n is

  • Option 1)

    any even integer

  • Option 2)

    any odd integer

  • Option 3)

    any integer

  • Option 4)

    zero

 

Answers (2)

As we learnt in 

Value of determinants of order 3 -

-

 

 \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+1}b& \left ( -1\right )^{n}c \end{vmatrix}= 0

\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}a& \left ( -1\right )^{n}b& \left ( -1\right )^{n}c \end{vmatrix}= 0

\begin{vmatrix} a &a+1&a-1 \\ -b & b+1 &b-1\\ c&c-1& c+1 \end{vmatrix}+(-1)^{n}\begin{vmatrix} a+1&b+1&c-1 \\ a-1& b-1 &c+1\\ a& -b& c \end{vmatrix}= 0

C_{1}\leftrightarrow C_{2}

\begin{vmatrix} a+1 &a&a-1 \\ b+1 & -b &b-1\\ c-1&c& c+1 \end{vmatrix}+(-1)^{n}\begin{vmatrix} a+1&b+1&c-1 \\ a-1& b-1 &c+1\\ a& -b& c \end{vmatrix}= 0

C_{2}\leftrightarrow C_{3}

\begin{vmatrix} a+1 &a-1&a \\ b+1 & b-1 &-b\\ c-1&c+1& c \end{vmatrix}+(-1)^{n}\begin{vmatrix} a+1&b+1&c-1 \\ a-1& b-1 &c+1\\ a& -b& c \end{vmatrix}= 0

\begin{vmatrix} a+1 &a-1&a \\ b+1 & b-1 &-b\\ c-1&c+1& c \end{vmatrix}\ (1+(-1)^{n})=0

\therefore 1+(-1)^{n}=0

So n is any odd integer.


Option 1)

any even integer

Incorrect Option

 

Option 2)

any odd integer

Correct Option

 

Option 3)

any integer

Incorrect Option

 

Option 4)

zero

Incorrect Option

 

N neha

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