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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

 

Posted by

Sabhrant Ambastha

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