Question

please answer it fast I will mark u as brainliest.... irrelevant answers will be reported....​

Answer 1

We're given the equation,

$\displaystyle\longrightarrow\left|\begin{array}{ccc}\sf{a}&\sf{a+1}&\sf{a-1}\\\sf{-b}&\sf{b+1}&\sf{b-1}\\\sf{c}&\sf{c-1}&\sf{c+1}\end{array}\right|+\left|\begin{array}{ccc}\sf{a+1}&\sf{b+1}&\sf{c-1}\\\sf{a-1}&\sf{b-1}&\sf{c+1}\\\sf{(-1)^{n+2}a}&\sf{(-1)^{n+1}b}&\sf{(-1)^nc}\end{array}\right|\sf{=0}$

Consider the second determinant in the LHS.

$\longrightarrow\left|\begin{array}{ccc}\sf{a+1}&\sf{b+1}&\sf{c-1}\\\sf{a-1}&\sf{b-1}&\sf{c+1}\\\sf{(-1)^{n+2}a}&\sf{(-1)^{n+1}b}&\sf{(-1)^nc}\end{array}\right|$

We can take its transpose,

$\longrightarrow\left|\begin{array}{ccc}\sf{a+1}&\sf{b+1}&\sf{c-1}\\\sf{a-1}&\sf{b-1}&\sf{c+1}\\\sf{(-1)^{n+2}a}&\sf{(-1)^{n+1}b}&\sf{(-1)^nc}\end{array}\right|=\left|\begin{array}{ccc}\sf{a+1}&\sf{a-1}&\sf{(-1)^{n+2}a}\\\sf{b+1}&\sf{b-1}&\sf{(-1)^{n+1}b}\\\sf{c-1}&\sf{c+1}&\sf{(-1)^nc}\end{array}\right|$

and can perform the operations $\sf{C_1\leftrightarrow C_2,}$

$\longrightarrow\left|\begin{array}{ccc}\sf{a+1}&\sf{b+1}&\sf{c-1}\\\sf{a-1}&\sf{b-1}&\sf{c+1}\\\sf{(-1)^{n+2}a}&\sf{(-1)^{n+1}b}&\sf{(-1)^nc}\end{array}\right|=\left|\begin{array}{ccc}\sf{a-1}&\sf{a+1}&\sf{(-1)^{n+2}a}\\\sf{b-1}&\sf{b+1}&\sf{(-1)^{n+1}b}\\\sf{c+1}&\sf{c-1}&\sf{(-1)^nc}\end{array}\right|$

and then $\sf{C_1\leftrightarrow C_3.}$

$\longrightarrow\left|\begin{array}{ccc}\sf{a+1}&\sf{b+1}&\sf{c-1}\\\sf{a-1}&\sf{b-1}&\sf{c+1}\\\sf{(-1)^{n+2}a}&\sf{(-1)^{n+1}b}&\sf{(-1)^nc}\end{array}\right|=\left|\begin{array}{ccc}\sf{(-1)^{n+2}a}&\sf{a+1}&\sf{a-1}\\\sf{(-1)^{n+1}b}&\sf{b+1}&\sf{b-1}\\\sf{(-1)^nc}&\sf{c-1}&\sf{c+1}\end{array}\right|$

So our equation becomes,

$\displaystyle\longrightarrow\left|\begin{array}{ccc}\sf{a}&\sf{a+1}&\sf{a-1}\\\sf{-b}&\sf{b+1}&\sf{b-1}\\\sf{c}&\sf{c-1}&\sf{c+1}\end{array}\right|+\left|\begin{array}{ccc}\sf{(-1)^{n+2}a}&\sf{a+1}&\sf{a-1}\\\sf{(-1)^{n+1}b}&\sf{b+1}&\sf{b-1}\\\sf{(-1)^nc}&\sf{c-1}&\sf{c+1}\end{array}\right|\sf{=0}$

Since $\sf{C_2}$ and $\sf{C_3}$ each are identical, we can combine both the determinants by adding $\sf{C_1}$ each as,

$\displaystyle\longrightarrow\left|\begin{array}{ccc}\sf{a+(-1)^{n+2}a}&\sf{a+1}&\sf{a-1}\\\sf{-b+(-1)^{n+1}b}&\sf{b+1}&\sf{b-1}\\\sf{c+(-1)^nc}&\sf{c-1}&\sf{c+1}\end{array}\right|\sf{=0}$

If $\sf{n}$ is an odd integer,

• $\sf{(-1)^{n+2}a=-a}$

• $\sf{(-1)^{n+1}b=b}$

• $\sf{(-1)^nc=-c}$

So,

$\displaystyle\longrightarrow\left|\begin{array}{ccc}\sf{a-a}&\sf{a+1}&\sf{a-1}\\\sf{-b+b}&\sf{b+1}&\sf{b-1}\\\sf{c-c}&\sf{c-1}&\sf{c+1}\end{array}\right|\sf{=0}$

$\displaystyle\longrightarrow\left|\begin{array}{ccc}\sf{0}&\sf{a+1}&\sf{a-1}\\\sf{0}&\sf{b+1}&\sf{b-1}\\\sf{0}&\sf{c-1}&\sf{c+1}\end{array}\right|\sf{=0}$

Therefore, $\sf{n}$ should be an odd integer. So (B) is the answer.