Question

Let $a,b,c$ be such that $b(a+c)\ne 0$. If $\left|\begin{array}{ccc}a& a+1& a-1\\ -b& b+1& b-1\\ c& c-1& c+1\end{array}\right|+\left|\begin{array}{ccc}a+1& b+1& c-1\\ a-1& b-1& c+1\\ (-1{)}^{n+2}a& (-1{)}^{n+1}b& (-1{)}^{n}c\end{array}\right|=0,$ then the value of $n$ is

• A. Zero
• B. Any even integer
• C. Any odd integer
• D. Any integer

Answer 1

Answer: (C)

Any odd integer

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

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

let $|A|=\left|\begin{array}{ccc}a& a+1& a-1\\ -b& b+1& b-1\\ c& c-1& c+1\end{array}\right|$

applying $C_2\to C_2+C_3$

$=2\left|\begin{array}{ccc}a& a& a-1\\ -b& b& b-1\\ c& c& c+1\end{array}\right|$

applying $C_1\to C_1-C_2$ and expanding gives

$\to 2\left|\begin{array}{ccc}0& a& a-1\\ -2b& b& b-1\\ 0& c& c+1\end{array}\right|$

$\therefore$ $|A|=4b(c+a)$

let $|B|=\left|\begin{array}{ccc}a+1& b+1& c-1\\ a-1& b-1& c+1\\ a& -b& c\end{array}\right|$

applying $R_1\leftrightarrow R3$ and $R_2\leftrightarrow R3$ gives

$\to \left|\begin{array}{ccc}a& -b& c\\ a+1& b+1& c-1\\ a-1& b-1& c+1\end{array}\right|$

clearly $B=A^T$

$\Rightarrow |B|=4b(c+a)$ ($\because |A^T|=|A|$)

$\therefore$ from (1)

$4b(c+a)[1+(-1{)}^n]=0$

but given that $b(c+a)\ne 0$

$\Rightarrow$ $(-1{)}^n=-1$

$\therefore$ $n$ is an odd integer