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. any integer
• B. zero
• C. any even integer
• D. any odd integer

Answer 1

The correct option is C any even integer

We know $|A|=|A^T|$
Transpose the second matrix and then $C_2\leftrightarrow C_3,C_2\leftrightarrow C_1$, we get

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

So, if $n$ is even, then the equation holds true.