Question

The value of $\left|\begin{array}{ccc}(-1{)}^{n}a& (-1{)}^{n+1}b& (-1{)}^{n+1}c\\ a+1& 1-b& 1+c\\ a-1& b+1& 1-c\end{array}\right|+\left|\begin{array}{ccc}(-1{)}^{n+1}a& a+1& a-1\\ (-1{)}^{n}b& 1-b& b+1\\ (-1{)}^{n+2}c& 1+c& 1-c\end{array}\right|$ is equal to ..............

Answer 1

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

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

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

Both determinant values is same, as one is transpose of other.

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