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

Answer: (D)

Any odd integer

For the sum of the given two determinants to be 0 the second determnant should be negative of the first one. hence we will be solving for the second determinant.

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

Now the value of the determinant remains unchanged if we take the transpose of the above matrix.Therefore the above determinant is equivalent to the following determinant

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

Now by performing the operations $C_1<->C_3$ and then $C_2<->C_3$ we get that the above determinant is equivalent to the following determinant.

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

The above determinant will be equal to negative of the first determinant only if n is any odd integer.