Question

$A_{n\times n}$ matrix formed using $0,1$ and $-1$ as its elements. The number of such matrices which are skew-symmetric is

• A. $\frac{n(n-1)}{2}$
• B. $(n-1{)}^2$
• C. $2^{\frac{n(n-1)}{2}}$
• D. $3^{\frac{n(n-1)}{2}}$

Answer 1

The correct option is D $3^{\frac{n(n-1)}{2}}$

We know that in a skew symmetric matrix, all the elements off principal diagonal are zero.

All the elements in upper triangle are additive inverses of elements in lower triangle.

$\Rightarrow a_{ij}=-a_{ji}$

So the $n$ elements of principal diagonal can be filled in $1$ way

Remaining elements are$=n^2-n=n(n-1)$

No. of elements in upper triangle leaving the principa diagonal $=\frac{n(n-1)}{2}$

And the elements in upper triangle can be filled in $3ways$ each by elements $1,0,-1$.

So the total no.of ways in which this matrix is possible is $=3^{\frac{n(n-1)}{2}}$