GATE 2022 EE Q.20: solution and answer

Official answer: BSource: Official answer key, GATE 2022 EE (IIT Kharagpur)

1. Swap the indices in the general term: $a_{j,i}=(j-i{)}^3={\left(-(i-j)\right)}^3=-(i-j{)}^3=-a_{i,j}$. The sign carries through because the power is odd.

2. So $a_{j,i}=-a_{i,j}$ for every $i,j$, which means $A^T=-A$. That is the definition of a skew-symmetric matrix. Setting $j=i$ gives $a_{i,i}=-a_{i,i}$, so every diagonal entry is $0$, as a skew-symmetric matrix needs.

3. Check by writing the matrix out. $a_{12}=(1-2{)}^3=-1$, $a_{13}=(1-3{)}^3=-8$, $a_{23}=(2-3{)}^3=-1$, the entries below the diagonal are their negatives, and every diagonal entry is $(i-i{)}^3=0$: $$A=\left[\begin{array}{ccc}0& -1& -8\\ 1& 0& -1\\ 8& 1& 0\end{array}\right]$$

4. It is not symmetric: $a_{12}=-1$ but $a_{21}=1$, so $A^T\ne A$. It is not null: it has non-zero entries.

5. It is not unitary. For a skew-symmetric matrix of odd order, $\det A=\det (A^T)=\det (-A)=(-1{)}^3\det A$, so $\det A=0$ and $A$ is singular. A unitary matrix is always invertible, since $|\det A|=1$. You can also see it directly: the first column $(0,1,8)$ has length $\sqrt{65}\ne 1$, but every column of a unitary matrix has length $1$.

6. So $A$ is skew-symmetric, which is option (B).