GATE 2025 IN Q.11: solution and answer

Official answer: ASource: Official answer key, GATE 2025 IN (IIT Roorkee)

1. Write the entry rule as one formula. $(-1{)}^{i+j}$ is $+1$ when $i+j$ is even and $-1$ when it is odd, so $a_{ij}=-\beta (-1{)}^{i+j}=-\beta (-1{)}^i(-1{)}^j$. Check: for $i+j$ even this gives $-\beta$, for $i+j$ odd it gives $\beta$, as printed.

2. Read that as an outer product. Let $u={\left[\begin{array}{ccccc}-1& 1& -1& \cdots & 1\end{array}\right]}^T$ with $u_i=(-1{)}^i$ for $i=1,\dots ,2n$. Then $a_{ij}=-\beta u_i{u}_j$, that is, $A=-\beta u{u}^T$.

3. Check the first rows. Row 1 is $\left[\begin{array}{ccccc}-\beta & \beta & -\beta & \cdots & \beta \end{array}\right]$ and row 2 is $\left[\begin{array}{ccccc}\beta & -\beta & \beta & \cdots & -\beta \end{array}\right]$, so $R_2=-R_1$. In general $a_{ij}/a_{1j}=(-1{)}^i/(-1{)}^1=(-1{)}^{i+1}$, so $R_i=(-1{)}^{i+1}{R}_1$.

4. Row-reduce with $R_i\to R_i-(-1{)}^{i+1}{R}_1$ for $i=2,\dots ,2n$. Every row below the first becomes zero.

5. One non-zero row is left, and it is non-zero because $\beta \ne 0$. So $\text{rank}(A)=1$ for every $n>2$ and every non-zero $\beta$. This is option (A).