GATE 2023 IN Q.63: solution and answer

Official answer: -8Source: Official answer key, GATE 2023 IN (IIT Kanpur)

1. Rank one means the row space is one-dimensional, so every row is a scalar multiple of any one nonzero row. (Equivalently, every $2\times 2$ minor is zero.) Row 1, $\left[\begin{array}{cc}1& 4\end{array}\right]$, is nonzero, so rows 2 and 3 must both be multiples of it.

2. Row 2: its first entry is $-3=-3\times 1$, so row 2 $=-3\times$ row 1. The second entries must agree too: $\alpha =-3\times 4=-12$. (The minor check says the same: $\left|\begin{array}{cc}1& 4\\ -3& \alpha \end{array}\right|=\alpha +12=0$.)

3. Row 3: its second entry is $6=\frac{3}{2}\times 4$, so row 3 $=\frac{3}{2}\times$ row 1. The first entries must agree too: $\beta =\frac{3}{2}\times 1=\frac{3}{2}$. (Minor check: $\left|\begin{array}{cc}1& 4\\ \beta & 6\end{array}\right|=6-4\beta =0$.)

4. Check the remaining minor with these values: $\left|\begin{array}{cc}-3& -12\\ \frac{3}{2}& 6\end{array}\right|=(-3)(6)-(-12)\left(\frac{3}{2}\right)=-18+18=0$. So the rank really is $1$ (it cannot be $0$, since $A$ has nonzero entries).

5. $$\frac{\alpha }{\beta }=\frac{-12}{3/2}=-8$$ The value is already an integer, so rounding changes nothing: the answer is $-8$.