GATE 2020 IN Q.3: solution and answer

Official answer: BSource: Official answer key, GATE 2020 IN (IIT Delhi)

1. Multiplying both sides of $Ax=b$ on the left by $A^T$ gives the normal equations $A^{T}Ax=A^{T}b$. Since $A^T$ is $4\times 2$ and $A$ is $2\times 4$, the matrix $A^{T}A$ is $4\times 4$.

2. The rank of a matrix cannot exceed its smaller dimension. $A$ is $2\times 4$, so $\rho (A)\le \min (2,4)=2$.

3. The rank of a product is at most the rank of either factor: $\rho (A^{T}A)\le \min (\rho (A^T),\rho (A))\le 2$. (For real $A$, in fact $\rho (A^{T}A)=\rho (A)$.)

4. A $4\times 4$ matrix is invertible only if its rank is $4$. Since $\rho (A^{T}A)\le 2<4$, $\det (A^{T}A)=0$ for every $2\times 4$ matrix $A$, so $(A^{T}A{)}^{-1}$ never exists and the proposed method cannot be carried out.

5. The same conclusion follows from the original system directly: $Ax=b$ has $4$ unknowns and a coefficient matrix of rank at most $2$, so there are at least $4-2=2$ free variables. If the system is consistent it has infinitely many solutions; it can never have a unique one.

6. Conditioning cannot rescue this. A singular matrix has an infinite condition number, so $A^{T}A$ can never be "well conditioned", which rules out (C). (D) fails for the same rank reason: however well behaved the entries of $A$ are, $A^{T}A$ is still singular and the system is still underdetermined. Hence the answer is (B).