JEE Advanced 2018 — Paper 2 — Question 30

Find the condition on $\vec{b}$ for the original system $A\vec{x}=\vec{b}$ to be consistent.

Coefficient matrix $A = \begin{bmatrix} -1 & 2 & 5 \\ 2 & -4 & 3 \\ 1 & -2 & 2 \end{bmatrix}$ has determinant $\det A = 0$, so rank is less than 3. Row reduce $A$: $R_2 \to R_2+2R_1$ gives $[0,0,13]$; $R_3 \to R_3+R_1$ gives $[0,0,7]$.

The second and third rows are multiples, so rank $=2$. For consistency, the augmented matrix must also have rank 2.

From the reduced rows: $13(b_3+b_1) = 7(b_2+2b_1)$. Simplify to get the condition: $b_1 + 7b_2 - 13b_3 = 0$. Thus $S = \{ (b_1,b_2,b_3) \in \mathbb{R}^3 \mid b_1+7b_2-13b_3=0 \}$. Now check each option: For a system to have at least one solution for every $\vec{b} \in S$, its coefficient matrix must either be invertible (so a unique solution exists for any RHS) or, if singular, the consistency condition must be automatically true for all $\vec{b} \in S$. Option A: Coefficient matrix $M_A = \begin{bmatrix} 1 & 2 & 3 \\ 0 & 4 & 5 \\ 1 & 2 & 6 \end{bmatrix}$. $\det M_A = 12 \neq 0$.

Hence it has a unique solution for every RHS, including all $\vec{b} \in S$. So A is correct. Option B: $M_B = \begin{bmatrix} 1 & 1 & 3 \\ 5 & 2 & 6 \\ -2 & -1 & -3 \end{bmatrix}$. $\det M_B = 0$ and the third equation is $-\frac13$ times the first plus $-\frac13$ times the second.

The consistency condition is $b_1+b_2+3b_3=0$. For a general $\vec{b} \in S$ (e.g., $b_1=1,b_2=0,b_3=1/13$), this fails. So B is not correct. Option C: $M_C = \begin{bmatrix} -1 & 2 & -5 \\ 2 & -4 & 10 \\ 1 & -2 & 5 \end{bmatrix}$.

All rows are multiples of $[-1,2,-5]$; rank 1. Consistency requires $b_2 = -2b_1$ and $b_3 = -b_1$. Not all $\vec{b} \in S$ satisfy these, so C is not correct. Option D: $M_D = \begin{bmatrix} 1 & 2 & 5 \\ 2 & 0 & 3 \\ 1 & 4 & -5 \end{bmatrix}$.

$\det M_D = 54 \neq 0$, so it has a unique solution for every RHS, including all $\vec{b} \in S$. Thus D is correct.

Therefore, the correct options are A and D.