JEE Advanced · 2008 · 13 Apr 2008 · Paper 1 · Matrices

Statement 1: For a unique solution, we need $ad-bc\ne 0$.

Total possible cases: $2^4=16$ (each of $a,b,c,d$ can be 0 or 1).

For $ad-bc\ne 0$:

• Either $ad=1$ and $bc=0$, or
• $ad=0$ and $bc=1$

Case 1: $ad=1,bc=0$

• $a=d=1$ (1 way)
• $bc=0$: either $b=0$ or $c=0$ (3 ways: $(0,0),(0,1),(1,0)$)
• Total: $1\times 3=3$ ways

Case 2: $ad=0,bc=1$

• $bc=1$: $b=c=1$ (1 way)
• $ad=0$: either $a=0$ or $d=0$ (3 ways: $(0,0),(0,1),(1,0)$)
• Total: $1\times 3=3$ ways

Total favorable: $3+3=6$

Probability = $\frac{6}{16}=\frac{3}{8}$. Statement 1 is TRUE. ✓

Statement 2: The system always has the trivial solution $(x,y)=(0,0)$. Therefore, the probability that the system has a solution is 1. Statement 2 is TRUE. ✓

However, Statement 2 (about having a solution) does not explain Statement 1 (about having a unique solution), as the trivial solution always exists regardless of uniqueness.

Answer: (B)