Consider the system of equations ax + by = 0, cx + dy = 0, where a, b, c, d ∈ {0, 1}.

STATEMENT - 1

Q: Wait, if Statement 2 is true, why isn't it the correct explanation for Statement 1?

This is a classic JEE trap! Statement 2 is a fundamental property of all homogeneous systems—they always have a solution (the trivial one). It is a universal truth. Statement 1, however, is a specific probabilistic calculation about the uniqueness of that solution. A universal property cannot explain a specific probabilistic outcome. Just because a system always has a solution doesn't mean that solution is always unique.

Q: Why do we only check $ad - bc \neq 0$?

In a system of linear equations $Ax=B$, a unique solution exists if and only if the matrix $A$ is invertible, which means its determinant is non-zero. Since our system is homogeneous ($B=0$), if the determinant is non-zero, the only solution is the trivial one ($x=0, y=0$). If the determinant is zero, the system has infinitely many solutions. Thus, the condition for a unique solution is strictly $\det(A) \neq 0$.

Q: Is it possible for the system to have no solution?

No. For a homogeneous system, the origin $(0,0)$ is always a solution. Therefore, the system is consistent by definition. The probability of having a solution is always 1, which confirms Statement 2. You never have to worry about inconsistency in a homogeneous system!

Question

Consider the system of equations $ax + by = 0, cx + dy = 0$, where $a, b, c, d \in \{0, 1\}$.

STATEMENT - 1 : The probability that the system of equations has a unique solution is $3/8$.

and

STATEMENT - 2 : The probability that the system of equations has a solution is 1.

Options:

(A) STATEMENT - 1 is True, STATEMENT - 2 is True, STATEMENT - 2 is a correct explanation for STATEMENT- 1

(B) STATEMENT - 1 is True, STATEMENT - 2 is True, STATEMENT - 2 is NOT a correct explanation for STATEMENT- 1

(C) STATEMENT - 1 is True, STATEMENT - 2 is False

(D) STATEMENT - 1 is False, STATEMENT - 2 is True

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing before a system of equations: $ax + by = 0$ and $cx + dy = 0$. At first glance, they look like any other pair of linear equations, but the right-hand side is zero. This is a **homogeneous system**.

In the world of linear algebra, this is a special place where the origin $(0,0)$ is always a welcome guest. Because $a(0) + b(0) = 0$ and $c(0) + d(0) = 0$, the point $(0,0)$ is always a solution.

This is what we call the **trivial solution**. Because this solution exists for any combination of $a, b, c, d$, the system is never inconsistent. It will always have at least one solution, which validates Statement 2.

### The Gatekeeper: The Determinant

Now, let us pivot to Statement 1. We are asked about the probability of a unique solution. A homogeneous system has a unique solution if and only if the lines are not coincident, meaning they intersect at only one point—the origin.

This occurs if and only if the determinant of the coefficient matrix $A$ is non-zero. We define the matrix and its determinant as:

$$A = \begin{bmatrix} a & b \ c & d \end{bmatrix}$$

$$\Delta = ad - bc$$

Our condition for a unique solution is $\Delta 
eq 0$, or $ad - bc 
eq 0$. If the determinant is zero, the lines are coincident, and we have infinitely many solutions. If it is non-zero, the origin is the only solution.

### The Combinatorial Dance

We have four variables: $a, b, c, d$. Each can be either $0$ or $1$, which gives us $2^4 = 16$ possible matrices. We must count the favorable cases where $ad - bc 
eq 0$.

Since $a, b, c, d \in \{0, 1\}$, the products $ad$ and $bc$ can only be $0$ or $1$. For their difference to be non-zero, they must be different. This leaves us with two scenarios:

**Case 1:** $ad = 1$ and $bc = 0$.
For $ad = 1$, we must have $a=1$ and $d=1$ ($1$ way). For $bc = 0$, the pairs $(b, c)$ can be $(0,0), (0,1), (1,0)$ ($3$ ways). Thus, $1 	imes 3 = 3$ favorable outcomes.

**Case 2:** $ad = 0$ and $bc = 1$.
For $bc = 1$, we must have $b=1$ and $c=1$ ($1$ way). For $ad = 0$, the pairs $(a, d)$ can be $(0,0), (0,1), (1,0)$ ($3$ ways). Thus, $1 	imes 3 = 3$ favorable outcomes.

Adding these together, we have $3 + 3 = 6$ favorable cases out of $16$ total possibilities. The probability is:

$$\frac{6}{16} = \frac{3}{8}$$

### The Final Verdict

We have proven that Statement 1 is true and Statement 2 is true. However, Statement 2 is a general property of homogeneous systems—a statement of existence.

Statement 1 is a specific calculation of uniqueness. Therefore, while both are true, **Statement 2 is not the correct explanation for Statement 1**. You have navigated the logic, counted the cases, and arrived at the truth.

Select Answer:

Visualized Solution

Homogeneous System

The Trivial Solution

Probability of Solution

Matrix Representation

Condition for Unique Solution

Total Possible Outcomes

Favorable Cases

Case 1: $a d = 1, b c = 0$

Case 2: $a d = 0, b c = 1$

Probability of Unique Solution

Final Verdict

The Sigma Insight: Solution of System of Linear Equations (Matrix Method and Cramer's Rule)

Analyzing the Setup

The Gatekeeper

The Combinatorial Dance

The Final Verdict

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