Let S1 and S2 be respectively the sets of all a ∈ R - {0} for which the system of linear equations a

Q: Wait, why do we reject the negative root here? Or do we?

Great question! In this problem, we aren't actually rejecting a root; we are proving that no real roots exist for the quadratic part. When we calculated the discriminant $D = -1259$, we found it to be negative. This means the quadratic $15a^2 + 31a + 37$ never crosses the x-axis. Since the leading coefficient (15) is positive, the parabola opens upwards and stays strictly above the x-axis for all real $a$. Thus, it can never be zero.

Q: Why can't we just use Cramer's rule to find $x, y, z$ directly?

You absolutely could, but that would be a massive amount of work! Cramer's rule requires calculating four determinants ($\Delta, \Delta_x, \Delta_y, \Delta_z$). The problem only asks for the *nature* of the solutions (unique vs. infinite), not the values of $x, y, z$. The determinant $\Delta$ is the master key—it tells you everything you need to know about the system's behavior without ever needing to solve for the variables.

Q: Is it possible for the system to be inconsistent?

Yes, in general, a system with $\Delta = 0$ can be inconsistent (no solution) or have infinitely many solutions. However, in this specific problem, we found that $\Delta$ is never zero for any $a \in \mathbb{R} - \{0\}$. Because $\Delta$ is never zero, the system is always consistent and always has a unique solution. Therefore, the case of 'no solution' or 'infinitely many solutions' simply never occurs here.

Question

Let $S_1$ and $S_2$ be respectively the sets of all $a \in \mathbb{R} - \{0\}$ for which the system of linear equations $ax + 2ay - 3az = 1$, $(2a + 1)x + (2a + 3)y + (a + 1)z = 2$, $(3a + 5)x + (a + 5)y + (a + 2)z = 3$ has unique solution and infinitely many solutions. Then

Options:

(A) $n(S_1) = 2$ and $S_2$ is an infinite set

(B) $S_1$ is an infinite set and $n(S_2) = 2$

(C) $S_1 = \Phi$ and $S_2 = \mathbb{R} - \{0\}$

(D) $S_1 = \mathbb{R} - \{0\}$ and $S_2 = \Phi$

Sigma Percentile · Accepted Answer

### The Gatekeeper of Linear Systems

Imagine you are standing before a complex system of three linear equations. With variables $x$, $y$, and $z$ dancing around a parameter $a$, it feels like the system could behave in any number of ways.

In the world of JEE Advanced, we don't guess; we analyze. The key to unlocking this mystery lies in the determinant of the coefficient matrix, which we call $\Delta$.

Think of $\Delta$ as the gatekeeper. If $\Delta 
eq 0$, the system is stable, predictable, and yields a unique solution. If $\Delta = 0$, the system enters a state of flux, potentially leading to infinitely many solutions or no solution at all.

Our mission is to find the values of $a$ that make this gatekeeper open or close.

### The Art of Simplification

We start by constructing our matrix from the coefficients of $x$, $y$, and $z$:

$$\Delta = \begin{vmatrix} a & 2a & -3a \ 2a+1 & 2a+3 & a+1 \ 3a+5 & a+5 & a+2 \end{vmatrix}$$

Many students would jump straight into expansion here, but that is a trap. Look at the first row: $a, 2a, -3a$. There is a common factor of $a$ waiting to be pulled out.

Let's extract it:

$$\Delta = a \begin{vmatrix} 1 & 2 & -3 \ 2a+1 & 2a+3 & a+1 \ 3a+5 & a+5 & a+2 \end{vmatrix}$$

Now, we have a much cleaner matrix. We want to create zeros to make the expansion trivial. By applying the column operations $C_2 	o C_2 - 2C_1$ and $C_3 	o C_3 + 3C_1$, we transform the first row into $1, 0, 0$.

The determinant becomes:

$$\Delta = a \begin{vmatrix} 1 & 0 & 0 \ 2a+1 & -2a+1 & 7a+4 \ 3a+5 & -5a-5 & 10a+17 \end{vmatrix}$$

### The Quadratic Mystery

Expanding along the first row is now a breeze. We are left with a two-by-two determinant calculation:

$$\Delta = a [ (-2a+1)(10a+17) - (-5a-5)(7a+4) ]$$

Let's expand these products carefully. The first term gives us $-20a^2 - 24a + 17$, and the second gives us $-35a^2 - 55a - 20$. Subtracting them, we get:

$$\Delta = a (15a^2 + 31a + 37)$$

Now, we must ask: when is $\Delta = 0$? Since the problem restricts $a$ to $\mathbb{R} - \{0\}$, the $a$ outside the bracket is never zero.

The only hope for $\Delta = 0$ is if the quadratic $15a^2 + 31a + 37$ equals zero. To check this, we look at the discriminant $D = b^2 - 4ac$:

$$D = (31)^2 - 4(15)(37) = 961 - 2220 = -1259$$

Since $D < 0$, the quadratic has no real roots. Because the leading coefficient is positive, the quadratic is always positive and never touches the $x$-axis.

### The Final Realization

We have reached the climax of our journey. We have proven that $\Delta$ is never zero for any $a \in \mathbb{R} - \{0\}$.

This means the system is always well-behaved, always consistent, and always yields a unique solution. Therefore, the set $S_1$ (unique solutions) is the entire set $\mathbb{R} - \{0\}$, and the set $S_2$ (infinitely many solutions) is the empty set, $\Phi$.

We have navigated the trap, simplified the chaos, and arrived at the elegant truth. The structure of this system is **perfectly unique**.

Select Answer:

Visualized Solution

System of Equations

The Determinant $Δ$

Factoring out $a$

Column Operations

Expanding $Δ$

Simplifying the Expression

Analyzing the Quadratic

Checking the Discriminant

Final Conclusion

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

The Gatekeeper of Linear Systems

The Art of Simplification

The Quadratic Mystery

The Final Realization

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Select Answer:

Visualized Solution

System of Equations

The Determinant .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Δ

Factoring out .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a

Column Operations

Expanding .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Δ