JEE MainMathematics

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Determinant condition for unique and infinite solutions of a linear system

Q: Can S_2 have elements if Δ = 0?

Yes, infinitely many solutions require Δ = 0 along with all sub-determinants Δ_x = Δ_y = Δ_z = 0. Since Δ ≠ 0 for all a ∈ ℝ - {0}, Δ = 0 never occurs, so S_2 is necessarily empty.

Let S_1 and S_2 be respectively the sets of all a ∈ 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

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AS_1 is an infinite set and n(S_2)=2BS_1=Φ and S_2=ℝ-{0}Cn(S_1)=2 and S_2 is an infinite setDS_1=ℝ-{0} and S_2=Φ

Step-by-step solutionView

Correct answer

The coefficient determinant simplifies to $Δ = - 5 a (a^{2} + 1)$ , which is non-zero for all $a \in R - {0}$ , meaning the system always has a unique solution and never has infinitely many solutions.

Option analysis

Why each option works or fails

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

Believing that solving $a^{2} + 1 = 0$ yields real roots ( $a = \pm 1$ or $a = \pm i$ ) that create non-unique solutions. Check whether solutions to polynomial factors are real numbers. Since $a^{2} + 1 = 0$ has no real roots and $a \neq 0$ , the determinant $Δ = - 5 a (a^{2} + 1)$ has no roots in $R - {0}$ .

B · S_1=Φ and S_2=ℝ-{0}

Confusing the condition for unique solutions ( $Δ \neq 0$ ) with that for infinitely many solutions, reversing the sets. A system of linear equations has a unique solution when its determinant is non-zero, not when it vanishes.

C · n(S_1)=2 and S_2 is an infinite set

Treating the solutions of $a^{2} + 1 = 0$ as points where the system has a unique solution, while assuming non-zero determinant implies infinitely many solutions. Recall that $Δ \neq 0$ implies a unique solution (Cramer's rule), which holds everywhere on $R - {0}$ , leaving no $a$ for infinitely many solutions.

D · S_1=ℝ-{0} and S_2=Φ

This correctly identifies that $Δ \neq 0$ for every non-zero real number $a$ . Correct: $Δ = - 5 a (a^{2} + 1) \neq 0$ for all real $a \neq 0$ , so $S_{1} = R - {0}$ and $S_{2} = Φ$ .

Reviewed route

Solution

StepWorking

01Given

The system of linear equations in $x, y, z$ : 1) $a x + 2 a y - 3 a z = 1$ 2) $(2 a + 1) x + (2 a + 3) y + (a + 1) z = 2$ 3) $(3 a + 5) x + (a + 5) y + (a + 2) z = 3$ with $a \in R - {0}$ . $S_{1}$ is the set of $a$ giving a unique solution, and $S_{2}$ is the set giving infinitely many solutions.

02Goal

Determine the sets $S_{1}$ and $S_{2}$ based on the coefficient determinant $Δ$ .

03Approach

A system of linear equations has a unique solution if $Δ \neq 0$ . If $Δ = 0$ , it has either no solution or infinitely many solutions. We compute $Δ = \det (A)$ and find its real roots.

04Execute

Factor out $a$ from the first row of $Δ$ : $Δ = ∣ \begin{array}{ccc} a & 2 a & - 3 a \\ 2 a + 1 & 2 a + 3 & a + 1 \\ 3 a + 5 & a + 5 & a + 2 \end{array} ∣ = a ∣ \begin{array}{ccc} 1 & 2 & - 3 \\ 2 a + 1 & 2 a + 3 & a + 1 \\ 3 a + 5 & a + 5 & a + 2 \end{array} ∣$ Perform column operations $C_{2} \to C_{2} - 2 C_{1}$ and $C_{3} \to C_{3} + 3 C_{1}$ : $∣ \begin{array}{ccc} 1 & 0 & 0 \\ 2 a + 1 & (2 a + 3) - 2 (2 a + 1) & (a + 1) + 3 (2 a + 1) \\ 3 a + 5 & (a + 5) - 2 (3 a + 5) & (a + 2) + 3 (3 a + 5) \end{array} ∣ = ∣ \begin{array}{ccc} 1 & 0 & 0 \\ 2 a + 1 & 1 - 2 a & 7 a + 4 \\ 3 a + 5 & - 5 a - 5 & 10 a + 17 \end{array} ∣$ Expanding along the first row: $Δ = a [(1 - 2 a) (10 a + 17) - (7 a + 4) (- 5 a - 5)]$ $= a [(- 20 a^{2} - 24 a + 17) - (- 35 a^{2} - 55 a - 20)]$ $= a [15 a^{2} + 31 a + 37]$

05Execute

Check the discriminant of the quadratic factor $15 a^{2} + 31 a + 37$ : $D = 31^{2} - 4 (15) (37) = 961 - 2220 = - 1259 < 0$ Since the discriminant is strictly negative, $15 a^{2} + 31 a + 37 > 0$ for all real $a$ . Thus, $Δ = 0 ⟺ a = 0$ . Since $a \in R - {0}$ , we have $Δ \neq 0$ for all valid $a$ . Therefore, the system has a unique solution for all $a \in R - {0}$ , meaning $S_{1} = R - {0}$ and $S_{2} = Φ$ .

✓Verify

At $a = 1$ , the first equation is $x + 2 y - 3 z = 1$ . The determinant is $1 \times (15 (1) + 31 (1) + 37) = 83 \neq 0$ , which gives a unique solution. This confirms $a = 1 \in S_{1}$ , eliminating options where $S_{1} = Φ$ or $n (S_{1}) = 2$ .

Hints that build this answer step by step

What is the primary condition on the coefficient matrix determinant $Δ$ to distinguish between a unique solution and non-unique solutions?
The system has a unique solution if $Δ \neq 0$ ; if $Δ = 0$ , it has either infinitely many solutions or no solution.
Evaluate the determinant of the coefficient matrix: $Δ = ∣ \begin{array}{ccc} a & 2 a & - 3 a \\ 2 a + 1 & 2 a + 3 & a + 1 \\ 3 a + 5 & a + 5 & a + 2 \end{array} ∣$
$Δ = - 5 a (a^{2} + 1)$
For which real values of $a \in R - {0}$ does $Δ = - 5 a (a^{2} + 1) = 0$ ?
There are no real solutions in $R - {0}$ , because $a^{2} + 1 > 0$ for all real $a$ .

✓ Source and academic review

Question type: Single correct
Exam relevance: JEE Main · Mathematics
Academic status: Answer verified against the official NTA key
Source: Previous-year question
Editorial review: 8 September 2026

Quick checks

Students also ask

Can

S_{2}

have elements if

Δ = 0

Yes, infinitely many solutions require $Δ = 0$ along with all sub-determinants $Δ_{x} = Δ_{y} = Δ_{z} = 0$ . Since $Δ \neq 0$ for all $a \in R - {0}$ , $Δ = 0$ never occurs, so $S_{2}$ is necessarily empty.

Answer

The coefficient determinant simplifies to $Δ = - 5 a (a^{2} + 1)$ , which is non-zero for all $a \in R - {0}$ , meaning the system always has a unique solution and never has infinitely many solutions.

Why each option works or fails

A: S_1 is an infinite set and n(S_2)=2 - Believing that solving $a^{2} + 1 = 0$ yields real roots ( $a = \pm 1$ or $a = \pm i$ ) that create non-unique solutions. Check whether solutions to polynomial factors are real numbers. Since $a^{2} + 1 = 0$ has no real roots and $a \neq 0$ , the determinant $Δ = - 5 a (a^{2} + 1)$ has no roots in $R - {0}$ .
B: S_1=Φ and S_2=ℝ-{0} - Confusing the condition for unique solutions ( $Δ \neq 0$ ) with that for infinitely many solutions, reversing the sets. A system of linear equations has a unique solution when its determinant is non-zero, not when it vanishes.
C: n(S_1)=2 and S_2 is an infinite set - Treating the solutions of $a^{2} + 1 = 0$ as points where the system has a unique solution, while assuming non-zero determinant implies infinitely many solutions. Recall that $Δ \neq 0$ implies a unique solution (Cramer's rule), which holds everywhere on $R - {0}$ , leaving no $a$ for infinitely many solutions.
D · correct: S_1=ℝ-{0} and S_2=Φ - This correctly identifies that $Δ \neq 0$ for every non-zero real number $a$ . Correct: $Δ = - 5 a (a^{2} + 1) \neq 0$ for all real $a \neq 0$ , so $S_{1} = R - {0}$ and $S_{2} = Φ$ .

Step-by-step solution

Given: The system of linear equations in $x, y, z$ : 1) $a x + 2 a y - 3 a z = 1$ 2) $(2 a + 1) x + (2 a + 3) y + (a + 1) z = 2$ 3) $(3 a + 5) x + (a + 5) y + (a + 2) z = 3$ with $a \in R - {0}$ . $S_{1}$ is the set of $a$ giving a unique solution, and $S_{2}$ is the set giving infinitely many solutions.
Goal: Determine the sets $S_{1}$ and $S_{2}$ based on the coefficient determinant $Δ$ .
Approach: A system of linear equations has a unique solution if $Δ \neq 0$ . If $Δ = 0$ , it has either no solution or infinitely many solutions. We compute $Δ = \det (A)$ and find its real roots.
Execute: Factor out $a$ from the first row of $Δ$ : $Δ = ∣ \begin{array}{ccc} a & 2 a & - 3 a \\ 2 a + 1 & 2 a + 3 & a + 1 \\ 3 a + 5 & a + 5 & a + 2 \end{array} ∣ = a ∣ \begin{array}{ccc} 1 & 2 & - 3 \\ 2 a + 1 & 2 a + 3 & a + 1 \\ 3 a + 5 & a + 5 & a + 2 \end{array} ∣$ Perform column operations $C_{2} \to C_{2} - 2 C_{1}$ and $C_{3} \to C_{3} + 3 C_{1}$ : $∣ \begin{array}{ccc} 1 & 0 & 0 \\ 2 a + 1 & (2 a + 3) - 2 (2 a + 1) & (a + 1) + 3 (2 a + 1) \\ 3 a + 5 & (a + 5) - 2 (3 a + 5) & (a + 2) + 3 (3 a + 5) \end{array} ∣ = ∣ \begin{array}{ccc} 1 & 0 & 0 \\ 2 a + 1 & 1 - 2 a & 7 a + 4 \\ 3 a + 5 & - 5 a - 5 & 10 a + 17 \end{array} ∣$ Expanding along the first row: $Δ = a [(1 - 2 a) (10 a + 17) - (7 a + 4) (- 5 a - 5)]$ $= a [(- 20 a^{2} - 24 a + 17) - (- 35 a^{2} - 55 a - 20)]$ $= a [15 a^{2} + 31 a + 37]$
Execute: Check the discriminant of the quadratic factor $15 a^{2} + 31 a + 37$ : $D = 31^{2} - 4 (15) (37) = 961 - 2220 = - 1259 < 0$ Since the discriminant is strictly negative, $15 a^{2} + 31 a + 37 > 0$ for all real $a$ . Thus, $Δ = 0 ⟺ a = 0$ . Since $a \in R - {0}$ , we have $Δ \neq 0$ for all valid $a$ . Therefore, the system has a unique solution for all $a \in R - {0}$ , meaning $S_{1} = R - {0}$ and $S_{2} = Φ$ .
Verify: At $a = 1$ , the first equation is $x + 2 y - 3 z = 1$ . The determinant is $1 \times (15 (1) + 31 (1) + 37) = 83 \neq 0$ , which gives a unique solution. This confirms $a = 1 \in S_{1}$ , eliminating options where $S_{1} = Φ$ or $n (S_{1}) = 2$ .

Determinant condition for unique and infinite solutions of a linear system

What feels right?

The coefficient determinant simplifies to $Δ = - 5 a (a^{2} + 1)$ , which is non-zero for all $a \in R - {0}$ , meaning the system always has a unique solution and never has infinitely many solutions.

Why each option works or fails

Solution

01Given

02Goal

03Approach

04Execute

05Execute

✓Verify

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