Can have elements if ?

Yes, infinitely many solutions require along with all sub-determinants . Since for all , never occurs, so is necessarily empty.

JEE MainMathematics

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Matrices and Determinants: Mathematics | JEE Main

Let

\mathrm{S}_1

and

\mathrm{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

Your answer stays private

What feels right?

No score. Commit to your first instinct. We’ll show what your mind noticed and what it missed.

\mathrm{S}_1

is an infinite set and

\mathrm{n}(\mathrm{S}_2)=2

\mathrm{S}_1=\Phi

and

\mathrm{S}_2=\mathbb{R}-\{0\}

\mathrm{n}(\mathrm{S}_1)=2

and

\mathrm{S}_2

is an infinite setD

\mathrm{S}_1=\mathbb{R}-\{0\}

and

\mathrm{S}_2=\Phi

Step-by-step solutionView

Correct answer

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

Option analysis

Why each option works or fails

A ·

\mathrm{S}_1

is an infinite set and

\mathrm{n}(\mathrm{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 $\Delta = -5a(a^2 + 1)$ has no roots in $\mathbb{R}-\{0\}$ .

B ·

\mathrm{S}_1=\Phi

and

\mathrm{S}_2=\mathbb{R}-\{0\}

Confusing the condition for unique solutions ( $\Delta \neq 0$ ) 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 ·

\mathrm{n}(\mathrm{S}_1)=2

and

\mathrm{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 $\Delta \neq 0$ implies a unique solution (Cramer's rule), which holds everywhere on $\mathbb{R}-\{0\}$ , leaving no $a$ for infinitely many solutions.

D ·

\mathrm{S}_1=\mathbb{R}-\{0\}

and

\mathrm{S}_2=\Phi

None. This correctly identifies that $\Delta \neq 0$ for every non-zero real number $a$ . Correct: $\Delta = -5a(a^2 + 1) \neq 0$ for all real $a \neq 0$ , so $\mathrm{S}_1 = \mathbb{R}-\{0\}$ and $\mathrm{S}_2 = \Phi$ .

Reviewed route

Solution

StepWorking

01given

The system of linear equations in $x, y, z$ : 1) $ax + 2ay - 3az = 1$ 2) $(2a+1)x + (2a+3)y + (a+1)z = 2$ 3) $(3a+5)x + (a+5)y + (a+2)z = 3$ with $a \in \mathbb{R} - \{0\}$ . $S_1$ 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 $\Delta$ .

03approach

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

04execute

Factor out $a$ from the first row of $\Delta$ : $\Delta = \left|\begin{matrix} a & 2a & -3a \\ 2a+1 & 2a+3 & a+1 \\ 3a+5 & a+5 & a+2 \end{matrix}\right| = a \left|\begin{matrix} 1 & 2 & -3 \\ 2a+1 & 2a+3 & a+1 \\ 3a+5 & a+5 & a+2 \end{matrix}\right|$ Perform column operations $C_2 \to C_2 - 2C_1$ and $C_3 \to C_3 + 3C_1$ : $\left|\begin{matrix} 1 & 0 & 0 \\ 2a+1 & (2a+3)-2(2a+1) & (a+1)+3(2a+1) \\ 3a+5 & (a+5)-2(3a+5) & (a+2)+3(3a+5) \end{matrix}\right| = \left|\begin{matrix} 1 & 0 & 0 \\ 2a+1 & 1-2a & 7a+4 \\ 3a+5 & -5a-5 & 10a+17 \end{matrix}\right|$ Expanding along the first row: $\Delta = a [(1-2a)(10a+17) - (7a+4)(-5a-5)]$ $= a [(-20a^2 - 24a + 17) - (-35a^2 - 55a - 20)]$ $= a [15a^2 + 31a + 37]$

05execute

Check the discriminant of the quadratic factor $15a^2 + 31a + 37$ : $D = 31^2 - 4(15)(37) = 961 - 2220 = -1259 < 0$ Since the discriminant is strictly negative, $15a^2 + 31a + 37 > 0$ for all real $a$ . Thus, $\Delta = 0 \iff a = 0$ . Since $a \in \mathbb{R} - \{0\}$ , we have $\Delta \neq 0$ for all valid $a$ . Therefore, the system has a unique solution for all $a \in \mathbb{R} - \{0\}$ , meaning $S_1 = \mathbb{R} - \{0\}$ and $S_2 = \Phi$ .

✓verify

At $a = 1$ , the first equation is $x + 2y - 3z = 1$ . 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 $S_1 = \Phi$ or $n(S_1) = 2$ .

Hints that build this answer step by step

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

✓ Source and academic review

Question type: Single correct
Exam relevance: JEE Main · Mathematics
Concepts assessed: Mathematics
Academic status: Reviewed by official_key
Source: pyq
Editorial review: 8 September 2026

Quick checks

Students also ask

Can

S_2

have elements if

\Delta = 0

Yes, infinitely many solutions require $\Delta = 0$ along with all sub-determinants $\Delta_x = \Delta_y = \Delta_z = 0$ . Since $\Delta \neq 0$ for all $a \in \mathbb{R} - \{0\}$ , $\Delta = 0$ never occurs, so $S_2$ is necessarily empty.

Answer

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

Why each option works or fails

A: $\mathrm{S}_1$ is an infinite set and $\mathrm{n}(\mathrm{S}_2)=2$ - 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 $\Delta = -5a(a^2 + 1)$ has no roots in $\mathbb{R}-\{0\}$ .
B: $\mathrm{S}_1=\Phi$ and $\mathrm{S}_2=\mathbb{R}-\{0\}$ - Confusing the condition for unique solutions ( $\Delta \neq 0$ ) 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: $\mathrm{n}(\mathrm{S}_1)=2$ and $\mathrm{S}_2$ 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 $\Delta \neq 0$ implies a unique solution (Cramer's rule), which holds everywhere on $\mathbb{R}-\{0\}$ , leaving no $a$ for infinitely many solutions.
D · correct: $\mathrm{S}_1=\mathbb{R}-\{0\}$ and $\mathrm{S}_2=\Phi$ - None. This correctly identifies that $\Delta \neq 0$ for every non-zero real number $a$ . Correct: $\Delta = -5a(a^2 + 1) \neq 0$ for all real $a \neq 0$ , so $\mathrm{S}_1 = \mathbb{R}-\{0\}$ and $\mathrm{S}_2 = \Phi$ .

Step-by-step solution

given: The system of linear equations in $x, y, z$ : 1) $ax + 2ay - 3az = 1$ 2) $(2a+1)x + (2a+3)y + (a+1)z = 2$ 3) $(3a+5)x + (a+5)y + (a+2)z = 3$ with $a \in \mathbb{R} - \{0\}$ . $S_1$ 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 $\Delta$ .
approach: A system of linear equations has a unique solution if $\Delta \neq 0$ . If $\Delta = 0$ , it has either no solution or infinitely many solutions. We compute $\Delta = \det(A)$ and find its real roots.
execute: Factor out $a$ from the first row of $\Delta$ : $\Delta = \left|\begin{matrix} a & 2a & -3a \\ 2a+1 & 2a+3 & a+1 \\ 3a+5 & a+5 & a+2 \end{matrix}\right| = a \left|\begin{matrix} 1 & 2 & -3 \\ 2a+1 & 2a+3 & a+1 \\ 3a+5 & a+5 & a+2 \end{matrix}\right|$ Perform column operations $C_2 \to C_2 - 2C_1$ and $C_3 \to C_3 + 3C_1$ : $\left|\begin{matrix} 1 & 0 & 0 \\ 2a+1 & (2a+3)-2(2a+1) & (a+1)+3(2a+1) \\ 3a+5 & (a+5)-2(3a+5) & (a+2)+3(3a+5) \end{matrix}\right| = \left|\begin{matrix} 1 & 0 & 0 \\ 2a+1 & 1-2a & 7a+4 \\ 3a+5 & -5a-5 & 10a+17 \end{matrix}\right|$ Expanding along the first row: $\Delta = a [(1-2a)(10a+17) - (7a+4)(-5a-5)]$ $= a [(-20a^2 - 24a + 17) - (-35a^2 - 55a - 20)]$ $= a [15a^2 + 31a + 37]$
execute: Check the discriminant of the quadratic factor $15a^2 + 31a + 37$ : $D = 31^2 - 4(15)(37) = 961 - 2220 = -1259 < 0$ Since the discriminant is strictly negative, $15a^2 + 31a + 37 > 0$ for all real $a$ . Thus, $\Delta = 0 \iff a = 0$ . Since $a \in \mathbb{R} - \{0\}$ , we have $\Delta \neq 0$ for all valid $a$ . Therefore, the system has a unique solution for all $a \in \mathbb{R} - \{0\}$ , meaning $S_1 = \mathbb{R} - \{0\}$ and $S_2 = \Phi$ .
verify: At $a = 1$ , the first equation is $x + 2y - 3z = 1$ . 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 $S_1 = \Phi$ or $n(S_1) = 2$ .

Matrices and Determinants: Mathematics | JEE Main

What feels right?

The coefficient determinant simplifies to $\Delta = -5a(a^2 + 1)$ , which is non-zero for all $a \in \mathbb{R}-\{0\}$ , 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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Matrices and Determinants: Mathematics | JEE Main

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