Is alone sufficient for infinitely many solutions?

No, can also mean no solution if any . However, here Eq(1) + Eq(2) gives . For this to be identically Eq(3) ( ), we need and RHS , which confirms infinite solutions.

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Matrices and Determinants: Mathematics | JEE Main

Let the system of linear equations

x+y+k z=2

2 x+3 y-z=1

3 x+4 y+2 z=k

have infinitely many solutions. Then the system

(\mathrm{k}+1) \mathrm{x}+(2 \mathrm{k}-1) \mathrm{y}=7

(2 \mathrm{k}+1) \mathrm{x}+(\mathrm{k}+5) \mathrm{y}=10

has:

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What feels right?

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

Ainfinitely many solutionsBunique solution satisfying

x-y=1

Cunique solution satisfying

x+y=1

Dno solution

Step-by-step solutionView

Correct answer

The original system has infinitely many solutions when $k = 3$ , giving a secondary system with the unique solution $x = 2, y = -1$ , which satisfies $x + y = 1$ .

Option analysis

Why each option works or fails

A · infinitely many solutions

Assuming that because the first $3 \times 3$ system has infinitely many solutions, the resulting $2 \times 2$ system in $k$ must also have infinitely many solutions without checking its determinant. Substitute the determined value of $k$ into the $2 \times 2$ system and evaluate its determinant; a non-zero determinant indicates a unique solution.

B · unique solution satisfying

x-y=1

Solving the $2 \times 2$ system correctly to find $x = 2$ and $y = -1$ , but making an arithmetic sign error when evaluating $x - y$ vs $x + y$ , obtaining $2 - (-1) = 1$ instead of $3$ . Check signs carefully when evaluating linear combinations of solutions: $x - y = 2 - (-1) = 3$ , whereas $x + y = 2 + (-1) = 1$ .

C · unique solution satisfying

x+y=1

None. The value $k = 3$ is correctly found from the consistency condition $R_1 + R_2 = R_3$ , which yields the system $4x + 5y = 7$ and $7x + 8y = 10$ , giving $(x, y) = (2, -1)$ satisfying $x + y = 1$ . None. This option is correct.

D · no solution

Conflating a non-zero determinant with inconsistency, or miscalculating the determinant of the $2 \times 2$ coefficient matrix as $0$ with inconsistent constants. Compute the determinant of the $2 \times 2$ system: $\Delta = 4(8) - 5(7) = 32 - 35 = -3 \neq 0$ . A non-zero determinant guarantees a unique solution, not no solution.

Reviewed route

Solution

StepWorking

01given

Given the 3-variable system: (1) $x + y + kz = 2$ (2) $2x + 3y - z = 1$ (3) $3x + 4y + 2z = k$ which has infinitely many solutions, and a secondary 2-variable system in $x$ and $y$ : $(k+1)x + (2k-1)y = 7$ $(2k+1)x + (k+5)y = 10$

02goal

Determine the value of $k$ from the first system, substitute it into the second system, and evaluate the nature and properties of its solution.

03approach

For a system of 3 linear equations to have infinitely many solutions, the coefficient determinant $\Delta$ must vanish. Find $k$ from $\Delta = 0$ , verify consistency with the constants, substitute $k$ into the $2 \times 2$ system, and find $(x, y)$ .

04execute

Evaluate $\Delta = \begin{vmatrix} 1 & 1 & k \\ 2 & 3 & -1 \\ 3 & 4 & 2 \end{vmatrix} = 1(6 - (-4)) - 1(4 - (-3)) + k(8 - 9) = 10 - 7 - k = 3 - k = 0 \implies k = 3$ . Notice also that adding equations (1) and (2) gives $(x+2x) + (y+3y) + (3-1)z = 3x + 4y + 2z = 2 + 1 = 3$ , which exactly matches equation (3) when $k = 3$ , ensuring infinitely many solutions.

05execute

Substitute $k = 3$ into the second system: $(3+1)x + (6-1)y = 7 \implies 4x + 5y = 7$ $(6+1)x + (3+5)y = 10 \implies 7x + 8y = 10$ Check the determinant: $D = 4(8) - 5(7) = 32 - 35 = -3 \neq 0$ , so the system has a unique solution.

06execute

Solve the $2 \times 2$ system: Multiply $4x + 5y = 7$ by 7 to get $28x + 35y = 49$ , and multiply $7x + 8y = 10$ by 4 to get $28x + 32y = 40$ . Subtracting gives $3y = 9 \implies y = 3$ . Then $4x + 5(3) = 7 \implies 4x = -8 \implies x = -2$ . Test the relation $x + y$ : $x + y = -2 + 3 = 1$ .

✓verify

Check: $x - y = -2 - 3 = -5 \neq 1$ , while $x + y = 1$ is satisfied. Thus, the system has a unique solution satisfying $x + y = 1$ .

✓ 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: 9 September 2026

Quick checks

Students also ask

\Delta = 0

alone sufficient for infinitely many solutions?

No, $\Delta = 0$ can also mean no solution if any $\Delta_x, \Delta_y, \Delta_z \neq 0$ . However, here Eq(1) + Eq(2) gives $3x + 4y + (k-1)z = 3$ . For this to be identically Eq(3) ( $3x + 4y + 2z = k$ ), we need $k-1 = 2 \implies k = 3$ and RHS $= 3 = k$ , which confirms infinite solutions.

Answer

The original system has infinitely many solutions when $k = 3$ , giving a secondary system with the unique solution $x = 2, y = -1$ , which satisfies $x + y = 1$ .

Why each option works or fails

A: infinitely many solutions - Assuming that because the first $3 \times 3$ system has infinitely many solutions, the resulting $2 \times 2$ system in $k$ must also have infinitely many solutions without checking its determinant. Substitute the determined value of $k$ into the $2 \times 2$ system and evaluate its determinant; a non-zero determinant indicates a unique solution.
B: unique solution satisfying $x-y=1$ - Solving the $2 \times 2$ system correctly to find $x = 2$ and $y = -1$ , but making an arithmetic sign error when evaluating $x - y$ vs $x + y$ , obtaining $2 - (-1) = 1$ instead of $3$ . Check signs carefully when evaluating linear combinations of solutions: $x - y = 2 - (-1) = 3$ , whereas $x + y = 2 + (-1) = 1$ .
C · correct: unique solution satisfying $x+y=1$ - None. The value $k = 3$ is correctly found from the consistency condition $R_1 + R_2 = R_3$ , which yields the system $4x + 5y = 7$ and $7x + 8y = 10$ , giving $(x, y) = (2, -1)$ satisfying $x + y = 1$ . None. This option is correct.
D: no solution - Conflating a non-zero determinant with inconsistency, or miscalculating the determinant of the $2 \times 2$ coefficient matrix as $0$ with inconsistent constants. Compute the determinant of the $2 \times 2$ system: $\Delta = 4(8) - 5(7) = 32 - 35 = -3 \neq 0$ . A non-zero determinant guarantees a unique solution, not no solution.

Step-by-step solution

given: Given the 3-variable system: (1) $x + y + kz = 2$ (2) $2x + 3y - z = 1$ (3) $3x + 4y + 2z = k$ which has infinitely many solutions, and a secondary 2-variable system in $x$ and $y$ : $(k+1)x + (2k-1)y = 7$ $(2k+1)x + (k+5)y = 10$
goal: Determine the value of $k$ from the first system, substitute it into the second system, and evaluate the nature and properties of its solution.
approach: For a system of 3 linear equations to have infinitely many solutions, the coefficient determinant $\Delta$ must vanish. Find $k$ from $\Delta = 0$ , verify consistency with the constants, substitute $k$ into the $2 \times 2$ system, and find $(x, y)$ .
execute: Evaluate $\Delta = \begin{vmatrix} 1 & 1 & k \\ 2 & 3 & -1 \\ 3 & 4 & 2 \end{vmatrix} = 1(6 - (-4)) - 1(4 - (-3)) + k(8 - 9) = 10 - 7 - k = 3 - k = 0 \implies k = 3$ . Notice also that adding equations (1) and (2) gives $(x+2x) + (y+3y) + (3-1)z = 3x + 4y + 2z = 2 + 1 = 3$ , which exactly matches equation (3) when $k = 3$ , ensuring infinitely many solutions.
execute: Substitute $k = 3$ into the second system: $(3+1)x + (6-1)y = 7 \implies 4x + 5y = 7$ $(6+1)x + (3+5)y = 10 \implies 7x + 8y = 10$ Check the determinant: $D = 4(8) - 5(7) = 32 - 35 = -3 \neq 0$ , so the system has a unique solution.
execute: Solve the $2 \times 2$ system: Multiply $4x + 5y = 7$ by 7 to get $28x + 35y = 49$ , and multiply $7x + 8y = 10$ by 4 to get $28x + 32y = 40$ . Subtracting gives $3y = 9 \implies y = 3$ . Then $4x + 5(3) = 7 \implies 4x = -8 \implies x = -2$ . Test the relation $x + y$ : $x + y = -2 + 3 = 1$ .
verify: Check: $x - y = -2 - 3 = -5 \neq 1$ , while $x + y = 1$ is satisfied. Thus, the system has a unique solution satisfying $x + y = 1$ .

Shortcut:

When to use it: When the coefficients of the first two equations add directly to the third.

given: Equations: $E_1: x + y + kz = 2$ , $E_2: 2x + 3y - z = 1$ , $E_3: 3x + 4y + 2z = k$ .

approach: Notice $E_1 + E_2$ has coefficients: $x(1+2) + y(1+3) = 3x + 4y$ , which matches the $x$ and $y$ coefficients of $E_3$ . For infinitely many solutions, $E_1 + E_2$ must equal $E_3$ .

execute: Equating the $z$ -coefficients and constant terms: $k - 1 = 2 \implies k = 3$ , and $2 + 1 = k \implies k = 3$ . Both consistently yield $k = 3$ .

execute: Substitute $k = 3$ into the system of two equations: $4x + 5y = 7$ and $7x + 8y = 10$ . Adding both: $11x + 13y = 17$ . Subtracting the first from the second: $3x + 3y = 3 \implies x + y = 1$ .

verify: Since $4/7 \neq 5/8$ , the system is non-parallel and consistent, meaning the solution is unique and directly gives $x + y = 1$ without finding individual values of $x$ and $y$ .

Matrices and Determinants: Mathematics | JEE Main

What feels right?

The original system has infinitely many solutions when $k = 3$ , giving a secondary system with the unique solution $x = 2, y = -1$ , which satisfies $x + y = 1$ .

Why each option works or fails

Solution

01given

02goal

03approach

04execute

05execute

06execute

✓verify

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Let…

Consider the system of linear equation…

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Students also ask

Matrices and Determinants: Mathematics | JEE Main

The original system has infinitely many solutions when k=3k = 3k=3, giving a secondary system with the unique solution x=2,y=−1x = 2, y = -1x=2,y=−1, which satisfies x+y=1x + y = 1x+y=1.

01given

02goal

03approach

04execute

05execute

06execute

✓verify

Let…

Consider the system of linear equation…

If…

Students also ask

The original system has infinitely many solutions when $k = 3$ , giving a secondary system with the unique solution $x = 2, y = -1$ , which satisfies $x + y = 1$ .