Do we also need to check and ?

Since row 1 and row 2 of the coefficient matrix are linearly independent (the normals are not parallel), setting and guarantees that the third plane belongs to the pencil of the first two planes, which automatically makes and .

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Matrices and Determinants: Mathematics | JEE Main

For

\alpha, \beta \in \mathbb{R}

, suppose the system of linear equations

\mathrm{x} - \mathrm{y} + \mathrm{z} = 5

2\mathrm{x} + 2\mathrm{y} + \alpha \mathrm{z} = 8

3\mathrm{x} - \mathrm{y} + 4\mathrm{z} = \beta

has infinitely many solutions. Then

\alpha

and

\beta

are the roots of

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\mathrm{x}^2 + 14\mathrm{x} + 24 = 0

\mathrm{x}^2 + 18\mathrm{x} + 56 = 0

\mathrm{x}^2 - 18\mathrm{x} + 56 = 0

\mathrm{x}^2 - 10\mathrm{x} + 16 = 0

Step-by-step solutionView

Correct answer

The system has infinitely many solutions when $\alpha = 4$ and $\beta = 14$ , making them the roots of $x^2 - 18x + 56 = 0$ .

Option analysis

Why each option works or fails

A ·

\mathrm{x}^2 + 14\mathrm{x} + 24 = 0

Believing that sign errors in solving for $\alpha$ or $\beta$ yield roots of opposite signs or different values such as $2$ and $12$ . Verify that $\alpha + \beta = 4 + 14 = 18$ and $\alpha\beta = 56$ , so the quadratic is $x^2 - (\alpha + \beta)x + \alpha\beta = 0$ , giving $x^2 - 18x + 56 = 0$ .

B ·

\mathrm{x}^2 + 18\mathrm{x} + 56 = 0

Formulating the quadratic equation with the wrong sign for the linear term as $x^2 + (\alpha+\beta)x + \alpha\beta = 0$ . Recall that a monic quadratic with roots $r_1, r_2$ is $x^2 - (r_1+r_2)x + r_1r_2 = 0$ , which yields $x^2 - 18x + 56 = 0$ .

C ·

\mathrm{x}^2 - 18\mathrm{x} + 56 = 0

None. The values $\alpha = 4$ and $\beta = 14$ give sum 18 and product 56, matching $x^2 - 18x + 56 = 0$ . Correctly determined $\alpha = 4$ and $\beta = 14$ and constructed the monic quadratic equation.

D ·

\mathrm{x}^2 - 10\mathrm{x} + 16 = 0

Miscalculating the linear combination coefficients, leading to values such as $\alpha = 2$ and $\beta = 8$ . Solve the coefficients of $x$ and $y$ carefully: $c_1(1, -1) + c_2(2, 2) = (3, -1)$ yields $c_1 = 2$ and $c_2 = 0.5$ .

Reviewed route

Solution

StepWorking

01given

The system of equations is: (1) $x - y + z = 5$ (2) $2x + 2y + \alpha z = 8$ (3) $3x - y + 4z = \beta$ It has infinitely many solutions.

02goal

Find the values of $\alpha$ and $\beta$ , then identify the quadratic equation whose roots are $\alpha$ and $\beta$ .

03approach

For a system of three linear equations to have infinitely many solutions, the main determinant $\Delta$ must equal 0, and all minor determinants $\Delta_x, \Delta_y, \Delta_z$ must also equal 0. We first find $\alpha$ from $\Delta = 0$ , then find $\beta$ from $\Delta_x = 0$ , and finally form the quadratic equation $x^2 - (\alpha + \beta)x + \alpha\beta = 0$ .

04execute

Compute $\Delta = \begin{vmatrix} 1 & -1 & 1 \\ 2 & 2 & \alpha \\ 3 & -1 & 4 \end{vmatrix} = 1(8 + \alpha) + 1(8 - 3\alpha) + 1(-2 - 6) = 8 + \alpha + 8 - 3\alpha - 8 = 8 - 2\alpha = 0 \implies \alpha = 4$ .

05execute

Substitute $\alpha = 4$ into $\Delta_x = \begin{vmatrix} 5 & -1 & 1 \\ 8 & 2 & 4 \\ \beta & -1 & 4 \end{vmatrix} = 0$ . Expanding: $5(8 + 4) + 1(32 - 4\beta) + 1(-8 - 2\beta) = 60 + 32 - 4\beta - 8 - 2\beta = 84 - 6\beta = 0 \implies \beta = 14$ .

06execute

The quadratic equation having roots $\alpha = 4$ and $\beta = 14$ is $x^2 - (\alpha + \beta)x + \alpha\beta = x^2 - (4 + 14)x + (4 \times 14) = x^2 - 18x + 56 = 0$ .

✓verify

Check linear combination: Note that $E_3 = a E_1 + b E_2$ . Looking at coefficients of $x$ and $y$ : $a + 2b = 3$ and $-a + 2b = -1 \implies 4b = 2 \implies b = 1/2, a = 2$ . Then coefficient of $z$ must satisfy $2(1) + (1/2)\alpha = 4 \implies \alpha = 4$ . The constant term must satisfy $2(5) + (1/2)(8) = 14 = \beta$ . Both match $\alpha = 4$ and $\beta = 14$ .

✓ 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

Do we also need to check

\Delta_y = 0

and

\Delta_z = 0

Since row 1 and row 2 of the coefficient matrix are linearly independent (the normals are not parallel), setting $\Delta = 0$ and $\Delta_x = 0$ guarantees that the third plane belongs to the pencil of the first two planes, which automatically makes $\Delta_y = 0$ and $\Delta_z = 0$ .

Does an infinite solution set always mean one equation is a linear combination of the other two?

Yes, the first two equations must represent non-parallel planes. Then, the system has infinitely many solutions if and only if the third plane passes through their line of intersection. This means $E_3 = c_1 E_1 + c_2 E_2$ .

Answer

The system has infinitely many solutions when $\alpha = 4$ and $\beta = 14$ , making them the roots of $x^2 - 18x + 56 = 0$ .

Why each option works or fails

A: $\mathrm{x}^2 + 14\mathrm{x} + 24 = 0$ - Believing that sign errors in solving for $\alpha$ or $\beta$ yield roots of opposite signs or different values such as $2$ and $12$ . Verify that $\alpha + \beta = 4 + 14 = 18$ and $\alpha\beta = 56$ , so the quadratic is $x^2 - (\alpha + \beta)x + \alpha\beta = 0$ , giving $x^2 - 18x + 56 = 0$ .
B: $\mathrm{x}^2 + 18\mathrm{x} + 56 = 0$ - Formulating the quadratic equation with the wrong sign for the linear term as $x^2 + (\alpha+\beta)x + \alpha\beta = 0$ . Recall that a monic quadratic with roots $r_1, r_2$ is $x^2 - (r_1+r_2)x + r_1r_2 = 0$ , which yields $x^2 - 18x + 56 = 0$ .
C · correct: $\mathrm{x}^2 - 18\mathrm{x} + 56 = 0$ - None. The values $\alpha = 4$ and $\beta = 14$ give sum 18 and product 56, matching $x^2 - 18x + 56 = 0$ . Correctly determined $\alpha = 4$ and $\beta = 14$ and constructed the monic quadratic equation.
D: $\mathrm{x}^2 - 10\mathrm{x} + 16 = 0$ - Miscalculating the linear combination coefficients, leading to values such as $\alpha = 2$ and $\beta = 8$ . Solve the coefficients of $x$ and $y$ carefully: $c_1(1, -1) + c_2(2, 2) = (3, -1)$ yields $c_1 = 2$ and $c_2 = 0.5$ .

Step-by-step solution

given: The system of equations is: (1) $x - y + z = 5$ (2) $2x + 2y + \alpha z = 8$ (3) $3x - y + 4z = \beta$ It has infinitely many solutions.
goal: Find the values of $\alpha$ and $\beta$ , then identify the quadratic equation whose roots are $\alpha$ and $\beta$ .
approach: For a system of three linear equations to have infinitely many solutions, the main determinant $\Delta$ must equal 0, and all minor determinants $\Delta_x, \Delta_y, \Delta_z$ must also equal 0. We first find $\alpha$ from $\Delta = 0$ , then find $\beta$ from $\Delta_x = 0$ , and finally form the quadratic equation $x^2 - (\alpha + \beta)x + \alpha\beta = 0$ .
execute: Compute $\Delta = \begin{vmatrix} 1 & -1 & 1 \\ 2 & 2 & \alpha \\ 3 & -1 & 4 \end{vmatrix} = 1(8 + \alpha) + 1(8 - 3\alpha) + 1(-2 - 6) = 8 + \alpha + 8 - 3\alpha - 8 = 8 - 2\alpha = 0 \implies \alpha = 4$ .
execute: Substitute $\alpha = 4$ into $\Delta_x = \begin{vmatrix} 5 & -1 & 1 \\ 8 & 2 & 4 \\ \beta & -1 & 4 \end{vmatrix} = 0$ . Expanding: $5(8 + 4) + 1(32 - 4\beta) + 1(-8 - 2\beta) = 60 + 32 - 4\beta - 8 - 2\beta = 84 - 6\beta = 0 \implies \beta = 14$ .
execute: The quadratic equation having roots $\alpha = 4$ and $\beta = 14$ is $x^2 - (\alpha + \beta)x + \alpha\beta = x^2 - (4 + 14)x + (4 \times 14) = x^2 - 18x + 56 = 0$ .
verify: Check linear combination: Note that $E_3 = a E_1 + b E_2$ . Looking at coefficients of $x$ and $y$ : $a + 2b = 3$ and $-a + 2b = -1 \implies 4b = 2 \implies b = 1/2, a = 2$ . Then coefficient of $z$ must satisfy $2(1) + (1/2)\alpha = 4 \implies \alpha = 4$ . The constant term must satisfy $2(5) + (1/2)(8) = 14 = \beta$ . Both match $\alpha = 4$ and $\beta = 14$ .

Shortcut:

When to use it: Fastest method when equations are simple: find scalar multipliers relating the equations directly without computing 3x3 determinants.

given: $E_1: x - y + z = 5$ , $E_2: 2x + 2y + \alpha z = 8$ , $E_3: 3x - y + 4z = \beta$ .

approach: For infinite solutions, equation $E_3$ must be a linear combination of $E_1$ and $E_2$ , so $E_3 = c_1 E_1 + c_2 E_2$ .

execute: Equating coefficients of $x$ and $y$ : $c_1 + 2c_2 = 3$ $-c_1 + 2c_2 = -1$ Adding gives $4c_2 = 2 \implies c_2 = 1/2$ . Then $c_1 = 3 - 2(1/2) = 2$ .

execute: Now match coefficient of $\Delta_z = 0$ and the constant term: $z$ : $c_1(1) + c_2(\alpha) = 4 \implies 2 + \frac{1}{2}\alpha = 4 \implies \alpha = 4$ . Constant: $c_1(5) + c_2(8) = \beta \implies 2(5) + \frac{1}{2}(8) = \beta \implies \beta = 14$ .

execute: The quadratic equation with roots $4$ and $14$ is $x^2 - (4+14)x + (4)(14) = x^2 - 18x + 56 = 0$ .

verify: Sum of roots is $18$ and product is $56$ , which matches Option (2).

Matrices and Determinants: Mathematics | JEE Main

What feels right?

The system has infinitely many solutions when $\alpha = 4$ and $\beta = 14$ , making them the roots of $x^2 - 18x + 56 = 0$ .

Why each option works or fails

Solution

01given

02goal

03approach

04execute

05execute

06execute

✓verify

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Matrices and Determinants: Mathematics | JEE Main

The system has infinitely many solutions when α=4\alpha = 4α=4 and β=14\beta = 14β=14, making them the roots of x2−18x+56=0x^2 - 18x + 56 = 0x2−18x+56=0.

01given

02goal

03approach

04execute

05execute

06execute

✓verify

Let…

Consider the system of linear equation…

If…

Students also ask

The system has infinitely many solutions when $\alpha = 4$ and $\beta = 14$ , making them the roots of $x^2 - 18x + 56 = 0$ .