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Values of the parameters for which a system of three equations has no solution

This type of question, asked 14 times: 2021 Paper 1, 2021 Paper 21, 2021 Paper 22, 2021 Paper 24, 14 Jun 2022, 25 Jun 2022, 26 Jun 2022, 27 Jun 2022, 28 Jun 2022, 25 Jul 2022, 10 Apr 2023, 23 Jan 2025, 21 Jan 2026 S2, 2 Apr 2026 S1. All 15 questions below, oldest first.

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  1. Question 1

    2021 Paper 1

    80
    The system of linear equations
    3x−2y−kz=103x- 2y-kz= 10
    2x−4y−2z=62x- 4y- 2z= 6
    x+2y−z=5mx+ 2y-z= 5m
    is inconsistent if:

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  2. Question 2

    2021 Paper 21

    70
    Let [λ]\lbrack\lambda\rbrack be the greatest integer less than or equal to λ\lambda. The set of all values of λ\lambda for which the system of linear equations x+y+z=4,3x+2y+5z=3x + y + z = 4,3x + 2y + 5z = 3, 9x+4y+(28+[λ])z=[λ]9x + 4y + (28 + \lbrack\lambda\rbrack)z = \lbrack\lambda\rbrack has a solution is:

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  3. Question 3

    2021 Paper 22

    73
    If the following system of linear equations
    2x+y+z=52x + y + z = 5
    x−y+z=3x - y + z = 3
    x+y+az=bx + y + az = b
    has no solution, then :

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  4. Question 4

    2021 Paper 24

    63
    Consider the system of linear equations
    −x+y+2z=0- x + y + 2z = 0
    3x−ay+5z=13x - ay + 5z = 1
    2x−2y−az=72x - 2y - az = 7
    Let S1S_{1} be the set of all a∈Ra\in R for which the system is inconsistent and S2S_{2} be the set of all a∈Ra\in R for which the system has infinitely many solutions. If n( S1)n\left( {\text{ }S}_{1} \right) and n(S2)n\left( S_{2} \right) denote the number of elements in S1S_{1} and S2S_{2} respectively, then

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  5. Question 5

    14 Jun 2022

    66
    The number of values of α\alpha for which the system of equations:
    x+y+z=αx + y + z =\alpha
    αx+2αy+3z=−1\alpha x+ 2\alpha y+ 3z= - 1
    x+3αy+5z=4x+ 3\alpha y+ 5z= 4
    is inconsistent, is

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  6. Question 6

    25 Jun 2022

    154
    The system of equations
    −kx+3y−14z=25- kx + 3y - 14z = 25
    −15x+4y−kz=3- 15x + 4y - kz = 3
    −4x+y+3z=4- 4x + y + 3z = 4
    is consistent for all kk in the set

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  7. Question 7

    26 Jun 2022

    64
    The ordered pair (a,b)(a,b), for which the system of linear equations
    3x−2y+z=b3x - 2y + z = b
    5x−8y+9z=35x - 8y + 9z = 3
    2x+y+az=−12x + y + az = - 1
    has no solution, is:

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  8. Question 8

    27 Jun 2022

    62
    Let the system of linear equations x+2y+z=2x+ 2y+z= 2, αx+3y−z=α,−αx+y+2z=−α\alpha x+ 3y-z=\alpha, -\alpha x+y+ 2z= -\alpha be inconsistent. Then α\alpha is equal to:

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  9. Question 9

    28 Jun 2022

    63
    If the system of linear equations
    2x+3y−z=−22x + 3y - z = - 2
    x+y+z=4x + y + z = 4
    x−y+∣λ∣z=4λ−4x - y + |\lambda|z = 4\lambda - 4
    where λ∈R\lambda \in R, has no solution, then

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  10. Question 10

    25 Jul 2022

    64
    The number of θ∈(0,4π)\theta \in (0,4\pi) for which the system of linear equations
    3(sin⁡3θ)x−y+z=23(\sin3\theta)x - y + z = 2
    3(cos⁡2θ)x+4y+3z=33(\cos2\theta)x + 4y + 3z = 3
    6x+7y+7z=96x + 7y + 7z = 9
    has no solution is:

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  11. Question 11

    25 Jul 2022

    152
    The number of real values λ\lambda, such that the system of linear equations
    2x−3y+5z=92x - 3y + 5z = 9
    x+3y−z=−18x + 3y - z = - 18
    3x−y+(λ2−∣λ∣)z=163x - y +\left( \lambda^{2}- |\lambda| \right)z = 16
    has no solution, is:-

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  12. Question 12

    10 Apr 2023

    172
    Let SS be the set of values of λ\lambda, for which the system of equations 6λx−3y+3z=4λ26\lambda x - 3y + 3z = 4\lambda^{2}, 2x+6λy+4z=12x + 6\lambda y + 4z = 1, 3x+2y+3λz=λ3x + 2y + 3\lambda z = \lambda has no solution. Then 12∑λ∈S∣λ∣12\sum_{\lambda \in S} |\lambda| is equal to

  13. Question 13

    23 Jan 2025

    128
    The system of equations
    x+y+z=6x+2y+5z=9x+5y+λz=μ\begin{matrix} & x + y + z = 6 \\ & x + 2y + 5z = 9 \\ & x + 5y + \lambda z = \mu \end{matrix}
    has no solution if

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  14. Question 14

    21 Jan 2026 S2

    68
    If the system of equations 3x+y+4z=33x + y + 4z = 3, 2x+αy−z=−32x + \alpha y - z = - 3, x+2y+z=4x + 2y + z = 4 has no solution, then the value of α\alpha is equal to:

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  15. Question 15

    2 Apr 2026 S1

    53
    Let α,β∈R\alpha,\beta\in R be such that the system of linear equations
    x+2y+z=5x+ 2y+z= 5
    2x+y+αz=52x+y+\alpha z= 5
    8x+4y+βz=188x+ 4y+\beta z= 18
    has no solution. Then βα\frac{\beta}{\alpha} is equal to :

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