PYQ Vault
EbooksTutor

Determinants and Matrices

MHT-CET Maths. Try each one first, then open its answer.

Top 10 MHT-CET PYQs to practise first

The questions and kinds of question asked most often come first. Each one says how often and in which papers.

  1. Question 1

    This type asked 11 times: 4 May 2023 S1, 4 May 2023 S2, 16 May 2023 S2, 12 May 2024 S1, 19 Apr 2025 S2, 26 Apr 2025 S1, 26 Apr 2025 S2, 17 Apr 2026 S1, 12 May 2026 S2, 14 May 2026 S1, 19 May 2026 S1. This one: 19 May 2026 S1 See all 11 of this type

    69
    Let A=[12−51]A = \begin{bmatrix} 1 & 2 \\ -5 & 1 \end{bmatrix} and A−1=xA+yIA^{-1} = xA + yI, then the pair (x,y)=⋯(x, y) = \cdots

    Tap an option to check your answer.

  2. Question 2

    This type asked 8 times: 9 May 2023 S1, 22 Apr 2025 S2, 13 Apr 2026 S1, 18 Apr 2026 S2, 19 Apr 2026 S2, 12 May 2026 S2, 14 May 2026 S2, 15 May 2026 S2. This one: 15 May 2026 S2 See all 8 of this type

    72
    If A=[0121233a1]A = \begin{bmatrix} 0 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & a & 1 \end{bmatrix} and A−1=[12−1212−43c52−3212]A^{-1} = \begin{bmatrix} \dfrac{1}{2} & -\dfrac{1}{2} & \dfrac{1}{2} \\ -4 & 3 & c \\ \dfrac{5}{2} & -\dfrac{3}{2} & \dfrac{1}{2} \end{bmatrix}, then ...

    Tap an option to check your answer.

  3. Question 3

    This type asked 8 times: 2 May 2023 S2, 3 May 2023 S2, 11 May 2023 S1, 9 May 2024 S2, 20 Apr 2025 S2, 11 Apr 2026 S2, 13 May 2026 S1, 15 May 2026 S1. This one: 15 May 2026 S1 See all 8 of this type

    70
    If [1112−17312][xyz]=[172]\begin{bmatrix} 1 & 1 & 1 \\ 2 & -1 & 7 \\ 3 & 1 & 2 \end{bmatrix}\begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 1 \\ 7 \\ 2 \end{bmatrix} then [xyz]=\begin{bmatrix} x \\ y \\ z \end{bmatrix} =

    Tap an option to check your answer.

  4. Question 4

    This type asked 6 times: 11 May 2023 S2, 9 May 2024 S1, 14 May 2024 S2, 17 Apr 2026 S2, 13 May 2026 S1, 19 May 2026 S2. This one: 19 May 2026 S2 See all 6 of this type

    67
    If A+I=(3−21−1)A + I = \begin{pmatrix} 3 & -2 \\ 1 & -1 \end{pmatrix}, then the inverse of the matrix (A−I)(A - I) is:

    Tap an option to check your answer.

  5. Question 5

    This type asked 6 times: 2 May 2023 S1, 3 May 2023 S1, 10 May 2023 S2, 11 Apr 2026 S1, 16 Apr 2026 S1, 20 Apr 2026 S2. This one: 20 Apr 2026 S2 See all 6 of this type

    64
    If A=[1−5−24]A = \begin{bmatrix} 1 & -5 \\ -2 & 4 \end{bmatrix}, then A−1=A^{-1} =

    Tap an option to check your answer.

  6. Question 6

    This type asked 5 times: May 2022 S1, 23 Apr 2025 S1, 15 Apr 2026 S2, 16 Apr 2026 S2, 14 May 2026 S1. This one: 14 May 2026 S1 See all 5 of this type

    69
    If A−1=[3−11−156−55−22]A^{-1} = \begin{bmatrix} 3 & -1 & 1 \\ -15 & 6 & -5 \\ 5 & -2 & 2 \end{bmatrix} and B=[12−2−1300−21]B = \begin{bmatrix} 1 & 2 & -2 \\ -1 & 3 & 0 \\ 0 & -2 & 1 \end{bmatrix}, then (AB)−1=(AB)^{-1} =

    Tap an option to check your answer.

  7. Question 7

    This type asked 5 times: 10 May 2023 S1, 13 May 2024 S2, 20 Apr 2025 S1, 22 Apr 2025 S1, 17 Apr 2026 S2. This one: 17 Apr 2026 S2 See all 5 of this type

    62
    If A=[1−tan⁡θ2tan⁡θ21]A = \begin{bmatrix} 1 & -\tan\dfrac{\theta}{2} \\ \tan\dfrac{\theta}{2} & 1 \end{bmatrix} and B=[1tan⁡θ2−tan⁡θ21]B = \begin{bmatrix} 1 & \tan\dfrac{\theta}{2} \\ -\tan\dfrac{\theta}{2} & 1 \end{bmatrix} then A−1BA^{-1}B is equal to

    Tap an option to check your answer.

  8. Question 8

    Asked 4 times: 15 May 2023 S2, 12 May 2024 S2*, 25 Apr 2025 S2, 16 Apr 2026 S1 (* = numbers or wording changed that year) See all 4 times it was asked

    65
    If A=[5a−b32]A = \begin{bmatrix} 5a & -b \\ 3 & 2 \end{bmatrix} and A (adj A)=AATA\,(\mathrm{adj}\,A) = AA^T, then 5a+b=5a + b =

    Tap an option to check your answer.

  9. Question 9

    This type asked 4 times: 20 Apr 2026 S1, 12 May 2026 S1, 19 May 2026 S1, 19 May 2026 S2. This one: 19 May 2026 S2 See all 4 of this type

    68
    The sum of the matrix [2−35−7]\begin{bmatrix} 2 & -3 \\ 5 & -7 \end{bmatrix} and its multiplicative inverse is

    Tap an option to check your answer.

  10. Question 10

    This type asked 4 times: 21 Apr 2025 S2, 18 Apr 2026 S1, 19 Apr 2026 S2, 12 May 2026 S1. This one: 12 May 2026 S1 See all 4 of this type

    68
    Let AA be an invertible matrix. Identify the correct statement(s) from the following. (I) A(adj A−1)=1∣A∣A2A(\mathrm{adj}\,A^{-1}) = \dfrac{1}{|A|}A^2 (II) (adj A−1)−1=1∣A∣A−1(\mathrm{adj}\,A^{-1})^{-1} = \dfrac{1}{|A|}A^{-1} (III) (adj A)=1∣A∣A−1(\mathrm{adj}\,A) = \dfrac{1}{|A|}A^{-1}

    Tap an option to check your answer.