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Determinant of nested adjoints and scalar multiples of A, given |A|

This type of question, asked 19 times: 2021 Paper 4, 2021 Paper 19, 26 Jun 2022, 28 Jun 2022, 27 Jul 2022, 24 Jan 2023, 25 Jan 2023, 8 Apr 2023, 10 Apr 2023, 13 Apr 2023, 15 Apr 2023, 31 Jan 2024, 5 Apr 2024, 6 Apr 2024, 9 Apr 2024, 22 Jan 2025, 2 Apr 2025, 3 Apr 2025, 7 Apr 2025. All 19 questions below, oldest first.

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  1. Question 1

    2021 Paper 4

    76
    Let AA be a 3×33 \times 3 matrix with det(A)=4det(A) = 4. Let RiR_{i} denote the ith i^{\text{th~}} row of AA. If a matrix BB is obtained by performing the operation R2→2R2+5R3R_{2}\rightarrow 2R_{2}+ 5R_{3} on 2A2A, then det(B)det(B) is equal to:

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  2. Question 2

    2021 Paper 19

    89
    Let AA be a 3×33 \times 3 real matrix. If det(2Adj(2Adj(Adj(2 A))))=241det(2Adj(2Adj(Adj(2\text{ }A)))) =2^{41}, then the value of det(A2)det\left( A^{2} \right) equal

  3. Question 3

    26 Jun 2022

    63
    Let AA be a 3×33 \times 3 invertible matrix. If ∣adj(24 A)∣=∣adj(3adj(2 A))∣|adj(24\text{ }A)| = |adj(3adj(2\text{ }A))|, then ∣A∣2|A|^{2} is equal to:

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  4. Question 4

    28 Jun 2022

    64
    Let AA be a matrix of order 3×33 \times 3 and det(A)=2det(A) = 2. Then det(det(A)adj(5adj(A3)))det\left( det(A)adj\left( 5adj\left( A^{3} \right) \right) \right) is equal to

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  5. Question 5

    27 Jul 2022

    171
    Consider a matrix A=[αβγα2β2γ2β+γγ+αα+β]A =\begin{bmatrix} \alpha & \beta & \gamma \\ \alpha^{2} & \beta^{2} & \gamma^{2} \\ \beta + \gamma & \gamma + \alpha & \alpha + \beta \end{bmatrix}, where α,β,γ\alpha,\beta,\gamma are three distinct natural numbers. If det(adj(adj(adj(adjA))))(α−β)16(β−γ)16(γ−α)16=232×316\frac{det(adj(adj(adj(adjA))))}{(\alpha - \beta)^{16}(\beta - \gamma)^{16}(\gamma - \alpha)^{16}}=2^{32}\times3^{16}, then the number of such 3 - tuples (α,β,γ)(\alpha,\beta,\gamma) is

  6. Question 6

    24 Jan 2023

    158
    Let AA be a 3×33 \times 3 matrix such that ∣adj(adj(adjA))∣=124|adj(adj(adjA))| =12^{4} Then ∣A−1adjA∣\left| A^{- 1}adjA \right| is equal to

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  7. Question 7

    25 Jan 2023

    78
    Let x,y,z>1x,y,z > 1 and A=[1log⁡xylog⁡xzlog⁡yx2log⁡yzlog⁡zxlog⁡zy3]A =\begin{bmatrix} 1 & \log_{x}y & \log_{x}z \\ \log_{y}x & 2 & \log_{y}z \\ \log_{z}x & \log_{z}y & 3 \end{bmatrix}. Then ∣adj(adjA2)∣\left| adj\left( adjA^{2} \right) \right| is equal to

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  8. Question 8

    8 Apr 2023

    74
    Let A=[21012−10−12]A =\begin{bmatrix} 2 & 1 & 0 \\ 1 & 2 & - 1 \\ 0 & - 1 & 2 \end{bmatrix}. If ∣adj(adj(adj2A))∣=(16)n|adj(adj(adj2A))| = (16)^{n}, then nn is equal to

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  9. Question 9

    10 Apr 2023

    69
    If AA is a 3×33 \times 3 matrix and ∣A∣=2|A| = 2, then ∣3adj(∣3A∣A2)∣\left| 3adj\left( |3A|A^{2} \right) \right| is equal to

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  10. Question 10

    13 Apr 2023

    165
    Let for A=[123a31112],∣A∣=2A =\begin{bmatrix} 1 & 2 & 3 \\ a & 3 & 1 \\ 1 & 1 & 2 \end{bmatrix},|A| = 2. If ∣2adj(2adj(2A))∣|2adj(2adj(2A))| =32n=32^{n}, then 3n+a3n + a is equal to

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  11. Question 11

    15 Apr 2023

    69
    Let the determinant of a square matrix AA of order mm be m−nm - n, where mm and nn satisfy 4m+n=224m + n = 22 and 17 m+4n=9317\text{ }m + 4n = 93. If det (nadj(adj(mA)))=(nadj(adj(mA))) = 3a5b6c3^{a}5^{b}6^{c}. then a+b+ca + b + c is equal to:

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  12. Question 12

    31 Jan 2024

    175
    Let AA be a 3×33 \times 3 matrix and det(A)=2det(A) = 2. If n=det⁡(adj(adj(…(adj A)…)))n = \det(\mathrm{adj}(\mathrm{adj}(\ldots(\mathrm{adj}\,A)\ldots))), where adj is applied 2024 times, then the remainder when nn is divided by 9 is equal to

  13. Question 13

    5 Apr 2024

    71
    Let AA and BB be two square matrices of order 3 such that ∣A∣=3|A| = 3 and ∣B∣=2|B| = 2. Then ∣ATA(adj(2 A))−1(adj(4 B))(adj(AB))−1AAT∣\left| A^{T}A(adj(2\text{ }A))^{- 1}(adj(4\text{ }B))(adj(AB))^{- 1}AA^{T} \right| is equal to :

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  14. Question 14

    6 Apr 2024

    170
    If A is a square matrix of order 3 such that det(A)=3det(A) = 3 and
    det(adj(−4adj(−3adj(3adj((2 A)−1)))))=2m3ndet\left( adj\left( - 4adj\left( - 3adj\left( 3adj\left( (2\text{ }A)^{- 1} \right) \right) \right) \right) \right)=2^{m}3^{n}
    then m+2nm + 2n is equal to:

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  15. Question 15

    9 Apr 2024

    87
    Let A be a non-singular matrix of order 3. If det(3adj(2adj((detA)A)))=3−13⋅2−10det(3adj(2adj((detA)A))) =3^{- 13}\cdot2^{- 10} and det⁡\det (3adj(2 A))=2m⋅3n(3adj(2\text{ }A)) =2^{m}\cdot3^{n}, then ∣3 m+2n∣|3\text{ }m + 2n| is equal to

  16. Question 16

    22 Jan 2025

    73
    Let AA be a square matrix of order 3 such that det(A)=−2det(A) = - 2 and det(3adj(−6adj(3A)))=2m+n⋅3mndet(3adj( - 6adj(3A))) =2^{m + n}\cdot3^{mn}, m>nm > n. Then 4 m+2n4\text{ }m + 2n is equal to _______\_\_\_\_\_\_\_ .

  17. Question 17

    2 Apr 2025

    69
    Let a∈Ra \in R and A be a matrix of order 3×33 \times 3 such that det(A)=−4det(A) = - 4 and A+I=[1a1210a12]A+I=\begin{bmatrix} 1 & a & 1 \\ 2 & 1 & 0 \\ a & 1 & 2 \end{bmatrix}, where II is the identity matrix of order 3×33 \times 3.If det((a+1)adj((a−1)A))det((a+ 1)adj((a- 1)A)) is 2m3n,m,n∈2^{m}3^{n},m,n\in {0,1,2,….20}\mathbf{\{ 0,1,2,\ldots.20\}}, then m+nm + n is equal to :

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  18. Question 18

    3 Apr 2025

    51
    Let AA be a matrix of order 3×33\times 3 and ∣A∣=5|A|=5. If ∣2 adj(3A adj(2A))∣=2α⋅3β⋅5γ, α,β,γ∈N|2\,\mathrm{adj}(3A\,\mathrm{adj}(2A))|=2^{\alpha}\cdot3^{\beta}\cdot5^{\gamma},\ \alpha,\beta,\gamma\in N, then α+β+γ\alpha+\beta+\gamma is equal to

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  19. Question 19

    7 Apr 2025

    65
    Let AA be a 3×33 \times 3 matrix such that ∣adj(adj(adjA))∣=81|adj(adj(adjA))| = 81. If
    S={n∈Z:(∣adj(adjA)∣)(n−1)22=∣A∣(3n2−5n−4)}S =\left\{ n \in\mathbb{Z}:(|adj(adjA)|)^{\frac{(n- 1)^{2}}{2}}= |A|^{\left( 3n^{2}- 5n- 4 \right)} \right\}
    then ∑n∈S∣A(n2+n)∣\sum_{n\in S}^{} \left| A^{\left( n^{2}+n \right)} \right| is equal to.

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