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

This type of question, asked 6 times: 27 Jun 2022, 31 Jan 2023, 10 Apr 2023, 22 Jan 2025, 23 Jan 2026 S1, 24 Jan 2026 S2. All 7 questions below, oldest first.

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  1. Question 1

    27 Jun 2022

    83
    The positive value of the determinant of the matrix A, whose Adj(Adj(A))=(1428−14−14142828−1414)Adj(Adj(A)) =\begin{pmatrix} 14 & 28 & - 14 \\ - 14 & 14 & 28 \\ 28 & - 14 & 14 \end{pmatrix}, is

  2. Question 2

    27 Jun 2022

    154
    Let AA and BB be two 3×33 \times 3 matrices such that AB=IAB = I and ∣A∣=18|A| =\frac{1}{8} then ∣adj(Badj(2 A))∣|adj(Badj(2\text{ }A))| is equal to

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  3. Question 3

    31 Jan 2023

    172
    Let AA be a n×nn \times n matrix such that ∣A∣=2|A| = 2. If the determinant of the matrix Adj(2⋅Adj(2 A−1))⋅Adj\left( 2 \cdot Adj\left( 2{\text{ }A}^{- 1} \right) \right)\cdot is 2842^{84}, then nn is equal to

  4. Question 4

    10 Apr 2023

    156
    If A=15!6!7![5!6!7!6!7!8!7!8!9!]A =\frac{1}{5!6!7!}\begin{bmatrix} 5! & 6! & 7! \\ 6! & 7! & 8! \\ 7! & 8! & 9! \end{bmatrix}, then ∣adj(adj(2A))∣|adj(adj(2A))| is equal to:

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  5. Question 5

    22 Jan 2025

    129
    For a 3×33 \times 3 matrix M , let trace (M)(M) denote the sum of all the diagonal elements of M . Let A be a 3×33 \times 3 matrix such that ∣A∣=12|A| =\frac{1}{2} and trace (A)=3(A) = 3. If B=adj(adj(2A))B = adj(adj(2A)), then the value of ∣B∣+trace(B)|B| + trace(B) equals:

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  6. Question 6

    23 Jan 2026 S1

    71
    Let ∣A∣=6|A| = 6, where A is a 3×33 \times 3 matrix. If ∣adj(3adj( A2.adj(2 A)))∣=2m.3n,m,n∈N\left| adj\left( 3adj\left( {\text{ }A}^{2}.adj(2\text{ }A) \right) \right) \right| = 2^{m}{.3}^{n},m,n \in N, then m+nm + n is equal to ____\_\_\_\_ .

  7. Question 7

    24 Jan 2026 S2

    58
    Let P=[pij]P =\left\lbrack p_{ij} \right\rbrack and Q=[qij]Q =\left\lbrack q_{ij} \right\rbrack be two square matrices of order 3 such that qij=2(i+j−1)pijq_{ij}=2^{(i + j - 1)}p_{ij} and det(Q)=210det(Q) =2^{10}. Then the value of det(adj(adjP))det(adj(adjP)) is :

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