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Determinant of adj A: |adj A| = |A|^(n - 1)

This type of question, asked 3 times: 15 Apr 2026 S1, 15 May 2026 S1, 19 May 2026 S1. All 3 questions below, oldest first.

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  1. Question 1

    15 Apr 2026 S1

    65
    Let A=[2k−11102k−11002k−1]A = \begin{bmatrix} 2k-1 & 1 & 1 \\ 0 & 2k-1 & 1 \\ 0 & 0 & 2k-1 \end{bmatrix} and B=[02k−111−2k0k−1−k0]B = \begin{bmatrix} 0 & 2k-1 & 1 \\ 1-2k & 0 & k \\ -1 & -k & 0 \end{bmatrix} where k is a real number. If det⁡(adjA)+det⁡(adjB)=116\det(\mathrm{adj}A) + \det(\mathrm{adj}B) = 11^6, then the value of k−5k - 5 is equal to...

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  2. Question 2

    15 May 2026 S1

    69
    If A=[1−1102−3210]A = \begin{bmatrix} 1 & -1 & 1 \\ 0 & 2 & -3 \\ 2 & 1 & 0 \end{bmatrix}, B=adj AB = \mathrm{adj}\ A and C=5AC = 5A, then the value of ∣adj B∣∣C∣\dfrac{|\mathrm{adj}\ B|}{|C|} is:

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  3. Question 3

    19 May 2026 S1

    68
    If A=[−12−3232−12]A = \begin{bmatrix} -\dfrac{1}{2} & -\dfrac{\sqrt{3}}{2} \\ \dfrac{\sqrt{3}}{2} & -\dfrac{1}{2} \end{bmatrix}, the value of ∣Adj(A2026)∣|\mathrm{Adj}(A^{2026})| is equal to:

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