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Inverse of a product: (AB)^-1 = B^-1 A^-1

This type of question, asked 5 times: May 2022 S1, 23 Apr 2025 S1, 15 Apr 2026 S2, 16 Apr 2026 S2, 14 May 2026 S1. All 5 questions below, oldest first.

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  1. Question 1

    May 2022 S1

    110
    If A=[121313]A = \begin{bmatrix} 1 & 2 & 1 \\ 3 & 1 & 3 \end{bmatrix} and B=[231212]B = \begin{bmatrix} 2 & 3 \\ 1 & 2 \\ 1 & 2 \end{bmatrix}, then (AB)−1=(AB)^{-1} =

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  2. Question 2

    23 Apr 2025 S1

    110
    If A and B are non-singular matrices of order 2 such that (AB)−1=16[−7−323](AB)^{- 1}=\frac{1}{6}\begin{bmatrix} - 7 & - 3 \\ 2 & 3 \end{bmatrix} and A−1=13[43−10]A^{- 1}=\frac{1}{3}\begin{bmatrix} 4 & 3 \\ - 1 & 0 \end{bmatrix} then B−1=B^{- 1}=

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  3. Question 3

    15 Apr 2026 S2

    67
    If A=[31−12],C=[7306]A = \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix}, C = \begin{bmatrix} 7 & 3 \\ 0 & 6 \end{bmatrix} and AB=CAB = C, then the inverse of matrix B is

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  4. Question 4

    16 Apr 2026 S2

    63
    If A=[235−2],B−1=[152525−15]A = \begin{bmatrix} 2 & 3 \\ 5 & -2 \end{bmatrix}, B^{-1} = \begin{bmatrix} \frac{1}{5} & \frac{2}{5} \\ \frac{2}{5} & -\frac{1}{5} \end{bmatrix}, then (AB)−1=(AB)^{-1} =

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  5. Question 5

    14 May 2026 S1

    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} =

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