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For any non-zero matrix A=[aij]3×3A = [a_{ij}]_{3\times3} and ∣A∣≠0|A| \neq 0, then

Match the LIST-I with LIST-II

LIST-ILIST-II
A.∣adj A∣\vert adj\,A \vertI.1∣A∣(adj A)\dfrac{1}{\vert A \vert}(adj\,A)
B.(adj A)A(adj\,A)AII.∣A∣2\vert A \vert^2
C.A−1A^{-1}III.1∣A∣\dfrac{1}{\vert A \vert}
D.∣A−1∣\vert A^{-1} \vertIV.∣A∣I\vert A \vert I

Choose the correct answer from the options given below:

Solution

✅ Correct Option: 4

For order 33, ∣adjA∣=∣A∣n−1=∣A∣2|adj A| = |A|^{n-1} = |A|^2. Also (adjA)A=∣A∣I(adj A)A = |A|I, A−1=1∣A∣(adjA)A^{-1} = \frac{1}{|A|}(adj A), and ∣A−1∣=1∣A∣|A^{-1}| = \frac{1}{|A|}. Hence A-II, B-IV, C-I, D-III.

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