Match List I with List II. "A is a non-singular matrix of order n."
| LIST I | LIST II | ||
| A. | $|adj A|$ | I. | $\frac{1}{|A|}A$ |
| B. | $(adj\, A)^{-1}$ | II. | $2^n|A|$ |
| C. | $adj(adj\, A)$ | III. | $|A|^{n-2}A$ |
| D. | $|2A|$ | IV. | $|A|^{n-1}$ |
Choose the correct answer from the options given below :
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → A-IV, B-II, C-III, D-II
(A) $|adj\, A|=|A|^{n-1}$ (IV)
(B) $(adj\, A)^{-1}$
$A\,adj\, A=|A|I$
so $(adj\, A)^{-1}=\frac{A}{|A|}$ (I)
(C) $adj(adj\, A)=|A|^{n-2}A$ (III)
(D) $|2A|=2^n|A|$ (II)