If A is an invertible matrix of order 2, then det ((adj A)-1) is equal to
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(\displaystyle\frac{1}{\det \mathrm{A}}\)
For a square matrix of order n, \(|\mathrm{adj}\,A| = |A|^{n-1}\). Here n = 2, so:
\(|\mathrm{adj}\,A| = |A|^{2-1} = |A|\)
Since A is invertible, \(|A| \ne 0\), so adj A is also invertible. The determinant of an inverse is the reciprocal of the determinant:
\( \displaystyle \det\left((\mathrm{adj}\,A)^{-1}\right) = \frac{1}{|\mathrm{adj}\,A|} = \frac{1}{\det A} \)