If A is an invertible matrix of order 3 and the determinant of A is 9, then the determinant of \(A^{-1}\) is:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 1/9
Since \(AA^{-1} = I\), the product rule for determinants gives
\(|A|\,|A^{-1}| = |I| = 1\)
\( \displaystyle |A^{-1}| = \frac{1}{|A|} = \frac{1}{9} \)
The result \(|A^{-1}| = 1/|A|\) does not depend on the order of the matrix.