If \(A = \begin{bmatrix}1 & 2 & 0\\ 0 & 3 & 1\\ 2 & 0 & 1\end{bmatrix}\), then \(|A^{-1}|\) is equal to
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(\displaystyle \frac{1}{7}\)
Since \(AA^{-1} = I\), the product rule gives \(|A|\,|A^{-1}| = 1\), so \(\displaystyle |A^{-1}| = \frac{1}{|A|}\).
Expanding \(|A|\) along the first row:
\(|A| = 1(3 - 0) - 2(0 - 2) + 0\)
\(|A| = 3 + 4 = 7\)
\(\displaystyle |A^{-1}| = \frac{1}{7}\)