If \(A = \begin{bmatrix}3 & 1 & 0\\ 2 & 4 & 1\\ 0 & 1 & 2\end{bmatrix}\), then \(|A^{-1}|\) is equal to
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(\displaystyle\frac{1}{17}\)
Since \(AA^{-1} = I\), taking determinants gives \(|A|\,|A^{-1}| = 1\). So the determinant of the inverse is the reciprocal of \(|A|\).
Expanding \(|A|\) along the first row:
\( |A| = 3(4 \cdot 2 - 1 \cdot 1) - 1(2 \cdot 2 - 1 \cdot 0) + 0 \)
\( = 3 \times 7 - 4 \)
\( = 17 \)
Hence
\( \displaystyle |A^{-1}| = \frac{1}{|A|} = \frac{1}{17} \)