If \(A = \begin{bmatrix}1 & 2 & 0\\ 0 & 1 & 3\\ 2 & 0 & 1\end{bmatrix}\), then the entry in the second row and third column of \(A^{-1}\) is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(\displaystyle -\frac{3}{13}\)
Expanding along the first row:
\( |A| = 1(1 - 0) - 2(0 - 6) + 0 = 1 + 12 = 13 \)
Since \(\displaystyle A^{-1} = \frac{1}{|A|}\,\mathrm{adj}\,A\) and the adjoint is the transpose of the cofactor matrix, the (2, 3) entry of \(A^{-1}\) is \(\displaystyle \frac{C_{32}}{|A|}\) (row and column swapped).
The minor of the (3, 2) element: delete row 3 and column 2.
\( \displaystyle M_{32} = \begin{vmatrix}1 & 0\\ 0 & 3\end{vmatrix} = 3 \)
\( C_{32} = (-1)^{3 + 2} \times 3 = -3 \)
\( \displaystyle \text{(2, 3) entry of } A^{-1} = -\frac{3}{13} \)
Check: row 2 of \(A^{-1}\) is \(\frac{1}{13}(6,\ 1,\ -3)\), and its product with column 3 of A is \(\frac{1}{13}(0 + 3 - 3) = 0\), with column 2 is \(\frac{1}{13}(12 + 1 + 0) = 1\), as it must be.
Why the other options are incorrect
Option (1) \(\frac{4}{13}\) — Incorrect. This is \(\frac{C_{23}}{|A|}\), the (3, 2) entry of \(A^{-1}\); the adjoint needs the transpose.
Option (2) \(\frac{3}{13}\) — Incorrect. This uses the minor 3 without the sign \((-1)^{3 + 2} = -1\).
Option (4) \(-\frac{4}{13}\) — Incorrect. This uses the minor \(M_{23} = -4\) of the wrong position and drops the sign.