If \(A = \begin{bmatrix}1 & 0 & 0\\ 2 & 1 & 0\\ 3 & 2 & 1\end{bmatrix}\), then \(A^{-1}\) is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(\begin{bmatrix}1 & 0 & 0\\ -2 & 1 & 0\\ 1 & -2 & 1\end{bmatrix}\)
A is lower triangular, so its determinant is the product of the diagonal entries:
\( |A| = 1 \times 1 \times 1 = 1 \)
The cofactors are
\( C_{11} = 1,\ C_{12} = -(2 - 0) = -2,\ C_{13} = 4 - 3 = 1 \)
\( C_{21} = 0,\ C_{22} = 1,\ C_{23} = -(2 - 0) = -2 \)
\( C_{31} = 0,\ C_{32} = 0,\ C_{33} = 1 \)
The adjoint is the transpose of the cofactor matrix, so cofactors of row i go into column i:
\( \displaystyle \mathrm{adj}\,A = \begin{bmatrix}1 & 0 & 0\\ -2 & 1 & 0\\ 1 & -2 & 1\end{bmatrix} \)
Since \(|A| = 1\), \(A^{-1} = \mathrm{adj}\,A\).
Check (third row of A times the columns): \(3 - 4 + 1 = 0\), \(2 - 2 = 0\), \(1\); the other rows also give the rows of I.
Why the other options are incorrect
Option (1) \(\begin{bmatrix}1 & 0 & 0\\ -2 & 1 & 0\\ -3 & -2 & 1\end{bmatrix}\) — Incorrect. Changing the signs below the diagonal does not give the inverse here: the third row of A times the first column gives \(3 - 4 - 3 = -4\), not 0.
Option (3) \(\begin{bmatrix}1 & -2 & 1\\ 0 & 1 & -2\\ 0 & 0 & 1\end{bmatrix}\) — Incorrect. This is the cofactor matrix itself; it has not been transposed.
Option (4) \(\begin{bmatrix}1 & 0 & 0\\ 2 & 1 & 0\\ -1 & 2 & 1\end{bmatrix}\) — Incorrect. The second row of A times the first column gives \(2 + 2 = 4\), not 0.