If \(A = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 3 & 2 \\ 0 & 4 & 3 \end{bmatrix}\), then \(A^{-1}\) is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(\begin{bmatrix} 1 & 0 & 0 \\ 0 & 3 & -2 \\ 0 & -4 & 3 \end{bmatrix}\)
Expanding along the first row:
\(|A| = 1 \times (3 \times 3 - 2 \times 4) = 1\)
The cofactors \(C_{ij}\) are:
\(C_{11} = 1,\quad C_{12} = 0,\quad C_{13} = 0\)
\(C_{21} = 0,\quad C_{22} = 3,\quad C_{23} = -4\)
\(C_{31} = 0,\quad C_{32} = -2,\quad C_{33} = 3\)
The adjoint is the transpose of the cofactor matrix, so row i of adj A holds \(C_{1i}, C_{2i}, C_{3i}\):
\( \displaystyle \mathrm{adj}\,A = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 3 & -2 \\ 0 & -4 & 3 \end{bmatrix} \)
\( \displaystyle A^{-1} = \frac{1}{|A|}\,\mathrm{adj}\,A = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 3 & -2 \\ 0 & -4 & 3 \end{bmatrix} \)
Check: the second row of A times the third column of this matrix is \(3 \times (-2) + 2 \times 3 = 0\), and times the second column is \(3 \times 3 + 2 \times (-4) = 1\).
Why the other options are incorrect
Option (1) \(\begin{bmatrix} 1 & 0 & 0 \\ 0 & 3 & -4 \\ 0 & -2 & 3 \end{bmatrix}\) — Incorrect. This is the cofactor matrix itself, not its transpose.
Option (2) \(\begin{bmatrix} 1 & 0 & 0 \\ 0 & -3 & 2 \\ 0 & 4 & -3 \end{bmatrix}\) — Incorrect. The signs of the 2 × 2 block are all reversed; A times it does not give I.
Option (4) \(\begin{bmatrix} 1 & 0 & 0 \\ 0 & 3 & 4 \\ 0 & 2 & 3 \end{bmatrix}\) — Incorrect. This is the transpose of A; the product of A with it has 13 in place (2, 2).