If \( \mathrm{adj}\,A=\begin{bmatrix}7 & -3 & 2\\3 & 0 & -3\\-1 & 3 & 1\end{bmatrix} \), then \( A^{-1} \) is:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( \pm\displaystyle\frac{1}{9}\begin{bmatrix}7 & -3 & 2\\3 & 0 & -3\\-1 & 3 & 1\end{bmatrix} \)
Since \( A^{-1}=\frac{1}{|A|}\,\mathrm{adj}\,A \), only |A| is needed. For a matrix of order 3, \( |\mathrm{adj}\,A|=|A|^{3-1}=|A|^2 \).
Expanding \( |\mathrm{adj}\,A| \) along R1:
\( \displaystyle |\mathrm{adj}\,A|=7(0\cdot 1-(-3)(3))-(-3)(3\cdot 1-(-3)(-1))+2(3\cdot 3-0\cdot(-1)) \)
\( \displaystyle =7(9)+3(0)+2(9) \)
\( \displaystyle =81 \)
\( \displaystyle |A|^2=81 \)
\( \displaystyle |A|=\pm 9 \)
\( \displaystyle A^{-1}=\frac{1}{|A|}\,\mathrm{adj}\,A \)
\( \displaystyle =\pm\frac{1}{9}\begin{bmatrix}7 & -3 & 2\\3 & 0 & -3\\-1 & 3 & 1\end{bmatrix} \)
Why the other options are incorrect
Option (1) \( \displaystyle\frac{1}{6}\begin{bmatrix}7 & -3 & 2\\3 & 0 & -3\\-1 & 3 & 1\end{bmatrix} \) — Incorrect. |A| is ±9, not 6.
Option (3) \( \pm\displaystyle\frac{1}{9}\begin{bmatrix}7 & 3 & -1\\-3 & 0 & 3\\2 & -3 & 1\end{bmatrix} \) — Incorrect. The adjoint is used as it is, not transposed; the transpose is already built into adj A.
Option (4) \( \displaystyle\frac{1}{9}\begin{bmatrix}7 & -3 & 2\\3 & 0 & -3\\-1 & 3 & 1\end{bmatrix} \) — Incorrect. The data give only \( |A|^2=81 \), so the sign of |A| cannot be fixed; A and −A have the same adjoint, so both signs are possible.