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If adjA=[7−3230−3−131]adjA=\begin{bmatrix}7 & -3 & 2\\3 & 0 & -3\\-1 & 3 & 1\end{bmatrix}, then A−1A^{-1} is:

Solution

✅ Correct Option: 2

For order 33, ∣adjA∣=∣A∣2|adjA|=|A|^{2}. Evaluating the given determinant gives 63+0+18=8163+0+18=81, so ∣A∣=±9|A|=\pm9.

Since A−1=1∣A∣adjAA^{-1}=\frac{1}{|A|}adjA, we get A−1=±19[7−3230−3−131]A^{-1}=\pm\frac{1}{9}\begin{bmatrix}7&-3&2\\3&0&-3\\-1&3&1\end{bmatrix}.

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