Question:

Since $M$ is invertible, we have $\det M \neq 0$. We are given that $M^{-1} = adj (adj M)$. Recall that $adj (adj M) = (\det M)^{n-2} M^{-1}$ for an $n \times n$ matrix. Therefore, we have $(\det M)^{3-2} M^{-1} = M^{-1}$, which implies $(\det M)^1 = 1$. Since $\det M \neq 0$, we must have $\det M = 1$. We now use the fact that $M^{-1} = adj (adj M)$. Recall that $adj M = \frac{1}{\det M} \cdot \text{cofactor matrix of } M$. Therefore, we have $M^{-1} = \frac{1}{\det M} \cdot \text{cofactor matrix of } adj M$. Since $\det M = 1$, we have $M^{-1} = \text{cofactor matrix of } adj M$. Now, recall that the cofactor matrix of $adj M$ is equal to $(adj M)^T$. Therefore, we have $M^{-1} = (adj M)^T$. Taking the transpose of both sides, we get $(M^{-1})^T = adj M$. Since $M$ is a $3 \times 3$ matrix, we have $(M^{-1})^T = (M^{-1})^T$. Therefore, we have $(M^{-1})^T = adj M = M^{-1}$.