Q.Find the adjoint and inverse of $A=\begin{pmatrix}1&0&2\\0&1&1\\1&1&0\end{pmatrix}$.

For a $3\times3$ matrix, the adjoint requires computing all nine cofactors first, arranging them into the cofactor matrix, and then transposing before dividing by the determinant. …

Step 1 — determinant: expand along row 1:

Since $|A|=-3\ne0$, $A$ is non-singular.

Step 2 — all nine cofactors:

$C_{11}=+\begin{vmatrix}1&1\\1&0\end{vmatrix}=0-1=-1$, $C_{12}=-\begin{vmatrix}0&1\\1&0\end{vmatrix}=-(0-1)=1$, $C_{13}=+\begin{vmatrix}0&1\\1&1\end{vmatrix}=0-1=-1$

$C_{21}=-\begin{vmatrix}0&2\\1&0\end{vmatrix}=-(0-2)=2$, $C_{22}=+\begin{vmatrix}1&2\\1&0\end{vmatrix}=0-2=-2$, $C_{23}=-\begin{vmatrix}1&0\\1&1\end{vmatrix}=-(1-0)=-1$

$C_{31}=+\begin{vmatrix}0&2\\1&1\end{vmatrix}=0-2=-2$, $C_{32}=-\begin{vmatrix}1&2\\0&1\end{vmatrix}=-(1-0)=-1$, $C_{33}=+\begin{vmatrix}1&0\\0&1\end{vmatrix}=1-0=1$

Step 3 — cofactor matrix and adjoint: the cofactor matrix is $\begin{pmatrix}-1&1&-1\\2&-2&-1\\-2&-1&1\end{pmatrix}$; transposing it gives

Step 4 — inverse: $A^{-1}=\dfrac{1}{-3}\begin{pmatrix}-1&2&-2\\1&-2&-1\\-1&-1&1\end{pmatrix}=\begin{pmatrix}1/3&-2/3&2/3\\-1/3&2/3&1/3\\1/3&1/3&-1/3\end{pmatrix}$ …

Forgetting to transpose the cofactor matrix before calling it the adjoint (using the cofactor matrix itself, unTransposed); sign errors in the alternating cofactor pattern, especially at $C_{21}, C_{23}$ …