Q.Find the inverse of $A=\begin{pmatrix}2&-1&1\\-1&2&-1\\1&-1&2\end{pmatrix}$ by the adjoint method.

For a $3\times3$ matrix, every cofactor must be computed individually before the adjoint (transpose of the cofactor matrix) can be formed and divided by the determinant.

Expanding along row 1, $|A|=2(4-1)-(-1)(-2+1)+1(1-2)=6-1-1=4$. All nine cofactors, arranged and transposed, give the adjoint shown; dividing by $4$ gives $A^{-1}$.

Step 1 — Determinant (expand along row 1)

Since $|A|=4\neq0$, $A$ is non-singular.

Step 2 — Cofactors

Step 3 — Cofactor matrix and adjoint

Cofactor matrix $=\begin{pmatrix}3&1&-1\\1&3&1\\-1&1&3\end{pmatrix}$. This particular cofactor matrix is symmetric, so its transpose (the adjoint) is identical:

Step 4 — Inverse

Check (independent recomputation): multiplying $A$ by $\text{adj}(A)$ directly, row 1 of $A$, $(2,-1,1)$, with column 1 of the adjoint, $(3,1,-1)$, gives $2(3)+(-1)(1)+1(-1)=6-1-1=4=|A|$ ✓; row 1 with column 2, $(1,3,1)$, gives $2(1)+(-1)(3)+1(1)=2-3+1=0$ ✓ (correctly zero, off the diagonal) — confirming $A\cdot\text{adj}(A)=|A|\cdot I$ as required.

Forgetting the alternating $(-1)^{i+j}$ sign when computing a cofactor (e.g. omitting the minus sign on $C_{12}$), and forgetting to TRANSPOSE the cofactor matrix before calling it the adjoint.