Computing the inverse with cofactors

Determinant. $\det A = 1(0-24) - 2(0-20) + 3(0-5) = -24 + 40 - 15 = 1$.

Cofactors.
- $C_{11}=+(1\cdot0 - 4\cdot6) = -24$.
- $C_{12}=-(0\cdot0-4\cdot5)=20$.
- $C_{13}=+(0\cdot6-1\cdot5)=-5$.
- $C_{21}=-(2\cdot0-3\cdot6)=18$.
- $C_{22}=+(1\cdot0-3\cdot5)=-15$.
- $C_{23}=-(1\cdot6-2\cdot5)=4$.
- $C_{31}=+(2\cdot4-3\cdot1)=5$.
- $C_{32}=-(1\cdot4-3\cdot0)=-4$.
- $C_{33}=+(1\cdot1-2\cdot0)=1$.

Cofactor matrix and transpose.
$$C = \begin{pmatrix} -24 & 20 & -5 \\ 18 & -15 & 4 \\ 5 & -4 & 1 \end{pmatrix}, \qquad \operatorname{com}(A) = C^T = \begin{pmatrix} -24 & 18 & 5 \\ 20 & -15 & -4 \\ -5 & 4 & 1 \end{pmatrix}.$$

Inverse. Since $\det A = 1$,
$$A^{-1} = \begin{pmatrix} -24 & 18 & 5 \\ 20 & -15 & -4 \\ -5 & 4 & 1 \end{pmatrix}.$$

Verification. $A\cdot A^{-1} = I$ (can be checked directly).