Matrix inverses

Invert the following matrices over fields \(\mathbb Z_3\) and \(\mathbb Z_5\)

Variant 1
\( \mathbf A= \begin{pmatrix} 1 & 0 & 1 & 1 \\ 2 & 0 & 1 & 1 \\ 2 & 1 & 0 & 0 \\ 1 & 2 & 1 & 0 \\ \end{pmatrix} \).
Result
Over \(\mathbb Z_3\): \(A^{-1}=\begin{pmatrix} 2 & 1 & 0 & 0 \\ 2 & 1 & 1 & 0 \\ 0 & 0 & 1 & 1 \\ 2 & 2 & 2 & 2 \\ \end{pmatrix}\), over \(\mathbb Z_5\): \(A^{-1}= \begin{pmatrix} 4 & 1 & 0 & 0 \\ 2 & 3 & 1 & 0 \\ 2 & 3 & 3 & 1 \\ 0 & 1 & 2 & 4 \\ \end{pmatrix}\).
Variant 2
\( \mathbf B= \begin{pmatrix} 0 & 2 & 2 & 1 \\ 1 & 0 & 2 & 0 \\ 2 & 1 & 0 & 2 \\ 2 & 2 & 1 & 1 \\ \end{pmatrix} \).
Result
\(\mathbf B\) is singular in both fields.
Variant 3
\( \mathbf C= \begin{pmatrix} 2 & 0 & 1 & 0 \\ 1 & 2 & 0 & 1 \\ 2 & 1 & 1 & 0 \\ 1 & 0 & 1 & 1 \end{pmatrix} \).
Result
Over \(\mathbb Z_3: \mathbf C^{-1}=\begin{pmatrix} 0 & 2 & 2 & 1 \\ 2 & 0 & 1 & 0 \\ 1 & 2 & 2 & 1 \\ 2 & 2 & 2 & 2 \end{pmatrix}\), over \(\mathbb Z_5: \mathbf C^{-1}= \begin{pmatrix} 4 & 3 & 4 & 2 \\ 4 & 0 & 1 & 0 \\ 3 & 4 & 2 & 1 \\ 3 & 3 & 4 & 3 \end{pmatrix}\).
Variant 4
\( \mathbf D= \begin{pmatrix} 1 & 1 & 0 & 0 \\ 1 & 2 & 1 & 1 \\ 1 & 1 & 2 & 0 \\ 0 & 1 & 2 & 1 \\ \end{pmatrix} \).
Result
\(\mathbf D\) is singular in both fields.
Variant 5
\( \mathbf E= \begin{pmatrix} 1 & 1 & 2 & 0 \\ 1 & 2 & 1 & 1 \\ 0 & 1 & 2 & 1 \\ 1 & 2 & 0 & 0 \\ \end{pmatrix} \).
Result
\(\mathbf E\) is singular over \(\mathbb Z_3\), but over \(\mathbb Z_5: \mathbf E^{-1}= \begin{pmatrix} 4 & 3 & 2 & 4 \\ 3 & 1 & 4 & 1 \\ 2 & 3 & 2 & 0 \\ 3 & 3 & 3 & 4 \\ \end{pmatrix}\).