$\left[ {\begin{array}{*{20}{c}}
{{a_{11}}}&{{a_{12}}}\\
{{a_{21}}}&{{a_{22}}}
\end{array}} \right],{a_{ij}} \in \left\{ {0,1,2} \right\},{a_{  11}} = {a_{12}}$ are

$\left[ {\begin{array}{*{20}{c}}
0&{0/1/2}\\
{0/1/2}&0
\end{array}} \right],\left[ {\begin{array}{*{20}{c}}
1&{0/1/2}\\
{0/1/2}&1
\end{array}} \right].\left[ {\begin{array}{*{20}{c}}
2&{0/1/2}\\
{0/1/2}&2
\end{array}} \right]$

At any place, $0/1/2$ means $0,1$ or $2$ will be the element at that place.

Hence therefore total $\left( {27 = 3 \times 3 + 3 \times 3 + 3 \times 3} \right)$

matrices of the above form. Out of which the matrices which are singular are

$\left[ {\begin{array}{*{20}{c}}
0&{0/1/2}\\
0&0
\end{array}} \right],\left[ {\begin{array}{*{20}{c}}
0&0\\
{1/2}&0
\end{array}} \right],\left[ {\begin{array}{*{20}{c}}
1&1\\
1&1
\end{array}} \right],\left[ {\begin{array}{*{20}{c}}
2&2\\
2&2
\end{array}} \right]$

Hence these are total $7=(3+2+1+1)$ singular matrices.

Therefore number of all non-singular matrices in the given form $=27-7=20$