\(A=1-3+3^2-3^3+...-3^{2021}+3^{2022}\)
\(\Rightarrow3A=3-3^2+3^3-3^4+...-3^{2022}+3^{2023}\)
\(\Rightarrow3A+A=4A\)
\(=\left(1-3+3^2-3^3+...-3^{2021}+3^{2022}\right)+\left(3-3^2+3^3-3^4+...-3^{2022}+3^{2023}\right)\)
\(=1+3^{2023}\)
\(\Rightarrow4A-3^{2023}=1+3^{2023}-3^{2023}=1\)