Example 2 Given points $A_{1}(-1,0), A_{2}(1, -1), A_{3}(2,0), A_{4}(3,3), A_{5}(0,5)$ on a plane, find the area of the pentagon $A_{1} A_{2} A_{3} A_{4} A_{5}$.
Solve: Since points $A_{1}, A_{2}, A_{3}, A_{4}, A_{5}$ are arranged counterclockwise, by Theorem 2, the area of pentagon $A_{1} A_{2} A_{3} A_{4} A_{5}$ is
$$\begin{array}{l}
S=\frac{1}{2}\left|\begin{array}{rrrrrr}
1 & 1 & 2 & 3 & 0 & -1 \\
0 & -1 & 0 & 3 & 5 & 0
\end{array}\right| \\
\quad=\frac{1}{2}[(-1) \times(-1)+1 \times 0+2 \times 3+3 \times 5+ \\
0 \times 0-1 \times 0-2 \times(-1)-3 \times 0-0 \times 3-(-1) \times 5] \\
\quad=14.5
\end{array}$$