For a \( 2 \times 2 \) matrix \( A = \begin{bmatrix} a & b \\ c & d \end{bmatrix} \), the adjugate matrix is:
\[
\text{adj } A = \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}.
\]
If \( \text{adj } A = A \), then:
\[
\begin{bmatrix} d & -b \\ -c & a \end{bmatrix} = \begin{bmatrix} a & b \\ c & d \end{bmatrix}.
\]
Equating elements:
\[
d = a, \quad -b = b, \quad -c = c, \quad a = d.
\]
From \( -b = b \), we get \( b = 0 \), and from \( -c = c \), we get \( c = 0 \). Thus:
\[
A = \begin{bmatrix} a & 0 \\ 0 & a \end{bmatrix}.
\]
The sum of the elements is:
\[
a + b + c + d = a + 0 + 0 + a = 2a.
\]
Final Answer: \( \boxed{2a} \)