I'm stuck in a problem where I'm asked to find the conjugate of the symmetric matrix A. $$A = \begin{bmatrix}0 & a & 0 & b\\ a & 0 & c & 0\\ 0 & c & 0 & d\\ b & 0 & d & 0\end{bmatrix}$$ I calculated the determinant of A: $$\begin{vmatrix}0 & a & 0 & b\\ a & 0 & c & 0\\ 0 & c & 0 & d\\ b & 0 & d & 0\end{vmatrix} \underset{l_{1} \Leftrightarrow l_{1}}{\rightarrow} \begin{vmatrix}a & 0 & c & 0\\0 & a & 0 & b\\ 0 & c& 0 & d\\ b & 0 & d & 0\end{vmatrix} \underset{l_{4} - \frac{b}{a} l_{1}}{\rightarrow} -\begin{vmatrix}a & 0 & c & 0\\0 & a & 0 & b\\ 0 & c & 0 & d\\ 0 & 0 & d - \frac{bc}{a} & 0\end{vmatrix} \underset{l_{3} - \frac{c}{a} l_{2}}{\rightarrow} -\begin{vmatrix}a & 0 & c & 0\\0 & a & 0 & b\\ 0 & 0 & 0 & d - \frac{bc}{a}\\ 0 & 0 & d - \frac{bc}{a} & 0\end{vmatrix}\underset{l_{3} \Leftrightarrow l_{4}}{\rightarrow} \begin{vmatrix}a & 0 & c & 0\\0 & a & 0 & b\\ 0 & 0 & d - \frac{bc}{a} & 0\\0 & 0 & 0 & d - \frac{bc}{a}\end{vmatrix}$$ The determinant is equal to $$ a^{2} \cdot \bigg(d- \frac{bc}{a}\bigg)^{2} $$
I used this row echelon form to calculate the adjugate of A by doing the following example: $$(adjA)_{1,2} = (adjA)_{2,1} = (-1)^{3} \cdot detA(1|2) = -\begin{vmatrix}0 & 0 & b\\ 0 & d-\frac{bc}{a} & 0\\ 0 & 0 & d-\frac{bc}{a}\end{vmatrix} = 0$$
What am I doing wrong here? Because the book solutions point that the adjugate I obtained is wrong...