matrix inverse of a 3 by 3 matrix

|A|= \begin{vmatrix} 1 & 3&-2 \\ -3& 0 &-5 \\ 2& 5 &0 \end{vmatrix}

Do the following operations, R₂→R₂+3R₁ and R₃→R₃-2R₁ and expand by C₁,

= \begin{vmatrix} 1 & 3 & -2\\ 0& 9 & -11 \\ 0& -1& 4 \end{vmatrix} = (36-11)=25

A^{-1}=\frac{adjA}{|A|}

adjA=\begin{pmatrix} A_{11} & A_{12} &A_{13} \\ A_{21} & A_{22} & A_{23} \\ A_{31} & A_{32} & A_{33} \end{pmatrix}^{T}

adjA=\begin{pmatrix} 25 & -7 & -17 \\ -13 & 5 & 1 \\ -13 & 11 & 10 \end{pmatrix}^{T}

A^{-1}=\frac{1}{25}\begin{pmatrix} 25 & -13 & -13 \\ -7 & 5 & 11 \\ -17 & 1 & 10 \end{pmatrix}

edits: I forgot to do the transpose.