<s>[INST] Inverting a $2 \times 2$ matrix $\mod 26$ I am trying to invert the matrix (in mod 26)
\begin{bmatrix}
     19 & 7\\
    19 & 0 \\
\end{bmatrix}
I compute the determinant 
((19*0)-(19*7)=23 (Reducing in mod 26)
\begin{bmatrix}
     0 & -7\\
    -19 & 19 \\
\end{bmatrix}
Switch a & d and negate b & c
Multiply by 23^-1 = 17
To get
\begin{bmatrix}
     0 & 11\\
    15 & 25 \\
\end{bmatrix}
But this can't be right
since wolfram gives me
\begin{bmatrix}
     0 & 15\\
    11 & 11\\
\end{bmatrix}
What could I be doing wrong?
 [/INST] Let $A$ be the given matrix. Then $AA^{-1}=I_2$. Hence, we must row reduce the augmented system:
$$\begin{bmatrix}19&7&|&1&0\\19&0&|&0&1\end{bmatrix}\to\begin{bmatrix}19&7&|&1&0\\0&-7&|&-1&1\end{bmatrix}\to \begin{bmatrix}19&0&|&0&1\\0&-7&|&-1&1\end{bmatrix}\to\begin{bmatrix}1&0&|&0&11\\0&1&|&-11&11\end{bmatrix}$$
using $19^{-1}=11$ and $-7=19$ in $\Bbb Z_{26}$.
Hence, $$A^{-1}=\begin{bmatrix}0&11\\-11&11\end{bmatrix}=\begin{bmatrix}0&11\\15&11\end{bmatrix}\qquad\text{in $\Bbb Z_{26}$ since $-11=15$}$$
</s>