Modular Inverse of a Matrix I have a 3 x 3 matrix M , I'm trying to find the inverse M mod (101). I currently know I can do this by finding M inverse and using x from Diophantine as a scalar and then taking mod 101 of each value (https://www.youtube.com/watch?v=LhBovzm4Ajs&t=2s). I'm having problems determining x by hand, so is there a different approach ?
Wolfram shows that it can be done with the function I'm just not understanding the notation
 A: It's a field. This means you put your matrix and a copy of the identity matrix side by side. Then use elementary row operations to take the left square to the identity matrix; this will require finding modular inverse of several numbers. However, once those numbers are found, doing the matrix operations is not bad, just keep reducing $\pmod {101}$ every time you do something. Let me think of a 2 by 2 example... 
$$
\left(
\begin{array}{rrrr}
5 & 3 & 1 & 0 \\
-3 & 5 & 0 & 1
\end{array}
\right)
$$
$$ \frac{1}{5} \equiv -20 \equiv 81 \pmod {101}  $$
mult first row by 81.
$$
\left(
\begin{array}{rrrr}
405 & 243 & 81 & 0 \\
-3 & 5 & 0 & 1
\end{array}
\right)
$$
reduce mod 101.
$$
\left(
\begin{array}{rrrr}
1 & 41 & 81 & 0 \\
-3 & 5 & 0 & 1
\end{array}
\right)
$$
add three times row one to row 2
$$
\left(
\begin{array}{rrrr}
1 & 41 & 81 & 0 \\
0 & 128 & 243 & 1
\end{array}
\right)
$$
$$
\left(
\begin{array}{rrrr}
1 & 41 & 81 & 0 \\
0 & 27 & 41 & 1
\end{array}
\right)
$$
$$ \frac{1}{27} \equiv 15  \pmod {101}  $$
second row by 15
$$
\left(
\begin{array}{rrrr}
1 & 41 & 81 & 0 \\
0 & 1 & 9 & 15
\end{array}
\right)
$$
add 60 times second to first
$$
\left(
\begin{array}{rrrr}
1 & 0 & 15 & 92 \\
0 & 1 & 9 & 15
\end{array}
\right)
$$
so thats it, inverse is
$$
\left(
\begin{array}{rr}
 15 & 92 \\
9 & 15
\end{array}
\right)
$$
