$C=AB$ I need to find A so $CB^{-1}=A$ working in modulo 10
and C=$ \begin{bmatrix}2 & 5 \\3 & 1 \end{bmatrix}$ mod 10
with B=$ \begin{bmatrix}3 & 4 \\2 & x \end{bmatrix}$ mod10
calculating $B^{-1}$
I wrote$ \begin{bmatrix}3 & 4 \\2 & x \end{bmatrix}$ $\begin{bmatrix}1&0 \\0 & 1 \\ \end{bmatrix}$
calculating the inverse of $3 (\mod10)$ and =$7$ and multiplying the first row
gives $ \begin{bmatrix}1 & 8 \\2 & x \end{bmatrix}$ $\begin{bmatrix}7&0 \\0 & 1 \\ \end{bmatrix}$
then R2-2R1
$ \begin{bmatrix}1 & 8 \\0 & x-16 \end{bmatrix}$ $\begin{bmatrix}7&0 \\6 & 1 \\ \end{bmatrix}$
but im not sure where to go from this, because I dont know how to calculate the inverse of $x-16(\mod10)$. I guess you could calculate the inverse of $x(\mod10)$ and multiply it through from the start, how would I go about finding A, namely $B^{-1}$
To find $B^{-1}$, you can start by analogy with inverses over $\mathbb{Q}$: $$B^{-1}=\frac{1}{3x-8} \begin{pmatrix} x&-4\\-2&3\end{pmatrix}$$
Note that if $x$ is even, then $det(B)$ is even, and therefore $B$ has no inverse modulo $10$. Further, if $x=1$, then $det(B)=5$, and therefore $B$ has no inverse modulo $10$. This leaves $x=3,5,-3,-1$.
For $x=5$, $det(B)=7$, and $7^{-1}\equiv 3\pmod{10}$. Hence $$B^{-1}=3\begin{pmatrix}7&-4\\-2&3\end{pmatrix}=\begin{pmatrix}1&-2\\4&-1\end{pmatrix}$$
The other three cases are solved similarly.