If I have $$B=\begin{pmatrix} 3&9\\9&x\end{pmatrix}\bmod 10$$
How would I find the inverse for all $x$ unknowns. I wrote
To find $B^{-1}$, $$B^{-1}=\frac{1}{3x-81} \begin{pmatrix} x&-9\\-9&3\end{pmatrix}$$
and that if $x$ is odd, $\det(b)$ is even and has no inverse modulo $10$
That leaves $x=-2,2,4,-4,6,-6,8,-8,10,-10$ I think
for $x=-8$ and $x=2$, $\det(b)=5$ and has no inverse modulo $10$
am I going right so far. some of the x values have equal determinants like -6,4 both have 1 as the determinant so would I have to evaluate them both and what x values would I need to evaluate because I'm pretty sure its not all of them.
I would need to find say for $x=-4 \det(b)=7$ and inverse $7$ mod $10$ is $3$. So would need also need to times the matrix like
$$=3\begin{pmatrix} 3&-9\\-9&3\end{pmatrix}\bmod 10$$ to find the matrix for $x=-4$
Edit:
For x=4 The inverse is 1 Hence for x=4
$B^{-1}$=$$=1\begin{pmatrix} 4&-9\\-9&3\end{pmatrix}\bmod 10$$ = $\begin{pmatrix} 4&-9\\-9&3\end{pmatrix}\bmod 10$
First modulo $10$, you only have the even values: $$x=\pm 2,\;\pm4.$$ There remains to check whether or not the corresponding determinant is a unit mod. $10$. Next, we can make a table of values: $$\begin{array}{r|rrrr} x&+2&-2&+4&-4 \\ \hline 3x-1& 5&3&1&-3 \\ \hline (3x-1)^{-1}& \text{—}& -3&1&3 \end{array}$$