Inverse of a matrix in $\mathbb{F}_5^{4\times4}$

Question

Let $f, \, g, \, h \in \mathbb{F}_5[X]$ where $$f=X^9+X^8+ \cdots +X^2+X+1,\\ g=X^4+X-2 = X^4+X+3, \\ h = 3X^2+4X+3.$$

$h$ is the greatest common divisor of $f$ and $g$.

It holds that $g/h=2X^2+4X+1$.

Show that $A=\begin{pmatrix} 1 & 4 & 1 & 3 \\ 3 & 1 & 1 & 1 \\ 1&1&2&1 \\ 2&1&2&2 \end{pmatrix}$ is a root of $g/h$ and with that, determine the inverse of $(f/h)(A)$.

Hint: Insert $A$ into $$(3X+1)\cdot f+(2X^6+X^5+X^4+4X^3+4X^2+2X+4) \cdot g =h=3X^2+4X+3.$$

You do not need to determine $f/h$ or $(f/h)(A)$.

So I have shown that $A$ is a root of $g/h$, but can someone please show me how to determine the inverse of $(f/h)(A)$ from here on, following the hint?

Following reuns answer ($f/g, \, g/h$ are coprime), I got:

$\exists \, x,\, y \in \mathbb{F}_5[X]:x \frac fh(A)+y \frac gh (A) = 1$, and because $(g/h)(A) = 0$: $$x \frac fh(A) = 1 \iff x(A) = ((f/h)(A))^{-1}.$$

But how does that help? Or is there a better way to solve it?

Thanks in advance!

Answer 1

The hint can be effectively used to determine the inverse:

Let $m = (3X+1), \, n = (2X^6+X^5+X^4+4X^3+4X^2+2X+4)$, so $mf + ng =h$.

Dividing by $h$ yields $(3X+1)(f/h)+n(g/h) = 1$.

Because of $(g/h)(A)=0$, it holds that $$\begin{align} (3A+1)(f/h)(A)+n(A)(g/h)(A) &= 1 &\Leftrightarrow \\ (3A+1)(f/h)(A)&=1 &\Leftrightarrow \\ (3A+1) &= ((f/h)(A))^{-1}. \end{align}$$

Answer 2

$g/h$ is coprime with $f/h$, there is $u,v$ such that $ug/h+vf/h=1$ so $v(A)$ is your inverse. Moreover you can reduce your computation modulo $ g/h=2X^2+aX+b$, searching for the inverse of $f/h$ in $\Bbb{F}_5[X]/(2X^2+aX+b)$