Left inverse of a matrix $3 \times 2$ in $\mathbb{F}_7[x]$

Question

Do you know a method to calculate inverse matrix in $\mathbb{F}_7[x]$?

I want to calculate left inverse the following matrix of $3 \times 2$ in $\mathbb{F}_7[x]$

\begin{bmatrix} x^2+1 & x-1 \\ 3(x-3)(x+3) & x-2 \\ 5(x-1)(x+1) & 2x-3 \end{bmatrix}

Answer 1

For this specific case, I would make an Ansatz that a left inverse of $$X:=\begin{bmatrix}x^2+1&x-1\\3x^2-27&x-2\\5x^2-5&2x-3\end{bmatrix}=\begin{bmatrix}\vert&\vert\\X_1&X_2\\\vert&\vert\end{bmatrix}$$ is of the form $$L=\begin{bmatrix}b_{i,j}x+c_{i,j}\end{bmatrix}_{i\in\{1,2\},j\in\{1,2,3\}}=Bx+C\,,$$ where $B:=\begin{bmatrix}b_{i,j}\end{bmatrix}_{i\in\{1,2\},j\in\{1,2,3\}}$ and $C:=\begin{bmatrix}c_{i,j}\end{bmatrix}_{i\in\{1,2\},j\in\{1,2,3\}}$ for some $b_{i,j},c_{i,j}\in\mathbb{F}_7$. I guess you can also try to find a left inverse of $X$ in the field $\mathbb{F}_7(x)$ and see how you can modify the calculations to get something in $\mathbb{F}_7[x]$.

For my approach, the first row $C_1$ of $C$ should annihilate the coefficient vectors $(1,3,5)$ and $(-1,-2,-3)$. This means $C_1$ is proportional to $\begin{bmatrix}1&-2&1\end{bmatrix}$. Since $C_1X_1$ should have the constant term $1$, we conclude that $$C_1=\begin{bmatrix}1&-2&1\end{bmatrix}\,.$$ The second row $C_2$ of $C$ should annihilate the coefficient vectors $(1,3,5)$ and $(1,-27,-5)=(1,1,2)$. Thus, $C_2$ is proportional to $\begin{bmatrix}1&3&-2\end{bmatrix}$. As $C_2X_2$ should have the constant term $1$, we obtain $$C_2=\begin{bmatrix}-1&-3&2\end{bmatrix}\,,$$ making $$C=\begin{bmatrix}1&-2&1\\-1&-3&2\end{bmatrix}\,.$$

Note that $$CX=\begin{bmatrix}1&x\\0&1\end{bmatrix}\,.$$ Therefore, we need that $$BX=\begin{bmatrix}0&-1\\0&0\end{bmatrix}\,.$$ The first row $B_1$ of $B$ should annihilate $(1,3,5)$ and $(1,1,2)$, but $B_1X_2=-1$. Thus, $B_1=\begin{bmatrix}1&3&-2\end{bmatrix}$. The second row $B_2$ of $B$ should annihilate $(1,3,5)$, $(1,1,2)$, and $(-1,-2,-3)$, so $B_2=\begin{bmatrix}0&0&0\end{bmatrix}$. Consequently, $$B=\begin{bmatrix}1&3&-2\\0&0&0\end{bmatrix}$$ and so $$L=Bx+C=\begin{bmatrix}x+1&3x-2&-2x+1\\-1&-3&2\end{bmatrix}\,.$$