I am trying to find the inverse matrix of $$\begin{pmatrix} \ln\left(x\right) & -1\\ \:\:1 & \ln\left(x\right) \end{pmatrix}$$ using the Gauss-Jordan method. Using a different method I could already find that the inverse matrix is:
$$\frac{1}{\ln ^2\left(x\right)+1}\begin{pmatrix}\ln \left(x\right)&-\left(-1\right)\\ -1&\ln \left(x\right)\end{pmatrix}=\begin{pmatrix}\frac{\ln \left(x\right)}{\ln ^2\left(x\right)+1}&\frac{1}{\ln ^2\left(x\right)+1}\\ -\frac{1}{\ln ^2\left(x\right)+1}&\frac{\ln \left(x\right)}{\ln ^2\left(x\right)+1}\end{pmatrix}$$
Suppose that the matrix $A$ is given by
$$ \begin{bmatrix} \ln(x) & -1 \\ 1 & \ln(x) \end{bmatrix} \tag{1} $$
using Gaussian Elimination $$ U=A, L=I \tag{2} $$ $$ \ell_{21} = \frac{u_{21}}{u_{11}} = \frac{1}{\ln(x)} \tag{3} $$ the point is to find the coefficient to zero the column $$ u_{2,1:2} = u_{2,1:2} -\frac{1}{\ln(x)}u_{1,1:2} \tag{4}$$ $$ u_{2,1:2} = \begin{bmatrix} 1 & \ln(x) \end{bmatrix} -\frac{1}{\ln(x)}\begin{bmatrix} \ln(x) & -1 \end{bmatrix} \tag{5}$$
which gives us
$$ u_{2,1:2} = \begin{bmatrix} 1 & \ln(x) \end{bmatrix} - \begin{bmatrix} 1 & -\frac{1}{\ln(x)}\end{bmatrix} \tag{6} $$
$$ u_{2,1:2} = \begin{bmatrix} 1 & \ln(x) \end{bmatrix} - \begin{bmatrix} 1 & -\frac{1}{\ln(x)}\end{bmatrix} \tag{7} $$
$$ u_{2,1:2} = \begin{bmatrix} 0 & \frac{\ln^{2}(x)+1}{\ln(x)} \end{bmatrix} \tag{8} $$
updating the matrix
$$ U = \begin{bmatrix} \ln(x) & -1 \\ 0 & \frac{\ln^{2}(x)+1}{\ln(x)} \end{bmatrix} \tag{9} $$
$$ L = \begin{bmatrix} 1 & 0 \\ \frac{1}{\ln(x)} & 1 \end{bmatrix} \tag{10} $$
$$ A = LU \tag{11} $$
So you'd find $U^{-1}L^{-1}$