Can mycotoxins and BPA induce oxidative stress and disrupt the intestinal barrier, increasing permeability and inflammation?

To find the inverse of the matrix $A = \begin{pmatrix} 1 & 3 & 1 \ 1 & 1 & 2 \ 2 & 1 & 2 \end{pmatrix}$ using Gaussian elimination, we set up an augmented matrix $[A | I]$, where $I$ is the identity matrix. We then perform elementary row operations to transform the left side into the identity matrix.

Step 1: Set up the augmented matrix

$$ \left(\begin{array}{ccc|ccc} 1 & 3 & 1 & 1 & 0 & 0 \ 1 & 1 & 2 & 0 & 1 & 0 \ 2 & 1 & 2 & 0 & 0 & 1 \end{array}\right) $$

Step 2: Create zeros in the first column below the first pivot

Subtract the first row from the second row ($R_2 \leftarrow R_2 - R_1$): $$ \left(\begin{array}{ccc|ccc} 1 & 3 & 1 & 1 & 0 & 0 \ 0 & -2 & 1 & -1 & 1 & 0 \ 2 & 1 & 2 & 0 & 0 & 1 \end{array}\right) $$

Subtract twice the first row from the third row ($R_3 \leftarrow R_3 - 2R_1$): $$ \left(\begin{array}{ccc|ccc} 1 & 3 & 1 & 1 & 0 & 0 \ 0 & -2 & 1 & -1 & 1 & 0 \ 0 & -5 & 0 & -2 & 0 & 1 \end{array}\right) $$

Step 3: Normalize the second row

Divide the second row by -2 ($R_2 \leftarrow -\frac{1}{2} R_2$): $$ \left(\begin{array}{ccc|ccc} 1 & 3 & 1 & 1 & 0 & 0 \ 0 & 1 & -1/2 & 1/2 & -1/2 & 0 \ 0 & -5 & 0 & -2 & 0 & 1 \end{array}\right) $$

Step 4: Create a zero in the second column below the second pivot

Add 5 times the second row to the third row ($R_3 \leftarrow R_3 + 5R_2$): $$ \left(\begin{array}{ccc|ccc} 1 & 3 & 1 & 1 & 0 & 0 \ 0 & 1 & -1/2 & 1/2 & -1/2 & 0 \ 0 & 0 & -5/2 & 1/2 & -5/2 & 1 \end{array}\right) $$

Step 5: Normalize the third row

Multiply the third row by $-2/5$ ($R_3 \leftarrow -\frac{2}{5} R_3$): $$ \left(\begin{array}{ccc|ccc} 1 & 3 & 1 & 1 & 0 & 0 \ 0 & 1 & -1/2 & 1/2 & -1/2 & 0 \ 0 & 0 & 1 & -1/5 & 1 & -2/5 \end{array}\right) $$

Step 6: Create zeros in the third column above the third pivot

Add $1/2$ times the third row to the second row ($R_2 \leftarrow R_2 + \frac{1}{2} R_3$): $$ \left(\begin{array}{ccc|ccc} 1 & 3 & 1 & 1 & 0 & 0 \ 0 & 1 & 0 & 2/5 & 0 & -1/5 \ 0 & 0 & 1 & -1/5 & 1 & -2/5 \end{array}\right) $$

Subtract the third row from the first row ($R_1 \leftarrow R_1 - R_3$): $$ \left(\begin{array}{ccc|ccc} 1 & 3 & 0 & 6/5 & -1 & 2/5 \ 0 & 1 & 0 & 2/5 & 0 & -1/5 \ 0 & 0 & 1 & -1/5 & 1 & -2/5 \end{array}\right) $$

Step 7: Create a zero in the second column above the second pivot

Subtract 3 times the second row from the first row ($R_1 \leftarrow R_1 - 3R_2$): $$ \left(\begin{array}{ccc|ccc} 1 & 0 & 0 & 0 & -1 & 1 \ 0 & 1 & 0 & 2/5 & 0 & -1/5 \ 0 & 0 & 1 & -1/5 & 1 & -2/5 \end{array}\right) $$

Final Result

The right side of the augmented matrix is the inverse of $A$: $$ A^{-1} = \begin{pmatrix} 0 & -1 & 1 \ 2/5 & 0 & -1/5 \ -1/5 & 1 & -2/5 \end{pmatrix} $$

Or, written with a common denominator: $$ A^{-1} = \frac{1}{5} \begin{pmatrix} 0 & -5 & 5 \ 2 & 0 & -1 \ -1 & 5 & -2 \end{pmatrix} $$