deletions | additions       diff --git a/Determinants.tex b/Determinants.tex  index 879bf3c..81d9a5b 100644  --- a/Determinants.tex  +++ b/Determinants.tex 
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The ordering of the column vecotrs has been chosen as an unravelling of the rank three tensors above. The determinant of this matrix is 1, thus it is unimodular, and is integer invertible, although the elements need not be integers, so any non-singular matrix will do. There exists a solution, which we knew as by inserting the identity matrix into the general problem we would retrieve an effective solution of the vector transform.  ..Insert solution  Approach the full general solution \begin{equation}  \begin{bmatrix}  \begin{bmatrix}1&1\\0&0\end{bmatrix}&  \begin{bmatrix}0&1\\0&1\end{bmatrix}\\  \begin{bmatrix}1&0\\1&0\end{bmatrix}&  \begin{bmatrix}0&0\\1&1\end{bmatrix}  \end{bmatrix}  \begin{bmatrix}x_1 \\ x_2\end{bmatrix}  =  \begin{bmatrix}  \begin{bmatrix}1&2\\0&1\end{bmatrix}\\  \begin{bmatrix}1&0\\2&1\end{bmatrix}  \end{bmatrix}  \end{equation}  where the solution is simply $x_1=1,x_2=1$. However the sucess of a gaussian elimination at least creating a row echelon form depends on the matrices subtracting from each other also, and would be impossible in this case. A more general equation would be \begin{equation}  \begin{bmatrix}  \begin{bmatrix}a_{1111}&a_{1112}\\a_{1121}&a_{1122}\end{bmatrix}&  \begin{bmatrix}a_{1211}&a_{1212}\\a_{1221}&a_{1222}\end{bmatrix}\\  \begin{bmatrix}a_{2111}&a_{2112}\\a_{2121}&a_{2122}\end{bmatrix}&  \begin{bmatrix}a_{2211}&a_{2212}\\a_{2221}&a_{2222}\end{bmatrix}  \end{bmatrix}  \begin{bmatrix}  \begin{bmatrix}x_{111}&x_{112}\\x_{121}&x_{122}\end{bmatrix}\\  \begin{bmatrix}x_{211}&x_{212}\\x_{221}&x_{222}\end{bmatrix}  \end{bmatrix}  =  \begin{bmatrix}  \begin{bmatrix}c_{111}&c_{112}\\c_{121}&c_{122}\end{bmatrix}\\  \begin{bmatrix}c_{211}&c_{212}\\c_{221}&c_{222}\end{bmatrix}  \end{bmatrix}  \end{equation}