Example SEEF Submatrices of extended echelon form

We illustrate Definition EEF with the matrix $A$, \begin{equation*} A= \begin{bmatrix} 1 & -1 & -2 & 7 & 1 & 6 \\ -6 & 2 & -4 & -18 & -3 & -26 \\ 4 & -1 & 4 & 10 & 2 & 17 \\ 3 & -1 & 2 & 9 & 1 & 12 \end{bmatrix} \end{equation*}

Augmenting with the $4\times 4$ identity matrix, \begin{equation*} M= \begin{bmatrix} 1 & -1 & -2 & 7 & 1 & 6 & 1 & 0 & 0 & 0 \\ -6 & 2 & -4 & -18 & -3 & -26 & 0 & 1 & 0 & 0 \\ 4 & -1 & 4 & 10 & 2 & 17 & 0 & 0 & 1 & 0 \\ 3 & -1 & 2 & 9 & 1 & 12 & 0 & 0 & 0 & 1 \end{bmatrix} \end{equation*} and row-reducing, we obtain \begin{equation*} N= \begin{bmatrix} \leading{1} & 0 & 2 & 1 & 0 & 3 & 0 & 1 & 1 & 1\\ 0 & \leading{1} & 4 & -6 & 0 & -1 & 0 & 2 & 3 & 0 \\ 0 & 0 & 0 & 0 & \leading{1} & 2 & 0 & -1 & 0 & -2 \\ 0 & 0 & 0 & 0 & 0 & 0 & \leading{1} & 2 & 2 & 1 \end{bmatrix} \end{equation*}

So we then obtain \begin{align*} B&= \begin{bmatrix} \leading{1} & 0 & 2 & 1 & 0 & 3 \\ 0 & \leading{1} & 4 & -6 & 0 & -1 \\ 0 & 0 & 0 & 0 & \leading{1} & 2 \\ 0 & 0 & 0 & 0 & 0 & 0 \end{bmatrix}\\ C&= \begin{bmatrix} \leading{1} & 0 & 2 & 1 & 0 & 3 \\ 0 & \leading{1} & 4 & -6 & 0 & -1 \\ 0 & 0 & 0 & 0 & \leading{1} & 2 \end{bmatrix}\\ J&= \begin{bmatrix} 0 & 1 & 1 & 1\\ 0 & 2 & 3 & 0 \\ 0 & -1 & 0 & -2 \\ \leading{1} & 2 & 2 & 1 \end{bmatrix}\\ K&= \begin{bmatrix} 0 & 1 & 1 & 1\\ 0 & 2 & 3 & 0 \\ 0 & -1 & 0 & -2 \end{bmatrix}\\ L&= \begin{bmatrix} \leading{1} & 2 & 2 & 1 \end{bmatrix} \end{align*}

You can observe (or verify) the properties of the following theorem with this example.