Let \begin{equation*} A= \begin{bmatrix} -2 & 3 & 0 & 1\\ 9 & -2 & 0 & 1\\ 1 & 3 & -2 & -1\\ 4 & 1 & 2 & 6 \end{bmatrix} \end{equation*}
Then expanding about the fourth row (Theorem DER with $i=4$) yields, \begin{align*} \detbars{A} &= (4)(-1)^{4+1} \begin{vmatrix} 3 & 0 & 1\\ -2 & 0 & 1\\ 3 & -2 & -1 \end{vmatrix} +(1)(-1)^{4+2} \begin{vmatrix} -2 & 0 & 1\\ 9 & 0 & 1\\ 1 & -2 & -1 \end{vmatrix}\\ &\quad\quad+(2)(-1)^{4+3} \begin{vmatrix} -2 & 3 & 1\\ 9 & -2 & 1\\ 1 & 3 & -1 \end{vmatrix} +(6)(-1)^{4+4} \begin{vmatrix} -2 & 3 & 0 \\ 9 & -2 & 0 \\ 1 & 3 & -2 \end{vmatrix}\\ &= (-4)(10)+(1)(-22)+(-2)(61)+6(46)=92 \end{align*}
Expanding about column 3 (Theorem DEC with $j=3$) gives \begin{align*} \detbars{A} &= (0)(-1)^{1+3} \begin{vmatrix} 9 & -2 & 1\\ 1 & 3 & -1\\ 4 & 1 & 6 \end{vmatrix} + (0)(-1)^{2+3} \begin{vmatrix} -2 & 3 & 1\\ 1 & 3 & -1\\ 4 & 1 & 6 \end{vmatrix} +\\ &\quad\quad(-2)(-1)^{3+3} \begin{vmatrix} -2 & 3 & 1\\ 9 & -2 & 1\\ 4 & 1 & 6 \end{vmatrix} + (2)(-1)^{4+3} \begin{vmatrix} -2 & 3 & 1\\ 9 & -2 & 1\\ 1 & 3 & -1 \end{vmatrix}\\ &=0+0+(-2)(-107)+(-2)(61)=92 \end{align*}
Notice how much easier the second computation was. By choosing to expand about the third column, we have two entries that are zero, so two $3\times 3$ determinants need not be computed at all!