\[m \times n\]
matrix \[A\]
is the order of the square matrix with the largest non zero determinant obtained by deleting rows and columns of \[A\]
The usual method to find the determinant rank is to reduce the matrix to echelon form, so that the leading term in each row is 1, then select the leading submatrix having 1s alonmg the diagonal. The order of this submatrix is the determinant rank of the original matrix.
Take the matrix
\[ \left( \begin{array}{cccc} 1 & 2 & 1 & 2 \\ 2 & 1 & 2 & 1 \\ 3 & 2 & 3 & 2 \\ 3 & 3 & 3 & 3 \\ 5 & 3 & 5 & 3 \end{array} \right) \]
To reduce this matrix to echelon form perform elementary row operations - adding or subtracting multiples of each row, interchanging rows, or scaling rows - until we have a matrix satisfying
1. All zero rows are at the bottom of the matrix.
2. The leading entry of each nonzero row after the first occurs to the right of the leading entry of the previous row.
3. The leading entry in any nonzero row is 1.
4. All entries in the column below a leading 1 are zero.
Subtract two times row 1 from row 2, subtract 3 times row 1 from rows 3 and 4, and subtract 5 times row 1 from row 5. We get
\[ \left( \begin{array}{cccc} 1 & 2 & 1 & 2 \\ 0 & -3 & 0 & -3 \\ 0 & -4 & 0 & -4 \\ 0 & -3 & 0 & -3 \\ 0 & -7 & 0 & -7 \end{array} \right) \]
Divide row 2 by -3
\[ \left( \begin{array}{cccc} 1 & 2 & 1 & 2 \\ 0 & 1 & 0 & 1 \\ 0 & -4 & 0 & -4 \\ 0 & -3 & 0 & -3 \\ 0 & -7 & 0 & -7 \end{array} \right) \]
Add four times row 2 to row 3, add three times row 2 to row 4 and add seven times row 2 to row 5.
\[ \left( \begin{array}{cccc} 1 & 2 & 1 & 2 \\ 0 & 1 & 0 & 1 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array} \right) \]
The largest submatrix satisfying the conditions is the 2 by 2 matrix in the upper left. The determinant rank is 2.