Evaluating rank of a matrix using determinanats I have been introduced to a method for calculating the rank of and $m\times n$ matrix using determinants. The method states that the rank of our matrix is equal to the order of the largest square submatrix that has non zero determinant. My problem with the method is that a matrix has too many submatrices, for example my professor asked me to evaluate the rank of a specific $5\times 5$ matrix,(I know from the solution that rank of the specific matrix is 3. But if I were to solve the problem using the method I have been introduced I would have to verify that the determinant of each of the $4\times 4$ submatrices is zero. But there are 25 such matrices hence the methis seems inefficient to me. So I am wondering are there any tricks to the method i.e is there a way to come to the conclusion without having to check the determinant for  each $4\times 4$ matrix?
Thanks for any help
 A: You can start by removing zero rows (columns), and any rows (columns) that are linear combinations of other rows (columns). Hopefully, this will trim the matrix down enough that you won't have to calculate determinants of large matrices. Here is an example of using this technique.
If this technique don't apply to your example, then you can just row reduce the matrix and count the number of pivots, as that is equal to the rank.
A: You can use the following fact.
Suppose $M_{i_1,\ldots,i_p,j_1,\ldots,j_p}$ is a non-zero minor of size $p$ constructed by taking the elements of the original matrix from the rows $i_1,\ldots,i_p$ and columns $j_1,\ldots,j_p$. If all the minors of size $p+1$
that inculude the rows $i_1,\ldots,i_p$ and columns $j_1,\ldots,j_p$ are equal to zero, then the rank of the matrix is $p$.
For example, suppose we have the matrix
$$
A=\left(\begin{array}{lllll}
     1&     0&     1&     0&     1\\
     0&     1&     1&     0&     1\\
     1&     0&     0&     1&     1\\
     1&     1&     2&     0&     2\\
     1&     1&     1&     1&     2\\
\end{array}\right);
$$
suppose we know that 
$$
M_{123,123}=\left|\begin{array}{lll}
1&     0&     1\\
     0&     1&     1   \\
     1&     0&     0  \\
\end{array}\right|=-1\ne 0.
$$
In this case, we should only check 4 (not 16) minors of size 4: $M_{1234,1234}$, $M_{1234,1235}$, $M_{1235,1234}$, $M_{1235,1235}$. This 4 minors are equal to zero, thus, ${\rm rank} A=3$.
