How to find the rank of a matrix

Question

$$ \begin{pmatrix} 1 & 3 & 0 & 5 & 0 \\ 2 & 0 & 2 & 0 & 2 \\ 4 & 4 & 4 & 2 & 1 \\ 5 & 0 & 4 & 0 & 1 \\ \end{pmatrix} $$

Answer 1

There is a $4\times 4$ invertible submatrix, since its determinant is a number of the form $4k+2\neq 0$.

If you divide the third column by $2$ and perform a reduction $\!\!\pmod{2}$ you get $$ \begin{pmatrix} 1 & 1 & 0 & 1 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 \\ 1 & 0 & 0 & 0 & 1 \\ \end{pmatrix} $$ where the matrix given by the second, third, fifth, first column, namely

$$ \begin{pmatrix} 1 & 0 & 0 & 1 \\ 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 1 & 1\\ \end{pmatrix} $$ has an odd determinant (hence it is invertible). It follows that the original matrix has rank $4$.

Answer 2

$(3, 5)$ and $(4, 2)$ are linearly independent, so it follows from inspection of the second and fourth columns that no nontrivial linear combination of rows 1 and 3 can give row 2, and (similarly) no nontrivial linear combination of rows 1, 2, and 3 can give row 4. So the matrix has full rank.

Answer 3

You can operate via Row echelon form or as an alternative search for the max dimension of the minor with $\det\neq0$.