Rank of a matrix by means of determinants

The rank of a matrix can also be calculated using determinants. We can define rank using what interests us now.

The rank of a matrix is the order of the largest non-zero square submatrix.

See the following example.

Example

$A = (\begin{array}{ccccc} 2 & 1 & 3 & 2 & 0 \\ 3 & 2 & 5 & 1 & 0 \\ - 1 & 1 & 0 & - 7 & 0 \\ 3 & - 2 & 1 & 17 & 0 \\ 0 & 1 & 1 & - 4 & 0 \end{array})$

1) Given $A$ , we eliminate rows or columns acording to the criterion to calculate the rank using the Gaussian elimination method. Thus,

Column $5$ can be discarded because all its elements are zero.

Column $3$ can be discarded because it is a linear combination of column $1$ and column $2$ . Specifically, $c 3 = c 1 + c 2$ .

$A = (\begin{array}{ccc} 2 & 1 & 2 \\ 3 & 2 & 1 \\ - 1 & 1 & - 7 \\ 3 & - 2 & 17 \\ 0 & 1 & - 4 \end{array})$

2) Is there any non-zero square submatrix of order $1$ ?

Any non-zero element is a non-zero square submatrix, therefore we will look at those of higher order.

Is there any non-zero square submatrix of order $2$ ?

$| \begin{array}{cc} 2 & 1 \\ 3 & 2 \end{array} | = 1 \neq 0$

Yes, there is, therefore we will look for higher orders.

4) Is there any non-zero square submatrix of order $3$ ?

$| \begin{array}{ccc} 2 & 1 & 2 \\ 3 & 2 & 1 \\ - 1 & 1 & - 7 \end{array} | = 0$

$| \begin{array}{ccc} 3 & 2 & 1 \\ - 1 & 1 & - 7 \\ 3 & - 2 & 17 \end{array} | = 0$

$| \begin{array}{ccc} - 1 & 1 & - 7 \\ 3 & - 2 & 17 \\ 0 & 1 & - 4 \end{array} | = 0$

No, there is not. Therefore, rank $(A) = 2$ , which is the order of the largest non-zero square submatrix.