Row echelon form

Transform the matrix A to row echelon form.

    ┌                ┐
    │  2   2   4  -8 │
    │ -1  -3  -4   5 │
A = │  2  -1  -1   0 │
    │  0  -2  -2   1 │
    └                ┘

How to solve this problem?

When a row of the matrix A is non-null, its first non-zero entry is the leading entry of the row. The matrix A is in row echelon form when any zero rows are below all non-zero rows, and for each non-zero row, the leading entry is in a column to the right of the leading entries of the previous rows. A convenient method consists of making zero all the entries that are below the leading entry (pivot) in each row, starting by the first row, until the matrix is in row echelon form.

• For this purpose, when the corresponding entry is non-zero (the one in the same column as the pivot), use elementary row operations of the third kind (add a multiple of one row to another row) replacing each row beneath the pivot row by itself minus the pivot row multiplied by quotient between the corresponding entry in the row and the pivot.
• Use the elementary row operations of the first kind (interchange two rows) to find a non-zero pivot or move the null-rows to the end.
• Use elementary row operations of the second kind (multiply a row through by a non-zero constant) to avoid working with fractional numbers by multiplying the row to be modified by a scalar, so that the entry that is below the pivot, be a multiple of the pivot.

Step 1: Make elementary row operations until all zero rows are below all non-zero rows, and for each non-zero row, the leading entry is in a column to the right of the leading entries of the previous rows.

r1 <———> r2
┌                ┐
│ -1  -3  -4   5 │
│  2   2   4  -8 │
│  2  -1  -1   0 │
│  0  -2  -2   1 │
└                ┘
r2 <———> r2 + 2•r1
r3 <———> r3 + 2•r1
┌                ┐
│ -1  -3  -4   5 │
│  0  -4  -4   2 │
│  0  -7  -9  10 │
│  0  -2  -2   1 │
└                ┘
r2 <———> r4
┌                ┐
│ -1  -3  -4   5 │
│  0  -2  -2   1 │
│  0  -7  -9  10 │
│  0  -4  -4   2 │
└                ┘
r3 <———> -2•r3
┌                 ┐
│ -1  -3  -4    5 │
│  0  -2  -2    1 │
│  0  14  18  -20 │
│  0  -4  -4    2 │
└                 ┘
r3 <———> r3 + 7•r2
r4 <———> r4 - 2•r2
┌                 ┐
│ -1  -3  -4    5 │
│  0  -2  -2    1 │
│  0   0   4  -13 │
│  0   0   0    0 │
└                 ┘

Step 2: Final solution.

    ┌                 ┐
    │ -1  -3  -4    5 │
    │  0  -2  -2    1 │
A ~ │  0   0   4  -13 │
    │  0   0   0    0 │
    └                 ┘