Rank of a matrix by reduction [3,1,4;0,5,8;…]

A = [ 3 1 4]
    [ 0 5 8]
    [−3 4 4]
    [ 1 2 4]

Step 1. R₁ ↔ R₄ to get a pivot 1:

[ 1 2 4]
[ 0 5 8]
[−3 4 4]
[ 3 1 4]

Step 2. R₃ → R₃ + 3R₁, R₄ → R₄ − 3R₁:

[1   2   4]
[0   5   8]
[0  10  16]
[0  −5  −8]

Step 3. R₃ → R₃ − 2R₂, R₄ → R₄ + R₂:

[1  2  4]
[0  5  8]
[0  0  0]
[0  0  0]

Two non-zero rows remain. Further column operations (C₂ → C₂ − 2C₁, C₃ → C₃ − 4C₁, then C₃ → C₃ − (8/5)C₂ and scaling) give the normal form

[I₂ 0]
[0  0]

Answer: rank(A) = 2.

(Check: the 3 × 3 minors are all zero, e.g. |3 1 4; 0 5 8; −3 4 4| = 3(20−32) − 1(0+24) + 4(0+15) = −36 − 24 + 60 = 0, while a 2 × 2 minor such as |1 2; 0 5| = 5 ≠ 0.)