Rank of a matrix by reduction [1,0,−5,6;3,−2,1,2;…]

A = [1  0 −5  6]
    [3 −2  1  2]
    [3 −2 −9 14]
    [4 −2 −4  8]

Step 1. R₂ → R₂ − 3R₁, R₃ → R₃ − 3R₁, R₄ → R₄ − 4R₁:

[1  0  −5    6]
[0 −2  16  −16]
[0 −2   6   −4]
[0 −2  16  −16]

Step 2. R₃ → R₃ − R₂, R₄ → R₄ − R₂:

[1  0  −5    6]
[0 −2  16  −16]
[0  0 −10   12]
[0  0   0    0]

This is an echelon form with three non-zero rows.

Normal form. Column operations (C₄ → C₄ − 6C₁, C₃ → C₃ + 5C₁, then clearing the remaining entries of rows 2 and 3 using the pivots −2 and −10 and dividing rows by the pivots) reduce the matrix to

[I₃ 0]
[0  0]

Answer: rank(A) = 3.