In normal form [Iᵣ 0; 0 0], the rank is:

The order r of the identity block

Why does R₄ become zero?

R₄ − R₂ makes (0,1,1,0) − (0,1,1,0) = 0

Short answer

Elementary row and column operations reduce the matrix to normal form [I3 0; 0 0], so rank = 3.

Find the rank of the following matrix by reducing it to normal form: [1 2 1 2; 1 3 2 2; 2 4 3 4; 3 7 4 6].

The rank is the order of the identity block in the normal form [Iᵣ 0; 0 0].

A = [1 2 1 2]
    [1 3 2 2]
    [2 4 3 4]
    [3 7 4 6]

Step 1. R₂ → R₂ − R₁, R₃ → R₃ − 2R₁, R₄ → R₄ − 3R₁:

[1 2 1 2]
[0 1 1 0]
[0 0 1 0]
[0 1 1 0]

Step 2. R₄ → R₄ − R₂:

[1 2 1 2]
[0 1 1 0]
[0 0 1 0]
[0 0 0 0]

Step 3 (column operations). C₂ → C₂ − 2C₁, C₃ → C₃ − C₁, C₄ → C₄ − 2C₁ clear the first row:

[1 0 0 0]
[0 1 1 0]
[0 0 1 0]
[0 0 0 0]

Then C₃ → C₃ − C₂ clears the second row, and the entries of R₃ are already isolated:

[1 0 0 0]
[0 1 0 0]   =  [I₃ 0]
[0 0 1 0]      [0  0]
[0 0 0 0]

The normal form contains I₃.

Answer: rank(A) = 3.

✓ Answer checked by Claude Opus 5.5 (high effort)