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

Short answer

Reduce matrix by row operations: R₂−3R₁, R₃−5R₁, R₄−4R₁, then R₃−R₂, R₄−R₂. Further column operations yield the normal form with I₂. The rank is 2.

Reduce the matrix [1 0 −5 6; 3 −2 1 2; 5 −2 −9 14; 4 −2 −4 8] into normal form and hence find its rank.

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

Step 1. R₂ → R₂ − 3R₁, R₃ → R₃ − 5R₁, R₄ → R₄ − 4R₁:

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

Step 2. R₃ → R₃ − R₂, R₄ → R₄ − R₂:

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

Step 3 (normal form). Divide R₂ by −2: R₂ = (0, 1, −8, 8). Now column operations C₃ → C₃ + 5C₁ + 8C₂ and C₄ → C₄ − 6C₁ − 8C₂ wipe out the entries in columns 3 and 4:

[1 0 0 0]
[0 1 0 0]   = [I₂ 0]
[0 0 0 0]     [0  0]
[0 0 0 0]

The normal form contains the identity matrix of order 2.

Answer: rank(A) = 2.

✓ Answer checked by Claude Opus 5.5 (high effort)

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