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Rank of a matrix by reduction [1,−1,−2,−4;2,3,−1,−1;…]

Short answer

After R2−2R1, R3−3R1, R4−6R1 and then 5R3−4R2, 5R4−9R2, the matrix reduces to three non-zero rows with one zero row, so rank = 3.

Find the rank of the matrix [1 −1 −2 −4; 2 3 −1 −1; 3 1 3 −2; 6 3 0 −7].

A = [1 −1 −2 −4]
    [2  3 −1 −1]
    [3  1  3 −2]
    [6  3  0 −7]

Step 1. R₂ → R₂ − 2R₁, R₃ → R₃ − 3R₁, R₄ → R₄ − 6R₁:

[1 −1 −2 −4]
[0  5  3  7]
[0  4  9 10]
[0  9 12 17]

Step 2. R₃ → 5R₃ − 4R₂, R₄ → 5R₄ − 9R₂:

R₃: (0, 20−20, 45−12, 50−28) = (0, 0, 33, 22)
R₄: (0, 45−45, 60−27, 85−63) = (0, 0, 33, 22)

[1 −1 −2 −4]
[0  5  3  7]
[0  0 33 22]
[0  0 33 22]

Step 3. R₄ → R₄ − R₃:

[1 −1 −2 −4]
[0  5  3  7]
[0  0 33 22]
[0  0  0  0]

Three non-zero rows remain, so the matrix is of rank 3 (|A| = 0).

Answer: rank(A) = 3.

✓ Answer checked by Claude Opus 5.5 (high effort)

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