Rank of a matrix by reduction [1,−1,2,−3;4,1,0,2;…]

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

Step 1. R₂ → R₂ − 4R₁:

[1 −1  2 −3]
[0  5 −8 14]
[0  3  0  4]
[0  1  0  2]

Step 2. Interchange R₂ and R₄ to get a pivot 1:

[1 −1  2 −3]
[0  1  0  2]
[0  3  0  4]
[0  5 −8 14]

Step 3. R₃ → R₃ − 3R₂, R₄ → R₄ − 5R₂:

[1 −1  2 −3]
[0  1  0  2]
[0  0  0 −2]
[0  0 −8  4]

Step 4. Interchange R₃ and R₄:

[1 −1  2 −3]
[0  1  0  2]
[0  0 −8  4]
[0  0  0 −2]

This is upper triangular with four non-zero diagonal entries (1, 1, −8, −2). Column operations (clearing the entries above each pivot) and then scaling reduce it to the normal form

[I₄]   (the 4 × 4 identity matrix)

Since |A| = 1·1·(−8)·(−2) = 16 ≠ 0, A is non-singular.

Answer: rank(A) = 4.