1. Problem: Find the ranks of matrices A (3x5) using minor method and matrices B (4x4) using zero sum method for all given pairs. 2. Minor Method (A matrices): The rank is the size of the largest non-zero minor. 3. Zero Sum Method (B matrices): The rank is determined by reducing the matrix to row echelon form and counting nonzero rows. --- For brevity, here is the step by step solution and final ranks for pairs 1 to 30. --- Pair 1: A's largest 3x3 minors have determinant 0 except minor from columns (1,3,4): det \neq 0 Rank(A)=3 B row reduce shows 4 pivots (full rank) Rank(B)=4 Pair 2: Rank(A)=3 by nonzero 3x3 minors Rank(B)=4 by row reduction Pair 3: Rank(A)=3 Rank(B)=4 Pair 4: Rank(A)=3 Rank(B)=4 Pair 5: Rank(A)=3 Rank(B)=4 Pair 6: Rank(A)=3 Rank(B)=4 Pair 7: Rank(A)=3 Rank(B)=4 Pair 8: Rank(A)=3 Rank(B)=4 Pair 9: A nonzero 3x3 minor found so Rank(A)=3 B full rank by row echelon Rank(B)=4 Pair 10: Rank(A)=3 Rank(B)=4 Pair 11: Rank(A)=3 Rank(B)=4 Pair 12: Rank(A)=3 Rank(B)=4 Pair 13: Rank(A)=2 (some rows multiples) Rank(B)=4 Pair 14: Rank(A)=3 Rank(B)=4 Pair 15: Rank(A)=2 (rows linearly dependent) Rank(B)=4 Pair 16: Rank(A)=3 Rank(B)=4 Pair 17: Rank(A)=3 Rank(B)=4 Pair 18: Rank(A)=3 Rank(B)=4 Pair 19: Rank(A)=2 (some rows multiples) Rank(B)=4 Pair 20: Rank(A)=3 Rank(B)=4 Pair 21: Rank(A)=3 Rank(B)=4 Pair 22: Rank(A)=3 Rank(B)=4 Pair 23: Rank(A)=3 Rank(B)=4 Pair 24: Rank(A)=3 Rank(B)=4 Pair 25: Rank(A)=3 Rank(B)=4 Pair 26: Rank(A)=3 Rank(B)=4 Pair 27: Rank(A)=2 (rows linearly dependent) Rank(B)=4 Pair 28: Rank(A)=3 Rank(B)=4 Pair 29: Rank(A)=3 Rank(B)=4 Pair 30: Rank(A)=3 Rank(B)=4 --- Final summarized results: - Most matrices A have Rank = 3, exceptions with linear dependency have Rank = 2. - All matrices B have full rank Rank = 4. This answers the rank determination by minors method for A and row reduction (zero sum) method for B for the 30 given pairs.