矩阵运算程序验证定理异常：AB≠BA等8条定理无法验证求助

Let's work through your problems one by one—you've already done great work cross-checking with online calculators, which helps rule out tool-specific errors!

1. Why does AB = BA here (contradicting the "AB ≠ BA" general rule)? #

First off, the statement "matrix multiplication is non-commutative" means it's not guaranteed to work for all matrix pairs, not that it never works. There are plenty of matrix pairs that do commute, and your A and B are one such example. Let's compute manually to confirm:

• AB = [[13 + 24, 12 + 25], [43 + 34, 42 + 35]] = [[11, 12], [24, 23]]
• BA = [[31 + 24, 32 + 23], [41 + 54, 42 + 53]] = [[11, 12], [24, 23]]

This is totally normal—commuting matrices include pairs like diagonal matrices, any matrix paired with the identity matrix, or special cases like your example. The general rule just means you can't assume commutativity unless you know the matrices have that property.

2. Fixes for Theorems 6-8 (Your Code Has Logical Errors) #

Your issues with verifying these theorems come from mismatched operations or incorrect order in your code, not the theorems themselves. Here's how to fix each:

Theorem 6: A(BC) = (AB)C (Associative Property) #

Your current code checks b.mult(c).mult(a).equals(a.mult(b).mult(c))—this is wrong because you're computing (BC)A instead of A(BC). The correct check should be:

System.out.print("6. A(BC) = (AB)C \t = "); 
System.out.println(a.mult(b.mult(c)).equals(a.mult(b).mult(c)));

Theorem 7: A(B+C) = AB + AC (Distributive Property) #

You used mult instead of add on the right-hand side, and also mixed up the order of addition. The correct code compares A(B+C) to AB + AC:

System.out.print("7. A(B+C) = AB + AC \t = "); 
System.out.println(a.mult(b.add(c)).equals(a.mult(b).add(a.mult(c))));

Theorem 8: (2A)B = 2(AB) = A(2B) (Scalar Multiplication Associativity) #

Your current code checks an unrelated expression. You need to split this into verifiable equality checks:

// Verify (2A)B = 2(AB)
boolean part1 = a.scalarMult(2).mult(b).equals(a.mult(b).scalarMult(2));
// Verify 2(AB) = A(2B)
boolean part2 = a.mult(b).scalarMult(2).equals(a.mult(b.scalarMult(2)));

System.out.print("8. (2A)B = 2(AB) = A(2B) = ");
System.out.println(part1 && part2);

3. Corrected Full Code Snippet #

Here's the revised code with all fixes (including a subtle mistake in Theorem 4):

Matrix a = new Matrix(new int[][]{{1,2},{4,3}}); 
Matrix b = new Matrix(new int[][]{{3,2},{4,5}}); 
Matrix c = new Matrix(new int[][]{{2,2},{-1,-1}}); 

System.out.print("1. (A^T)^T=A \t\t = "); 
System.out.println(a.transpose().transpose().equals(a)); 

System.out.print("2. (A+B)^T = A^T+B^T = "); 
Matrix added = a.add(b).transpose(); 
System.out.println(added.equals(a.transpose().add(b.transpose()))); 

System.out.print("3. (2A)^T = 2A^T \t = "); 
System.out.println(a.scalarMult(2).transpose().equals(a.transpose().scalarMult(2))); 

System.out.print("4. (AB)^T = B^T A^T \t = "); 
Matrix mult = b.transpose().mult(a.transpose()); // Fixed order: (AB)^T is B^T A^T, not A^T B^T
System.out.println(mult.equals(a.mult(b).transpose())); 

System.out.println("5. AB != BA \t\t = " + !a.mult(b).equals(b.mult(a))); // Check for inequality explicitly

System.out.print("6. A(BC) = (AB)C \t = "); 
System.out.println(a.mult(b.mult(c)).equals(a.mult(b).mult(c)));

System.out.print("7. A(B+C) = AB + AC \t = "); 
System.out.println(a.mult(b.add(c)).equals(a.mult(b).add(a.mult(c))));

// Verify Theorem 8
boolean part1 = a.scalarMult(2).mult(b).equals(a.mult(b).scalarMult(2));
boolean part2 = a.mult(b).scalarMult(2).equals(a.mult(b.scalarMult(2)));
System.out.print("8. (2A)B = 2(AB) = A(2B) = ");
System.out.println(part1 && part2);

With these fixes, all theorem checks should align with the expected matrix properties.

内容的提问来源于stack exchange，提问作者arcticlights