Let $A$ and $B$ be two $n×n$ matrices over real numbers. Let rank($M$) and det($M$) denote the…

Question

Let $A$ and $B$ be two $n×n$ matrices over real numbers. Let rank($M$) and det($M$) denote the rank and determinant of a matrix $M$, respectively. Consider the following statements. I. rank($AB$) = rank($A$)*rank ($B$) II. det($AB$) = det($A$)*det($B$) III. rank($A+B$) ≤ rank($A$) + rank($B$) IV. det($A+B$) ≤ det($A$) + det($B$) Which of the above statements are TRUE ?

Accepted Answer

Correct answer: C — Answer: Statements II and III are true; Statements I and IV are false. Explanation and justification: Statement II (det(AB) = det(A)det(B)) This is a standard property of determinants for square matrices: the determinant is a multiplicative map on the ring of n×n matrices, so det(AB)=det(A)det(B). Statement III (rank(A+B) ≤ rank(A) + rank(B)) Each column of A+B is the sum of the corresponding columns of A and B, so the column space of A+B is contained in the sum of the column spaces of A and B. The dimension of a sum of subspaces is at most the sum of their dimensions, hence rank(A+B) ≤ rank(A)+rank(B). Statement I (rank(AB) = rank(A)·rank(B)) — False In general, rank(AB) ≤ min(rank(A),rank(B)), so equality with the product of ranks is false. A simple counterexample in 2×2: take A = [[1,0],[0,0]] and B = [[0,0],[0,1]]. Then rank(A)=rank(B)=1, but AB is the zero matrix with rank 0, so rank(AB) ≠ rank(A)·rank(B). Statement IV (det(A+B) ≤ det(A) + det(B)) — False Determinant is not subadditive in this sense and can grow faster under addition. For example in 2×2 take A = 2I and B = I. Then det(A)=4, det(B)=1 and det(A+B)=det(3I)=9, so det(A+B)=9 > 4+1=5, contradicting the proposed inequality. Therefore the only true statements are the multiplicativity of determinant and the rank subadditivity.

Answer

A. I and II only

Answer

B. I and IV only

Answer

C. II and III only

Answer

D. III and IV only

Let \(A\) and \(B\) be two \(n×n\) matrices over real numbers. Let rank(\(M\))…