Let A be the set of all non-singular n × n real matrices, for some fixed n ≥…
1994
Let A be the set of all non-singular n × n real matrices, for some fixed n ≥ 2, and let * be the matrix multiplication operation. Then ⟨A, *⟩ is:
- A.
A is closed under * but ⟨A, *⟩ is not a semigroup.
- B.
⟨A, *⟩ is a semigroup but not a monoid.
- C.
⟨A, *⟩ is a monoid but not a group.
- D.
⟨A, *⟩ is a group but not an abelian group.
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Correct answer: D
Concept
A set with a binary operation forms an algebraic-structure ladder: it is a semigroup if the operation is closed and associative; a monoid if, in addition, an identity element exists; a group if, further, every element has an inverse; and an abelian (commutative) group if, on top of the group axioms, the operation also commutes: a * b = b * a for all a, b in the set.
Application
Here A = the set of all n × n non-singular real matrices for a fixed n ≥ 2 (det ≠ 0), and * = matrix multiplication. Check each axiom in order:
Closure: if det(M1) ≠ 0 and det(M2) ≠ 0, then det(M1 * M2) = det(M1) × det(M2) ≠ 0, so M1 * M2 ∈ A.
Associativity: matrix multiplication is associative for all matrices, so (M1 * M2) * M3 = M1 * (M2 * M3).
Identity: the identity matrix I has det(I) = 1 ≠ 0, so I ∈ A, and M * I = I * M = M for every M ∈ A.
Inverse: every M ∈ A has det(M) ≠ 0, so M-1 exists with det(M-1) = 1/det(M) ≠ 0, so M-1 ∈ A too.
Commutativity: matrix multiplication is not commutative in general, so M1 * M2 ≠ M2 * M1 for most pairs in A.
All four group axioms hold, but commutativity fails — so ⟨A, *⟩ is a group that is not abelian.
Cross-check
A concrete pair confirms non-commutativity directly. Illustrate with the smallest valid case, n = 2 (the same non-commutativity persists for every larger fixed n, since an n = 2 block can always be embedded inside a larger non-singular matrix). Take:
M1 = [[1, 1], [0, 1]] and M2 = [[1, 0], [1, 1]] (both have determinant 1, so both are non-singular).
M1 * M2 = [[2, 1], [1, 1]] while M2 * M1 = [[1, 1], [1, 2]] — the two products differ, so M1 * M2 ≠ M2 * M1, confirming the group is non-abelian.
Contrast with the other options: A is closed and associative (rules out “not a semigroup”), the identity matrix exists in A (rules out “not a monoid”), and every element has an inverse in A (rules out “not a group”) — leaving “group but not abelian” as the only structure that fits every axiom check.
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