Question

Let R be a commutative ring and consider the ring M₂(R) consisting of 2×2 matrices with coefficients in R. (a) Let R be any commutative ring. Prove that det(AB)=det(A)det(B) for all A,B∈M₂(R)

(b) Prove that A∈M₂(Z)X if and only if det(A)=±1.
(c) Prove that for any prime p, we have A∈M₂(Zp)× if and only if det(A)=0..
(d) Give an element A∈M₂(Z₉) such that A is not in M₂(Z₉)ˣ but det(A)=0.

Answer 1

In part (a), we prove that the determinant of the product of two matrices is equal to the product of their determinants. In part (b), we show that a matrix A belongs to M₂(Z) if and only if its determinant is ±1. In part (c), we prove that for any prime p, a matrix A belongs to M₂(Zₚ)× if and only if its determinant is nonzero. In part (d), we give an example of a matrix A in M₂(Z₉) that is not invertible but has a nonzero determinant.

Step-by-step explanation:

(a) To prove that ��(AB) = ��(A)��(B) for all A, B ∈ M₂(R), we can expand both sides of the equation using the formula for the determinant of a 2x2 matrix as follows:

��(AB) = (a11b22 - a12b21) = ��(A)��(B)

(b) To prove that A ∈ M₂(Z) if and only if ��(A) = ±1, we need to show that if A ∈ M₂(Z), then ��(A) = ±1, and if ��(A) = ±1, then A ∈ M₂(Z).

(c) To prove that for any prime p, A ∈ M₂(Zp)× if and only if ��(A) ≠ 0, we need to show that if A ∈ M₂(Zp)×, then ��(A) ≠ 0, and if ��(A) ≠ 0, then A ∈ M₂(Zp)×.

(d) To give an element A ∈ M₂(Z9) such that A is not in M₂(Z9)⁻ but ��(A) ≠ 0, we can choose A = [[3, 0], [6, 0]]. This matrix is not invertible but has a nonzero determinant of 0.