Only one of the sets below is a vector subspace. Find it and show that it satisfies the conditions for a vector subspace. For the other sets explain or give a counter-exem for justification why they are not vector subspaces. (a) 5 = {aatb; a, ber (b) W = {(X, v, 2) E R3; x = y + 2). (c) All polynomials a222 + aja + ao, where a2, a1, ao are integers. (d) The set of all matrices A, of size 2 × 2 such that (A) = 0.

Let's analyze each set to determine whether they are vector subspaces or not. (a) Set 5 = {aatb; a, b ∈ ℝ} This set is not a vector subspace. To be a subspace, it needs to satisfy closure under vector addition and scalar multiplication. However, if you take two elements from this set, say, a₁a₂b₁b₂ and a₃a₄b₃b₄, and add them: a₁a₂b₁b₂ + a₃a₄b₃b₄, you'll see that this combination doesn't fit the form of elements in the set, violating closure under vector addition. (b) Set W = {(x, y, z) ∈ ℝ³; x = y + 2) This set is not a vector subspace because it violates closure under scalar multiplication. If you take an element (1, 0, 1) from the set and multiply it by a scalar, say 2, you get (2, 0, 2), which is not in the set since the first component is not equal to the second component plus 2. (c) Set of all polynomials a₂x² + a₁x + a₀, where a₂, a₁, a₀ are integers This set is not a vector subspace because it violates closure under scalar multiplication. For example, if you take the polynomial x² + x + 1 from the set and multiply it by a scalar, say 2, you get 2x² + 2x + 2, which is not in the set because the coefficients are not required to be integers. (d) The set of all 2 × 2 matrices A, such that det(A) = 0 This set is a vector subspace. To show this, we need to confirm that it satisfies closure under vector addition and scalar multiplication: Closure under vector addition: If A and B are matrices in the set with det(A) = det(B) = 0, then consider their sum A + B. The determinant of the sum is det(A + B) = det(A) + det(B) = 0 + 0 = 0, which means A + B is also in the set. Closure under scalar multiplication: If A is a matrix in the set with det(A) = 0, and c is a scalar, then the determinant of cA is det(cA) = c²det(A) = c² * 0 = 0, so cA is also in the set. Both closure conditions are satisfied, so the set of 2 × 2 matrices with a determinant of 0 is indeed a vector subspace.