Prove that follo ing set forms a group under multiplioation.[1 00 0] , [-1 00 1] , [1 00 -1] , [-1 00 -1]}

To prove that a set forms a group under a certain operation, we need to show that it satisfies the four group axioms: closure, associativity, identity, and invertibility.

1. Closure: For any two elements in the set, their product (under the operation) is also in the set.

2. Associativity: For any three elements in the set, the product of the first two, multiplied by the third, is the same as the product of the last two, multiplied by the first.

3. Identity: There is an element in the set such that, when any element is multiplied by it, the result is the original element.

4. Invertibility: For each element in the set, there is another element such that, when the two are multiplied, the result is the identity element.

Let's denote the elements of the set as follows: A = [1 00 0], B = [-1 00 1], C = [1 00 -1], D = [-1 00 -1].

1. Closure:

   • A * A = A, A * B = B, A * C = C, A * D = D
   • B * A = B, B * B = A, B * C = D, B * D = C
   • C * A = C, C * B = D, C * C = A, C * D = B
   • D * A = D, D * B = C, D * C = B, D * D = A So, the set is closed under multiplication.

2. Associativity: This is a bit more complex to show, but it can be done by checking all possible combinations of three elements. For example, (A * B) * C = A * (B * C), (A * C) * B = A * (C * B), etc.

3. Identity: The element A = [1 00 0] is the identity element because when any element is multiplied by A, the result is the original element.

4. Invertibility:

   • The inverse of A is A itself.
   • The inverse of B is B.
   • The inverse of C is C.
   • The inverse of D is D.

Therefore, the set {[1 00 0], [-1 00 1], [1 00 -1], [-1 00 -1]} forms a group under multiplication.