Let 𝐺 = [𝑎 00 0] : 𝑎 𝑖𝑠 𝑎 𝑛𝑜𝑛 − 𝑧𝑒𝑟𝑜 𝑟𝑒𝑎𝑙 𝑛𝑢𝑚𝑏𝑒𝑟 } . hho that 𝐺 is a oommutativegroup under multiplioation

The question seems to be asking to prove that the set G, consisting of matrices of the form [a 00 0] where a is a non-zero real number, forms a commutative group under multiplication. However, the question is not clearly written and contains several typographical errors. Here's a step-by-step guide to prove it:

1. Identity: The identity element in the group under multiplication is the matrix [1 00 0]. Any matrix in G when multiplied with this identity matrix gives the same matrix, hence the identity exists.

2. Closure: If we take any two matrices [a 00 0] and [b 00 0] in G and multiply them, we get another matrix [ab 00 0] which is also in G. Hence, the set G is closed under multiplication.

3. Inverse: For every matrix [a 00 0] in G, there exists an inverse [1/a 00 0] such that their multiplication gives the identity matrix [1 00 0]. Hence, every element has an inverse in G.

4. Commutativity: For any two matrices [a 00 0] and [b 00 0] in G, the order of multiplication does not matter. [a 00 0] * [b 00 0] = [ab 00 0] = [b 00 0] * [a 00 0]. Hence, the group is commutative.

Therefore, G is a commutative group under multiplication.