The group GL(2, Z5) consists of all invertible 2x2 matrices with entries from the set Z5 = {0, 1, 2, 3, 4}.

For a matrix A = [[a, b], [c, d]] in GL(2, Z5) to be invertible, the determinant ad - bc must be nonzero in Z5. In other words, ad - bc must not be congruent to 0 modulo 5.

To find all such matrices, we can consider each entry in the matrix individually and find all possible values for each entry such that ad - bc is not congruent to 0 modulo 5.

Let's consider the possible values for each entry:

For the first entry a, since it can be any element in Z5, there are 5 possible values: a = 0, 1, 2, 3, 4.

For the second entry b, since ad - bc must not be congruent to 0 modulo 5, we need to consider all possible values for b such that (ad - bc) is not congruent to 0 modulo 5 for each value of a. Let's consider each case:

- If a = 0, then (ad - bc) = 0 - b(0) = 0, which is congruent to 0 modulo 5. Therefore, b cannot be 0.

- If a = 1, then (ad - bc) = 1d - bc. We need to find all possible values for b such that 1d - bc is not congruent to 0 modulo 5 for each value of d. Since a and d can be any element in Z5, we need to find b such that (1d - bc) is not congruent to 0 modulo 5 for any combination of d and b. This requires some trial and error, but we can find that b = 1, 2, 3, 4 are all valid values.

- Similarly, for a = 2, 3, 4, we can find that b = 1, 2, 3, 4 are all valid values.

Therefore, for the second entry b, there are 4 possible values: b = 1, 2, 3, 4.

For the third entry c, we can use the same logic as above to find that c = 1, 2, 3, 4 are all valid values.

For the fourth entry d, since it can be any element in Z5, there are 5 possible values: d = 0, 1, 2, 3, 4.

Therefore, the group GL(2, Z5) for multiplication of matrix operation consists of all matrices of the form:

[[a, b], [c, d]]

where a, b, c, and d can each take on any element in Z5.

Note that the number of elements in the group GL(2, Z5) is 5^4 = 625, since there are 5 possible values for each of the 4 entries in the matrix.