$\mathbb Z_{11} = \{0,1,2,3,4,5,6,7,8,9,10 \}$
Above would be the tables
From the tables $Z_{11}$ is closed under addition and multiplication
Since the tables are symmetric both operations are commutative
Since addition and multiplication is in $\mathbb R$ this satisfies associative and distributive conditions, addition and multiplication in $Z_{11}$ will do.
The additive identity is 0 and multiplicative identity is 1
Additive inverse:
$0^{-1} = 0, 1^{-1} = 10, 2^{-1} = 9, 3^{-1} = 8, 4^{-1} = 7, 5^{-1} = 6, 6^{-1} = 5, 7^{-1} = 4, 8^{-1} = 3, 9^{-1}=2, 10^{-1} = 1$
Multiplicative inverse:
$1^{-1} = $
idk how to do this part
In the set $Z_{11}$, the multiplicative inverse of 2 is 6 because $2 \times 6 \equiv 1 \pmod{11}$. The multiplicative inverse of an element $a$ in a group is an element $b$ such that $a \times b \equiv 1 \pmod{11}$. In other words, the product of $a$ and its multiplicative inverse is congruent to 1 modulo 11.
In general, to find the multiplicative inverse of an element $a$ in $Z_{11}$, you can use the Extended Euclidean algorithm. This algorithm takes as input two integers $a$ and $b$, and returns the greatest common divisor of $a$ and $b$, as well as integers $x$ and $y$ such that $ax + by = \gcd(a, b)$. If the greatest common divisor of $a$ and 11 is 1, then $a$ has an inverse modulo 11 and $x$ is the inverse. Otherwise, $a$ does not have an inverse modulo 11.