Linear Algebra: A Modern Introduction

$\mathbb{Z}_3$: $2$ $\mathbb{Z}_4$: $0$ $\mathbb{Z}_5$: $3$.
First perform the computation in $\mathbb{Z}$: $$2+1+2+2+1=8$$ The result of the computation in $\mathbb{Z}_n$ is the remainder of $8$ divided by $n$. So in $\mathbb{Z}_3$, the result is $2$; in $\mathbb{Z}_4$, $0$; and in $\mathbb{Z}_5$, $3$.