This exercise is inspired from an exercise found in Hatcher's book, p. 156, 13:
Compute the homology of the space $X$ obtained from the $1$-sphere $\mathbb{S}^1$ by attaching two $2$-cells via attaching maps $\mathbb{S}^1 \to \mathbb{S}^1$ of degree $2$ and $3$.
My result is: $$\begin{align*}H_k(X) = \begin{cases} \mathbb{Z} & k = 0,\\ \mathbb{Z}/(2\mathbb{Z} \oplus 3\mathbb{Z}) & k = 1,\\ 0 & k >1 \end{cases}\end{align*}$$
is this correct?
Edit. Giving it a second look, writing $\mathbb{Z}/(2\mathbb{Z} \oplus 3\mathbb{Z})$ is blatantly stupid. Now I understand the "matrix notation" provided in the answer and can apply it to the cellular chain complex of the space $X$: $$0 \to \mathbb{Z} \oplus \mathbb{Z} \to \mathbb{Z} \to \mathbb{Z} \to 0$$ If we write $s$ for the $2$-cell with degree $2$ and $t$ for the $2$-cell with degree $3$, the cellular boundary formula yields $$\partial s = 2a \qquad \text{and} \qquad \partial t = 3 a$$ where $a$ denotes the $1$-cell of the cell-complex.
The cellular chain group is given by $C_2=\Bbb Z^2$, $C_1=\Bbb Z$ and $C_0=\Bbb Z$. The map from $C_2$ to $C_1$ is given by the matrix $(2\ 3)$ and that from $C_1$ to $C_0$ is zero. The $C_2\to C_1$ map is surjective, but has kernel $\cong \Bbb Z$, so $H_2(X)\cong \Bbb Z$ while $H_1(X)=0$.