We have a ring
$R=\begin{pmatrix} \mathbb{Z} & 0 \\ \mathbb{Z} & \mathbb{Z} \end{pmatrix}$
Let $I=\begin{pmatrix} 12\mathbb{Z} & 0 \\ 3\mathbb{Z} & 3\mathbb{Z} \end{pmatrix}$
How do I show that $R/I$ is artinian ring? $R/I=I$, the only ideal that isn't equal to I is $\begin{pmatrix} 0 & 0 \\ 3\mathbb{Z} & 3\mathbb{Z} \end{pmatrix}$ and $\begin{pmatrix} 12\mathbb{Z} & 0 \\ 3\mathbb{Z} & 0 \end{pmatrix}$. These are both DCC condition is satisfied with them.
So then you just conclude it's aritian. Also, the niradical elements are just the bottom left corner of the matrix.