Question: > 
> Let $R$ be an arbitrary ring and $N \vartriangleleft R$ nilpotent such that $R /N$ is semisimple. Show that $N = rad(R)$.
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Any hints to get me started? I know that $rad(R/N)=N$ since $R /N$ is semisimple. Can I show that $rad(R)=rad(R/N)$?

Answer: 
**Hint:** $2(2,-3,1)-(4,1,1)=(0,-7,1).$

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Alternatively, if the [determinant](https://en.wikipedia.org/wiki/Linear_independence#Alternative_method_using_determinants) of the matrix with the vectors as rows or columns is $0$,

then they do not span $\mathbb R^3$.
Stack the vectors vertically and reduce the matrix to the echelon form (through row-wise elementary transformations only) as follows:
\begin{align\*}
\begin{pmatrix}
2 & -3 & 1 \\
4 & 1 & 1 \\
0 & -7 & 1 \\
\end{pmatrix} \to 
\begin{pmatrix}
2 & -3 & 1 \\
0 & 7 & -1 \\
0 & -7 & 1 \\
\end{pmatrix} \to
\begin{pmatrix}
2 & -3 & 1 \\
0 & 7 & -1 \\
0 & 0 & 0 \\
\end{pmatrix}.
\end{align\*}

Can you identify what transformation I used in each step and reach the conclusion?