How can I generate $\mathrm{SL}(n,\mathbb Z)$ by the subgroup $\mathrm{SL}(n-1,\mathbb Z)$ and another Element of $\mathrm{SL}(n,\mathbb Z)$?

Let $\{z_1,...,z_n\}$ be the canonical Basis of $\mathbb{Z}^n$, such that $z_i$ equals the vector $(0,\dotsc,0,i,0,\dotsc,0)$ with a 1 in the $i$th position. I want to show that the $\mathrm{SL}(n,\mathbb Z)$ is generated by the subgroup $\mathrm{SL}(n-1,\mathbb Z)$ corresponding to the subbasis $\{z_1,\dotsc,z_n-1\}$ and the product of the following two matrices:

1. The Matrix A which transpose the two vectors $z_{n-1}$ and $z_n$.

2. The Matrix B which changes $z_n$ into the element $-z_n$.

It's clear that $\det(AB)=1$, since $\det(A)=\det(B)=-1$. So it follows that $\{\mathrm{SL}(n1,\mathbb Z)\backslash\mathrm{SL}(n-1,\mathbb Z)\}$.

So how can I show that?

Thanks for help.

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You'll need to modify your conjecture, because there is the intermediate subgroup of things of the form $\pmatrix{A&b\cr 0& d}$ where $A$ is in $SL(n-1,\mathbb Z)$ and $b$ is $(n-1)\times 1$, and $d\in \{\pm1\}$. –  paul garrett Jan 18 '12 at 19:39
Thanks. I modified my question. –  Peter Jan 18 '12 at 20:19