Here is a proof of the above fact.

Let be the commutator subgroup of the general linear group ; i.e.,

.

First, it is clear that is contained in the special linear group , since for any . Next, we claim that contains all matrices

.

This follows from noting that

.

By taking transposes, it also follows that contains all matrices

.

Further, contains all matrices

since

for any .

Now let

.

Then . Using the above results,

if , and

if , and the latter since

for any . Thus , i.e., .

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Your matrix product for has a typo in it (set to see this quickly). See https://math.stackexchange.com/a/2762248/448 for a variant that works.

Thanks for pointing that out!