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Prove that the product of two matrices in $SL_2(\mathbb{R} \!\,)$ has determinant one.

I don't know how I can rigorously prove this, but all matrices in $SL_2(\mathbb{R} \!\,)$ have determinant 1, and the determinant of the product of 2 matrices is equal to the product of the determinants, and 1*1=1. Is there a way to prove this more formally?

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There is nothing wrong with your proof. The only step that is in any doubt is the fact that the determinant is multiplicative, i.e. $\det(AB) = \det A\det B$ for square matrices $A,B$. Presumably your class notes or textbook have already proven this fact, but if not, you can easily prove it for $2\times 2$ matrices by direct computation.

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