 Solution

For an invertible $A$, factor $\det (A+tB)=\det A\det (I+t{A}^{-1}B)$. The determinant's permutation expansion gives $\det (I+tC)=1+t\mathrm{tr}C+O(t^2)$. Differentiating at zero yieldsAt a determinant-one matrix choose $B=A/n$; the derivative is one. Thus one is a regular value of the smooth determinant map on $GL(n)$. The regular level set theorem makes $SL(n)$ a submanifold of codimension of a submanifold one, withThis includes $n=1$, when $SL(1)=\{1\}$ has dimension zero.