= Solution
An embedded smooth $d$-dimensional <manifold> $X\subset\mathbb R^N$ is locally carried by a smooth ambient coordinate change to $\mathbb R^d\times\{0\}$. A map between such <manifolds> is a <smooth map between manifolds> if its coordinate representatives are smooth. Its <differential> $df_x:T_xX\to T_{f(x)}Y$ sends the tangent vector represented by a curve $\gamma$ to $(f\circ\gamma)'(0)$; equivalently it is the <derivative> in charts.
A <regular value> $y$ is one for which $df_x$ is surjective at every $x\in f^{-1}(y)$, including vacuous regularity when the fibre is empty. For example zero is not a <regular value> of $f:\mathbb R\to\mathbb R$, $f(x)=x^2$, because $df_0=0$.
For an invertible <matrix> $A$, the <derivative of the determinant> is
$$
(d\det)_A(H)=\det(A)\operatorname{Tr}(A^{-1}H).
$$
At a determinant-one <matrix> it is nonzero, since taking $H=A/n$ gives value one. The <regular level set theorem> therefore makes $SL_n(\mathbb R)=\det^{-1}(1)$ an embedded <manifold> of dimension $n^2-1$, also for $n=1$ when it is a single point. <Matrix> multiplication has <polynomial> ambient coordinates and restricts to the <group>, so it is smooth. At the identity,
$$
\boxed{T_I SL_n(\mathbb R)=\{H:\operatorname{Tr}H=0\},\qquad\dim SL_n(\mathbb R)=n^2-1}.
$$
Finally suppose the singular-matrix <set> were an embedded smooth <manifold> near zero for $n\ge2$. Every curve $t\mapsto tE_{ij}$ lies in it, since its rank is at most one. Thus every <matrix> unit belongs to its <tangent space> at zero. These span $\mathbb R^{n^2}$, forcing its local dimension to be $n^2$. An embedded <manifold> of full ambient dimension is locally open, but arbitrarily small scalar multiples of the identity are invertible. This contradiction proves \b[the singular-matrix <set> is not an embedded smooth <manifold>]. The argument addresses the subset-manifold convention specified at the start.
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