Let $\mathrm{GL}^+(n)$ be the $n\times n$ real matrices with positiv determinant, and let
$i\colon \mathrm{GL}^+(n)\mapsto \mathrm{GL}^+(n+1)$, $i(A)=\begin{pmatrix} 1 & 0 \\ 0 & A \end{pmatrix}$.
Is $i$ a homotopy equivalence for $n>2$?
If not, is $i_*\colon \pi_1(\mathrm{GL}^+(n))\rightarrow \pi_1(\mathrm{GL}^+(n+1))$ an isomorphism for $n>2$?
Can you describe the homogeneous space $\mathrm{GL}^+(n+1)/\mathrm{GL}^+(n)$, with the subgroup being seen inside the larger group using the map $i$?
If so, then you can use the long exact sequence for the fibration $$\mathrm{GL}^+(n)\to \mathrm{GL}^+(n+1)\to \mathrm{GL}^+(n+1)/\mathrm{GL}^+(n).$$
Later. I thought the quotient would be easier to handle! One way to do this is the following, I think.
Let $V=\mathbb R^{n+1}$, let $X$ be the set of all pairs $(u,U)$ with $u$ a nonzero element in $V$, $U$ a subspace of codimension $1$ in $V$ and $\mathbb Ru\oplus U=V$. The natural action of $\mathrm{GL}^+(n+1)$ on $X$ is transitive if $n>0$. and the stabilizer of the element $(e_1,U_0)$ with $U_0$ the subspace of vectors of $V$ with vanishing first coordinate is $\mathrm{GL}^+(n)$. It follows that $X=\mathrm{GL}^+(n+1)/\mathrm{GL}^+(n)$.
Now $X$ is slightly more complicated that what might be wished. It has a subspace $Y$ of those pairs $(u,V)$ such that $u$ is orthogonal to $V$ and has norm $1$. It should not be difficul to show that $Y$ is a deformation retract of $X$ («first move the vector so that it is orthogonal and then normalize it»). Moreover, since an element of $Y$ is completely determined by its first component, we see that $Y$ is homeomorphic to a sphere $S^{n+1}$.
Assuming one can check this, this should get you going!