If $c:[0,1]\to \mathbb{R}^{n^2}$ is continuous function and each $(c^1(t),\ldots,c^n(t))$ is a basis of of $\mathbb{R}^n$, show that $$[c^1(0),\ldots,c^n(0)]=[c^1(1),\ldots,c^n(1)]$$
Proof: Let $c^i(0)=\sum_{j=1}^na_{ij}(t)c^j(t)$ for $t\in [0,1]$. Now $$ \det[c^1(0),\ldots,c^n(0)]=\det(a_{ij}(t))\det[c^1(t),\ldots,c^n(t)] .$$ Since, $c$ is continuous so $\det(a_{ij}(t))$ is continuos in $t$ and it is non-zero $\implies \det$ will not change the sign. After that I stuck. Please help.