Prove $A_1, A_2, \ldots,A_{\ell-1}, A_{\ell+1}\ldots \ldots, A_{m} $ in $\mathbb R^m$ are linearly independent viewed as vectors in $\mathbb R^{m-1}$? Suppose I have linearly independent vectors $A_1, A_2, \ldots, A_m$ in $\mathbb R^m$

Consider the matrix $B = [A_1, A_2, \ldots  A_m]$ consisting of these vectors and suppose $B^{-1}[A_1, A_2, \ldots,A_{\ell-1}, A_{\ell+1}\ldots \ldots, A_{m}]$ has the $\ell$-row consisting only of $0$'s.

Is it then neccesarily true that $A_1, A_2, \ldots,A_{\ell-1}, A_{\ell+1}\ldots \ldots, A_{m}$ are linearly independent in $\mathbb R^{m-1}$ considered as vectors with the $\ell$-component removed ?

I've tried to prove that $A_1, A_2, \ldots,A_{\ell-1}, A_{\ell+1}\ldots \ldots, A_{m}$, with the $\ell$-component removed, span $\mathbb R^{m-1}$ or are linearly independent, however I've not yet come to a final result.
This is not true. Let $m=3$ and $A_1=(0,2,3),~ A_2=(1,1,1),~ A_3=(0,1,1)$. Then

\begin{align}
\det(A_1,A_2,A_3)=\begin{vmatrix}
0 &amp; 1 &amp; 0\\
2 &amp; 1 &amp; 1\\
3 &amp; 1 &amp; 1
\end{vmatrix}
=-1\neq 0.
\end{align}
If we delete the first components of $A_2$ and $A_3$ the system $\{A_2,A_3\}$ wont be linearly independent in $\mathbb R^2$

Let $A_1=(1,2,3),~A_2=(0,1,1),~A_3=(-1,1,1)$. Then 
$$B=\begin{bmatrix}
1&amp;0&amp;-1\\ 2&amp;1&amp;1\\ 3&amp;1&amp;1
\end{bmatrix}\quad \text{and}\quad 
B^{-1}=\begin{bmatrix}
0&amp;-1&amp;1\\ 1&amp;4&amp;-3\\ -1&amp;-1&amp;1
\end{bmatrix}.$$
The first row of the matrix $B^{-1}[A_2,A_3]$ is only 0 and the system $\{A_2,A_3\}$ after deleted the first components is not linearly independent.
It is necessary that any subset is linearly independent so that a set is linearly independent (by definition).