Let $AB = I_n$. $A$ and $B$ are nonsingular, square matrices of size $n$.
Let $A_{r1}$ be the first row of $A$. The products $A_{r1} B_j = 0, j \in \{2,\dots,N\}$. $B_j$ is the $j^{th}$ column of $B$.
Does this means that $A_{r1}$ is not in the span of $B_j, j \in \{2,\dots,N\}$ ? Yes.
Let $C$ be matrix whose rows are $B_j^T$,$ j \in \{2,\dots,N\}$. $A_{r1}$ is in the nullspace of $C$.
Since nullspace and rowspace of $C$ are orthogonal, $\{A_{r1}^T,B_j,j \in \{2,\dots,N\}\}$ form a linearly independent set, hence a basis for $R^n$.
$\{B_j\}$ $,j \in \{1,\dots,N\}$ forms a basis for $R^n$.
Is $A_{r1}^T$ a scalar multiple of $B_1$? Need not be.
Does $\{A_{r1}^T,B_j\}$,$j \in \{1,\dots,N\}$ form a basis or linearly independent set. No.
Because no basis set of $R^n$ can contain more than $n$ members of $R^n$.
What is wrong with this chain of arguments?