All the matrices in this statement are in the field $\mathbb{F}\_2$. Let $I$ be the identity matrix of size $10 \times 10$. What are all the possible $n$ ($\geq 6$) for which
there exists a matrix $X$ of size $n \times 10$, such that any $6$ rows of $X$ jointly with any $4$ rows of $I$ form a $10 \times 10$ invertible matrix?



Existence of a matrix in $\mathbb{F}_2$ with some invertibility properties
There is no such $n$. Assume that it exists, then $n=6$ also satisfies this property. Denote the rows of $X$ by $r\_1,\ldots,r\_6$, the columns of $X$ by $c\_1,\ldots,c\_{10}$, the rows of $I$ by $e\_1,\ldots,e\_{10}$. Assume that, say, $c\_1,\ldots,c\_{6}$ are linearly dependent. Then the minor formed by the first 6 rows and columns of $X$ equals to 0, this yields that $r\_1,\ldots,r\_6,e\_7,e\_8,e\_9,e\_{10}$ are linearly dependent. So any 6 vectors in the set $\{c\_1,\ldots,c\_{10}\}$ are linearly independent. But $c\_1,\ldots,с\_7$ are linearly dependent. Thus their linear dependence is $c\_1+\ldots+c\_7=0$. Analogously $c\_8+c\_1+c\_2+\ldots+c\_6=0$ and we get $c\_7=c\_8$, a contradiction.


