Does this matrix of ones and zeros have a nonzero maximal minor? Choose $n\ge 5$. We define a ${n\choose 3}\times {n\choose 2}$ matrix $A$ in the following way:
Each row is identified with a triplet $\{i,j,k\}$ and each column identified with a pair $\{s,t\}$.
Define $A_{\{i,j,k\},\{s,t\}}=1$ if $\{s,t\}\subseteq \{i,j,k\}$ and zero otherwise. 
Does $A$ have a nonzero maximal minor? 
Edit: the determinant of the square case $n=5$ is $-48$.
 A: Before we start, recall that for any real matrix $X$ the products $X^{T}X$ and $XX^{T}$ are symmetric and so diagonalisable; and that their non-zero eigenvalues coincide, even having the same multiplicities.
Recall also that if $A$ and $B$ are sets of size $r$ then $|A\cup B|=r+1$ if and only if $|A\cap B|=r-1$. 
Assume throughout that $n>2r$.
Let $V_r$ denote the real vector space spanned by the $r$-subsets of $\{1,2,\dots,n\}$; and let $D_r:V_r\to V_{r+1}$ be the linear map that carries each $r$-subset to the sum of the $(r+1)$-subsets which it lies in. Our intention is to show that $D_r$ has rank $n \choose r$.
Consider the map $D_r^{T} D_r$; the $(R_1,R_2)$ entry is precisely the number of $(r+1)$-subsets containing both $R_1$ and $R_2$. Similarly  the $(R_1,R_2)$ entry of $D_{r-1} D_{r-1}^{T}$ is precisely the number of $(r-1)$-subsets contained in both $R_1$ and $R_2$.
Note that if $R_1$ and $R_2$ are distinct $r$-subsets then one of two things happens. Either (i) there exists no $(r+1)$ subset containing both $R_1$ and $R_2$, because $R_1\cup R_2$ is too large, and in this case there is no common $(r-1)$ subset, because $R_1\cap R_2$ is too small; or else (ii) $|R_1\cup R_2|=r+1$ and $|R_1\cap R_2|=r-1$, and so $R_1$ and $R_2$ lie in exactly one $(r+1)$-subset and contain exactly one $(r-1)$-subset. 
When $R_1=R_2$ then $R_1$ lies in $(n-r)$ subsets of size $(r+1)$, and contains $r$ subsets of size $(r-1)$. 
Putting these together we see that $D_r^{T} D_r=D_{r-1} D_{r-1}^{T}+(n-2r)I$.
We can now build up the eigenvalues of the $D_r^{T} D_r$ recursively.
(i) The eigenvalue of $D_0^{T} D_0$ is $n$ with multiplicity $n\choose 0$.
(ii) Hence the eigenvalues of $D_0 D_0^{T}$ are $n$ with multiplicity $n\choose 0$ and $0$ with multiplicity 
${n\choose 1} -  {n\choose 0}$.
(iii) Hence the eigenvalues of $D_1^{T} D_1$ are $2n-2$ with multiplicity $n\choose 0$ and $n-2$ with multiplicity 
${n\choose 1} -  {n\choose 0}$. 
(iv) Hence the eigenvalues of $D_1 D_1^{T}$ are $2n-2$ with multiplicity $n\choose 0$,  $n-2$ with multiplicity 
${n\choose 1} -  {n\choose 0}$ and $0$ with multiplicity 
${n\choose 2} -  {n\choose 1}$.  
(v) Hence the eigenvalues of $D_2^{T} D_2$ are $3n-6$ with multiplicity $n\choose 0$,  $2n-6$ with multiplicity 
${n\choose 1} -  {n\choose 0}$ and $n-4$ with multiplicity 
${n\choose 2} -  {n\choose 1}$.  
It is now clear that $D_2^{T} D_2$ is non-singular and so $D_2$ is of full rank.
I dedicate this answer to the memory of Dr Jack Hammersley : he set an undergraduate examination question on mean recurrence times of random walks on some highly symmetric graphs which inspired my interest in problems like this. 
