Can william do the matrix successfully?

The answer is YES.

Suppose $A$ is a $(0,1)-matrix$ of size $m×n$ $(m,n\ge 3)$ with property $\Phi$ . I'll show that $A$ can be reduced to $\left( \begin{array}{cc} 0 & 1 \\ 1 & 0 \\ \end{array} \right)$ $①$ or $\left( \begin{array}{cc} 1 & 0 \\ 0 & 1 \\ \end{array} \right)$ $②$ by induction.

(1) Any $2×n(n\ge 3)$ matrix $B$ with property $\Phi$ can be reduces to $①$ or $②$ .

WLOG the first column of $B$ is $\left( \begin{array}{c} 1 \\ 0 \\ \end{array} \right)$ ,by $\Phi$ there exists another column of $B$ which is $\left( \begin{array}{c} 0 \\ 1 \\ \end{array} \right)$ (Otherwise the first line has no 0,which is a contradiction). Save the two columns and delete other elements in $B$ to get matrix $②$ .

(2) Do induction on $m$ . Case $m=2$ has been done. Assume the conclusion when m=k is true.

For m=k+1,we first delete the $(k+1)_{th}$ line of A to get a $k×n$ matrix $A'$ .

( i ) If $A'$ has the property $\Phi$ ,then delete the $(k+1)_{th}$ line of A and apply the inductive assumption to get $①$ or $②$ .

( ii ) If A' doesn't have the property $\Phi$ ,then there exists $j$ such that the $j_{th}$ column of A' consists only of 0 or consists

only of 1. WLOG the former case. Use $a(x,y)$ to represent the element at the $x^{th}$ line and $y^{th}$ column of A.

Then $a(k+1,j)=1$ . Since $A$ has property $\Phi$ , there exists $h (h≠j)$ such that $a(k+1,h)=0$ , and there exists $l (l≠ k+1)$

such that $a(l,h)=1$ . We're happy to see that $a(l,j)=0$ because it's an element of the $j_{th}$ column of A'.

Now we save $a(k+1,j)$ , $a(k+1,h)$ , $a(l,h)$ and $a(l,j)$ , and delete the other elements of $A$ by deleting all the lines except the $(k+1)^{th}$ line

and the $l^{th}$ line and deleting all the columns except the $j^{th}$ line and the $h^{th}$ line .So we get $①$ or $②$ as desired.

$\blacksquare$