证明不存在 \(01\) 方阵 \(A\) 使得：
\(A^TA=\begin{pmatrix}7&2&\dots &2\\2&7&\dots&2\\ \vdots&\vdots&\ddots&\vdots\\ 2&2&\dots&7\end{pmatrix}_{22\times22}\)
且
\(A\begin{pmatrix}1\\1\\ \vdots\\ 1\end{pmatrix}_{22\times1}=\begin{pmatrix}7\\7\\ \vdots\\ 7\end{pmatrix}_{22\times1}\)
证明：
若 \(\exists A\) 满足上述条件。
\(\because A^TA=\begin{pmatrix}7&2&\dots &2\\2&7&\dots&2\\ \vdots&\vdots&\ddots&\vdots\\ 2&2&\dots&7\end{pmatrix}_{22\times22}\)
\(\therefore \begin{aligned} |A^TA| & = \begin{vmatrix}7&2&\dots &2\\2&7&\dots&2\\ \vdots&\vdots&\ddots&\vdots\\ 2&2&\dots&7\end{vmatrix}_{22\times22} \\ & = ( 7+(22-1)\times 2 )(7-2)^{22-1} \\ &= 7^2\times 5^{21}\end{aligned}\)
\(\because |A|^2=|A^T||A|=|A^TA|\)
\(\therefore |A|=\pm\sqrt{|A^TA|}=\pm7\times5^{10}\sqrt{5}\)
\(\because A\) 为 \(01\) 矩阵。
\(\therefore |A|\in \mathbb{Z}\)
\(\because \pm7\times5^{10}\sqrt{5}\notin \mathbb{Z}\)
\(\therefore\) 假设不成立，即 \(\nexists A\) 满足上述条件，原命题得证。
原文：https://www.cnblogs.com/zhangshaojia/p/15202217.html