设n阶矩阵A = (a[i,j]),A^(-1) = (b[i,j]),其中1 ≤ i,j ≤ n.
由A^(-1)·A = E,有i ≠ j时∑{1 ≤ k ≤ n} b[i,k]·a[k,j] = 0,i = j时∑{1 ≤ k ≤ n} b[i,k]·a[k,j] = 1.
因此1 = ∑{1 ≤ j ≤ n} ∑{1 ≤ k ≤ n} b[i,k]·a[k,j] = ∑{1 ≤ k,j ≤ n} b[i,k]·a[k,j]
= ∑{1 ≤ k ≤ n} ∑{1 ≤ j ≤ n} b[i,k]·a[k,j] = ∑{1 ≤ k ≤ n} b[i,k]·∑{1 ≤ j ≤ n} a[k,j].
而A的各行元素之和均为a ≠ 0,即∑{1 ≤ j ≤ n} a[k,j] = a对任意1 ≤ k ≤ n成立.
代入得1 = ∑{1 ≤ k ≤ n} b[i,k]·a,即1/a = ∑{1 ≤ k ≤ n} b[i,k]对任意1 ≤ i ≤ n成立.
也即A^(-1)的各行元素之和均为1/a = a^(-1).