“…However, while performing the Gaussian elimination, it is necessary to perform column pivoting since otherwise the Gaussian elimination may not be stable [24]. Assume that after column pivoting the linear system reads B ã = 0, (21) and that we have a bijective permutation map σ from the set {1, 2, 3, 4} onto itself such that ãi = a σ(i) where i ∈ {1, 2, 3, 4}; and Bi,j = B i,σ(j) where j ∈ {1, 2, 3, 4}. (22) Note that the map σ is invertible, i.e., we can write ãσ −1 (i) = a i and Bi,σ −1 (j) = B i,j .…”