The matrix \( \mathbf{A} \) which rephrases a second derivative in a discretized form is much simpler than the general matrix $$ \mathbf{A} = \begin{bmatrix} 2 & -1 & 0 & 0 &0 & 0\\ -1 & 2 & -1 &0 &0 &0 \\ 0 & -1 & 2 & -1 & 0& 0 \\ 0 & \dots & \dots & \dots &\dots & \dots \\ 0 &0 &0 &-1 &2& -1 \\ 0 & 0 &0 &0 &-1 & 2 \\ \end{bmatrix}. $$ This matrix fulfills the condition of a weak dominance of the diagonal, with \( |b_1| > |c_1| \), \( |b_n| > |a_n| \) and \( |b_k| \ge |a_k|+|c_k| \) for \( k=2,3,\dots,n-1 \). This is a relevant but not sufficient condition to guarantee that the matrix \( \mathbf{A} \) yields a solution to a linear equation problem. The matrix needs also to be irreducible. A tridiagonal irreducible matrix means that all the elements \( a_i \) and \( c_i \) are non-zero. If these two conditions are present, then \( \mathbf{A} \) is nonsingular and has a unique LU decomposition.