The matrix is a triangular matrix. Definition: An n x n upper triangular matrix A is a matrix with the property that <math>A_(ij) = 0 for j > i</math> . The entries forming the triangle can be be zero or nonzero. We can use backward substitution to solve such a matrix-vector equation.
The Echelon matrix is a generalization of triangular matrices.
Triangular linear systems as a matrix-vector dot product <MATH> \begin{array}{l|l} \text{The following triangular linear systems} & \textrm{can be expressed as} \\ \hline \\ \begin{array}{rrrrrrrr} [ & 1, & 0.5, & -2, & 4 & ] & . & v & = & -8 \\ [ & 0, & 3, & 3, & 2 & ] & . & v & = & 3 \\ [ & 0, & 0, & 1, & 5 & ] & . & v & = & -4 \\ [ & 0, & 0, & 0, & 2 & ] & . & v & = & 6 \end{array} & \begin{bmatrix} 1 & 0.5 & -2 & 4 \\ 0 & 3 & 3 & 2 \\ 0 & 0 & 1 & 5 \\ 0 & 0 & 0 & 2 \end{bmatrix} * [-8,3,-4,-6] \end{array} </MATH>
Solving a triangular linear system with Backward substitution in Python:
# rowlist is a list of vector Ex.: [{2,3,-4},{1,2},{5}]
# b are the value. Ex: [10,3,15]
# label list define the domain {A,B,C} of {1,2,4}. Ex: [0,1,2]
def triangular_solve(rowlist, label_list, b):
# Initialize vector x to zero vector
x = zero_vec(set(label_list))
for r in reversed(range(len(rowlist))):
c = label_list[r]
# Procedure populate x entry by entry.
x[c] = (b[r] - x*rowlist[r])/rowlist[r][c]
return x