# Coding Back-Substitution¶

In [1]:
import numpy as np

Here's an upper-triangular matrix $A$ and two vectors $x$ and $b$ so that $Ax=b$.

See if you can find $x$ by computation.

In [2]:
n = 5

A = np.random.randn(n, n) * np.tri(n).T
print(A)

x = np.random.randn(n)
print(x)

b = A @ x
[[ 0.53359808 -1.66129881  0.31267643 -0.07384466  1.20957795]
[-0.         -1.1204435  -1.5348203   1.38270361 -0.34971611]
[ 0.          0.         -0.47187693  0.9763103   0.55054242]
[ 0.         -0.         -0.         -0.16929913  0.21209806]
[-0.          0.          0.         -0.         -0.52165269]]
[-0.9170418   1.40215838  1.41534372 -0.53305575 -1.02625922]
In [3]:
xcomp = np.zeros(n)

for i in range(n-1, -1, -1):
tmp = b[i]
for j in range(n-1, i, -1):
tmp -= xcomp[j]*A[i,j]

xcomp[i] = tmp/A[i,i]

Now compare the computed $x$ against the reference solution.

In [4]:
print(x)
print(xcomp)
[-0.9170418   1.40215838  1.41534372 -0.53305575 -1.02625922]
[-0.9170418   1.40215838  1.41534372 -0.53305575 -1.02625922]

Questions/comments:

• Can this fail?
• What's the operation count?