Chapter 3 Solution of Linear Systems - Math/CS

52 CHAPTER 3. SOLUTION OF LINEAR SYSTEMS
Row-Oriented Column-Oriented
for i = 3 downto 1 do for j = 3 downto 1 do
for j = i + 1 to 3 xj := bj/aj,j
bi := bi − ai,jxj
for i = 1 to j − 1
next j bi := bi − ai,jxj
xi := bi/ai,i
next i
next i next j
Figure 3.6: Pseudocode Row- and Column-Oriented Backward Substitution
Observe that the right hand sides of these equations can again be divided naturally into rows and
columns. Row–oriented backward substitution evaluates the right hand side one row at a time,
while column–oriented backward substitution evaluates the right hand side one column at a time.
Pseudocodes implementing row– and column–oriented backward substitution for systems of size
n = 3 are presented in Fig. 3.6.
Problem 3.1.19. Why is a diagonal linear system also an upper triangular linear system?
Problem 3.1.20. Modify the row– and column–oriented pseudocodes in Fig. 3.6 for backward substitution
so they apply to upper triangular linear systems of order n for any n.
Problem 3.1.21. Consider a general upper triangular linear system of order n. What are the
conditions on the entries of the coefficient matrix so that its solution can be determined by each of
the backward substitution pseudocodes?
Problem 3.1.22. Illustrate the operation of column–oriented backward substitution when used to
solve the upper triangular linear system in Fig. 3.1(c). Hint: Show the value of b each time it has
been modified by the for-loop and the value of each entry of x as it is computed.
Problem 3.1.23. Use the row–oriented version of backward substitution to solve the following linear
system of order 4.
2x1 + 2x2 + 3x3 + 4x4 = 20
0x1 + 5x2 + 6x3 + 7x4 = 34
0x1 + 0x2 + 8x3 + 9x4 = 25
0x1 + 0x2 + 0x3 + 10x4 = 10
Problem 3.1.24. Repeat the previous problem but this time using the column–oriented version of
backward substitution.
Problem 3.1.25. Modify the row– and column–oriented pseudocodes for backward substitution so
that the solution x is written over b.
Problem 3.1.26. Consider a general upper triangular linear system of order n. Show that row–
n(n − 1)
n(n − 1)
oriented backward substitution costs multiplications, subtractions, and n divi-
2
2
sions. Hint: Draw a picture of A and place on the entry ai,j the symbol ⊗ every time that entry is
involved in a multiply, the symbol ⊖ every time that entry is involved in a subtract, and the symbol
⊘ every time that entry is involved in a divide. How many symbols are attached to each entry of A?
Hint: A more general approach observes that there are (n − i) multiplies and subtracts in the inner
loop and that this loop is executed for each value of i = n, (n − 1), · · · , 1. So, the total cost is
multiplies and subtracts.
0 + 1 + · · · + (n − 1) =
n
n−1 n−1
(i − 1) = i = i
i=1
i=0
i=1