Linear Algebra and Its Applications, 5th Edition
Authors: David C. Lay, Steven R. Lay, Judi J. McDonald
ISBN-13: 978-0321982384
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See our solution for Question 32E from Chapter 1.2 from Lay's Linear Algebra and Its Applications, 5th Edition.
Problem 32E
Chapter:
Problem:
0
Step-by-Step Solution
Given Information
We are given with when an $n \times (n + 1)$ matrix is row reduced to reduced echelon form. We have to find approximately what fraction of the total number of operations (flops) is involved in the backward phase of the reduction when $n =30$ and when $n = 300$?
Step-1:
For an $n \times (n + 1)$ matrix the reduction to echelon form can take approximately $\dfrac { 2 n ^ { 3 } } { 3 }$ flops.
Step-1: Number of flops for $n=30$:
\[ \begin{aligned} \dfrac { 2 n ^ { 3 } } { 3 } & = \dfrac { 2 ( 30 ) ^ { 3 } } { 3 } \\ & = \dfrac { 2 \times 30 \times 30 \times 30 } { 3 } \\ & = 2 \times 30 \times 30 \times 10 \\ & = 18000 \text { flops } \end{aligned} \] Also, the reduction to reduced echelon form needs at most $n ^ { 2 } = ( 30 ) ^ { 2 } = 900$ flops.
The fraction of the total number of operations (flops) involved in the backward phase of the reduction \[\begin{array}{l} F = \dfrac{{900}}{{18000}} \times 100\% \\ = 0.05 \times 100\% \\ = 5\% \end{array}\]
The Fraction is 5%
Step-2: Number of flops for $n=300$:
\[ \begin{aligned} \dfrac { 2 n ^ { 3 } } { 3 } & = \dfrac { 2 ( 300 ) ^ { 3 } } { 3 } \\ & = \dfrac { 2 \times 300 \times 300 \times 300 } { 3 } \\ & = 2 \times 300 \times 300 \times 100 \\ & = 18000000 \text { flops } \end{aligned} \] Also, the reduction to reduced echelon form needs at most $n ^ { 2 } = ( 300 ) ^ { 2 } = 90000$ flops.
The fraction of the total number of operations (flops) involved in the backward phase of the reduction \[\begin{array}{l} F = \dfrac{{90000}}{{18000000}} \times 100\% \\ = 0.005 \times 100\% \\ = 0.5\% \end{array}\]
The Fraction is 0.5%
We are given with when an $n \times (n + 1)$ matrix is row reduced to reduced echelon form. We have to find approximately what fraction of the total number of operations (flops) is involved in the backward phase of the reduction when $n =30$ and when $n = 300$?
Step-1:
For an $n \times (n + 1)$ matrix the reduction to echelon form can take approximately $\dfrac { 2 n ^ { 3 } } { 3 }$ flops.
Step-1: Number of flops for $n=30$:
\[ \begin{aligned} \dfrac { 2 n ^ { 3 } } { 3 } & = \dfrac { 2 ( 30 ) ^ { 3 } } { 3 } \\ & = \dfrac { 2 \times 30 \times 30 \times 30 } { 3 } \\ & = 2 \times 30 \times 30 \times 10 \\ & = 18000 \text { flops } \end{aligned} \] Also, the reduction to reduced echelon form needs at most $n ^ { 2 } = ( 30 ) ^ { 2 } = 900$ flops.
The fraction of the total number of operations (flops) involved in the backward phase of the reduction \[\begin{array}{l} F = \dfrac{{900}}{{18000}} \times 100\% \\ = 0.05 \times 100\% \\ = 5\% \end{array}\]
The Fraction is 5%
Step-2: Number of flops for $n=300$:
\[ \begin{aligned} \dfrac { 2 n ^ { 3 } } { 3 } & = \dfrac { 2 ( 300 ) ^ { 3 } } { 3 } \\ & = \dfrac { 2 \times 300 \times 300 \times 300 } { 3 } \\ & = 2 \times 300 \times 300 \times 100 \\ & = 18000000 \text { flops } \end{aligned} \] Also, the reduction to reduced echelon form needs at most $n ^ { 2 } = ( 300 ) ^ { 2 } = 90000$ flops.
The fraction of the total number of operations (flops) involved in the backward phase of the reduction \[\begin{array}{l} F = \dfrac{{90000}}{{18000000}} \times 100\% \\ = 0.005 \times 100\% \\ = 0.5\% \end{array}\]
The Fraction is 0.5%