## Linear Algebra and Its Applications, exercise 1.3.8

Exercise 1.3.8. Given a system of equations of order n = 600, how long would it take to solve in terms of the number of multiplication-subtractions? In seconds, on a PC capable of 8,000 operations per second? On a VAX system capable of 80,000 operations per second? On a Cray X-MP/2 capable of 12 million operations per second?

Answer: As discussed in section 1.3, subsection The Cost of Elimination, for large n the number of operations is approximately $\frac{1}{3}n^3$. For n = 600 the number of operations is therefore approximately

$\frac{1}{3} \cdot 600^3 = \frac{1}{3} \cdot 6^3 \cdot 100^3 = \frac{1}{3} \cdot 2^3 \cdot 3^3 \cdot (10^2)^3 = 2^3 \cdot 3^2 \cdot 10^6 = 8 \cdot 9 \cdot 1,000,000 = 72,000,000$

At 8,000 operations per second this would take 72,000,000 / 8,000 = 9,000 seconds (two and a half hours). At 80,000 operations per second this would take 900 seconds (15 minutes), At 12 million operations per second this would take six seconds.

NOTE: This continues a series of posts containing worked out exercises from the (out of print) book Linear Algebra and Its Applications, Third Edition by Gilbert Strang.

If you find these posts useful I encourage you to also check out the more current Linear Algebra and Its Applications, Fourth Edition, Dr Strang’s introductory textbook Introduction to Linear Algebra, Fourth Edition and the accompanying free online course, and Dr Strang’s other books.

This entry was posted in linear algebra. Bookmark the permalink.