## Linear Algebra and Its Applications, Exercise 1.7.9

Exercise 1.7.9. Given the matrix $A = \begin{bmatrix} .001&0 \\ 1&1000\end{bmatrix}$

compare the pivots in standard elimination with those used in partial pivoting.

Answer: In the standard elimination process we would use a multiplier $l_{21} = 1000$ (1/.001) to multiply the first row and subtract it from the second: $\begin{bmatrix} .001&0 \\ 1&1000\end{bmatrix} \rightarrow \begin{bmatrix} .001&0 \\ 0&1000 \end{bmatrix}$

Here the pivots are .001 and 1000, a difference of six orders of magnitude.

With partial pivoting we would first do a row exchange: $\begin{bmatrix} .001&0 \\ 1&1000 \end{bmatrix} \rightarrow \begin{bmatrix} 1&1000 \\ .001&0 \end{bmatrix}$

and then use the multiplier $l_{21} = .001$ (.001/1) to multiply the first row and subtract it from the second: $\begin{bmatrix} 1&1000 \\ .001&0 \end{bmatrix} \rightarrow \begin{bmatrix} 1&1000 \\ 0&-1 \end{bmatrix}$

In this case the two pivots are 1 and -1 and are of the same order of magnitude.

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.