## 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.

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