Linear Algebra and Its Applications, Exercise 1.7.7

Posted on July 9, 2011 by hecker

Exercise 1.7.7. Take the 3 by 3 Hilbert matrix from the previous exercise

A = \begin{bmatrix} 1&\frac{1}{2}&\frac{1}{3} \\ \frac{1}{2}&\frac{1}{3}&\frac{1}{4} \\ \frac{1}{3}&\frac{1}{4}&\frac{1}{5} \end{bmatrix}

and compute b assuming that x = (1, 1, 1) and x = (0, 6, -3.6) are solutions to Ax = b.

Answer: We first multiply A times x = (1, 1, 1):

b = Ax = \begin{bmatrix} 1&\frac{1}{2}&\frac{1}{3} \\ \frac{1}{2}&\frac{1}{3}&\frac{1}{4} \\ \frac{1}{3}&\frac{1}{4}&\frac{1}{5} \end{bmatrix} \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix} = \begin{bmatrix} 1 + \frac{1}{2} + \frac{1}{3} \\ \frac{1}{2} + \frac{1}{3} + \frac{1}{4} \\ \frac{1}{3} + \frac{1}{4} + \frac{1}{5} \end{bmatrix} = \begin{bmatrix} \frac{11}{6} \\ \frac{13}{12} \\ \frac{47}{60} \end{bmatrix} \approx \begin{bmatrix} 1.83 \\ 1.08 \\ .783 \end{bmatrix}

expressing the final result to three significant digits. We then multiply A by x = (0, 6, -3.6) again expressing the final result to three significant digits:

b = Ax = \begin{bmatrix} 1&\frac{1}{2}&\frac{1}{3} \\ \frac{1}{2}&\frac{1}{3}&\frac{1}{4} \\ \frac{1}{3}&\frac{1}{4}&\frac{1}{5} \end{bmatrix} \begin{bmatrix} 0 \\ 6 \\ -3.6 \end{bmatrix} = \begin{bmatrix} 0 + 3 - 1.2 \\ 0 + 2 - 0.9 \\ 0 + 1.5 - .72 \end{bmatrix} = \begin{bmatrix} 1.80 \\ 1.10 \\ .780 \end{bmatrix}

Note that the values of b are almost identical, though the values of the solutions x are very different. This indicates that the solution x to Ax = b is very sensitive to small differences in b.

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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