## Linear Algebra and Its Applications, Exercise 1.7.7

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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