Linear Algebra and Its Applications · Problem 7.2.17

Computations with Matrices — Gilbert Strang

Mathematics · condition-number-residual

Problem

(Suggested by Moler and Van Loan) Compute

b - Ay

and

b - Az

when

b = \begin{bmatrix} .217 \\ .254 \end{bmatrix} \qquad A = \begin{bmatrix} .780 & .563 \\ .913 & .659 \end{bmatrix} \qquad y = \begin{bmatrix} .341 \\ -.087 \end{bmatrix} \qquad z = \begin{bmatrix} .999 \\ -1.0 \end{bmatrix}.

y

closer than

z

to solving

Ax = b

? Answer in two ways: Compare the residual

b - Ay

b - Az

. Then compare

y

and

z

to the true

x = (1,-1)

. Sometimes we want a small residual, sometimes a small

\delta x