Linear Algebra and Its Applications · Problem 7.2.25

Computations with Matrices — Gilbert Strang

Mathematics · partial-pivoting-lu

Problem

Choosing the largest available pivot in each column (partial pivoting), factor each

A

into

PA = LU

A = \begin{bmatrix} 1 & 0 \\ 2 & 2 \end{bmatrix} \qquad \text{and} \qquad A = \begin{bmatrix} 1 & 0 & 1 \\ 2 & 2 & 0 \\ 0 & 2 & 0 \end{bmatrix}.