## Linear Algebra and Its Applications, Exercise 1.5.17

Exercise 1.5.17. If LPU order is used then rows are exchanged only at the end of elimination: $A = \begin{bmatrix} 1&1&1 \\ 1&1&3 \\ 2&5&8 \end{bmatrix} \rightarrow \begin{bmatrix} 1&1&1 \\ 0&0&2 \\ 0&3&6 \end{bmatrix} = PU = \begin{bmatrix} 1&0&0 \\ 0&0&1 \\ 0&1&0 \end{bmatrix} \begin{bmatrix} 1&1&1 \\ 0&3&6 \\ 0&0&2 \end{bmatrix}$

Specify what L is in the above case.

Answer: Since no row exchanges are done during elimination proper, the multipliers in L stay in their original places. In this case the multipliers are 1 in the (2,1) position and 2 in the (3,1) position, so we have $L = \begin{bmatrix} 1&0&0 \\ 1&1&0 \\ 2&0&1 \end{bmatrix}$

We can check this by computing LPU: $LPU = \begin{bmatrix} 1&0&0 \\ 1&1&0 \\ 2&0&1 \end{bmatrix} \begin{bmatrix} 1&0&0 \\ 0&0&1 \\ 0&1&0 \end{bmatrix} \begin{bmatrix} 1&1&1 \\ 0&3&6 \\ 0&0&2 \end{bmatrix} = \begin{bmatrix} 1&0&0 \\ 1&1&0 \\ 2&0&1 \end{bmatrix} \begin{bmatrix} 1&1&1 \\ 0&0&2 \\ 0&3&6 \end{bmatrix} = \begin{bmatrix} 1&1&1 \\ 1&1&3 \\ 2&5&8 \end{bmatrix}$

We thus have LPU = A.

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 .

This entry was posted in linear algebra. Bookmark the permalink.