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Linear Algebra and Its Applications, Exercise 1.5.17

January 18, 2011 12:00 am

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.

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