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Exercise 2.7.11

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Notice that

PA = [\begin{matrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{matrix}] [\begin{matrix} 0 & 0 & 6 \\ 1 & 2 & 3 \\ 0 & 4 & 5 \end{matrix}] = [\begin{matrix} 1 & 2 & 3 \\ 0 & 4 & 5 \\ 0 & 0 & 6 \end{matrix}]

is upper triangular. Multiplying $A$ on the right by a permutation matrix $P_{2}$ exchanges the columns of $A$ . To make this $A$ lower triangular, we also need $P_{1}$ to exchange rows 2 and 3:

P_{1} A P_{2} = [\begin{matrix} 1 \\ 1 \\ 1 \end{matrix}] A [\begin{matrix} 1 \\ 1 & 1 \\ 1 \end{matrix}] = [\begin{matrix} 6 & 0 & 0 \\ 5 & 4 & 0 \\ 3 & 2 & 1 \end{matrix}] .

mathchakra

2022-01-16 12:17

Exercise 2.7.11

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