Question 3.7.2: Find an LU-decomposition of B = [2 1 -1 -4 3 3 6 8 -3].

By row reducing, we get

\left [ \begin{matrix} 2 & 1 & -1 \\ -4 & 3 & 3 \\ 6 & 8 & -3 \end{matrix} \right ] \begin{matrix} \\ R_{2} + 2R_{1} \\R_{3} – 3 R_{1} \end{matrix} \sim \left [ \begin{matrix} 2 & 1 & -1 \\ 0 & 5 & 1 \\ 0 & 5 & 0 \end{matrix} \right ] \begin{matrix} \\ \\R_{3} – R_{2} \end{matrix} \sim \left [ \begin{matrix} 2 & 1 & -1 \\ 0 & 5 & 1 \\ 0 & 0 & -1 \end{matrix} \right ] = U

We used the multiplier 2 to get a 0 in the 2, 1-entry of U, so (L)_{21} = −2.
We used the multiplier −3 to get a 0 in the 3, 1-entry of U, so (L)_{31} = 3.
We used the multiplier −1 to get a 0 in the 3, 2-entry of U, we (L)_{32} = 1.
Thus,

L = \left [ \begin{matrix} 1 & 0 & 0 \\ -2 & 1 & 0 \\ 3 & 1 & 1 \end{matrix} \right ]

We can easily verify that B = LU.