## Subsection5.1.1Of Gaussian elimination and LU factorization

###### Homework5.1.1.1.

Reduce the appended system

\begin{equation*} \begin{array}{r r r | r} 2 \amp -1 \amp 1 \amp 1 \\ -2 \amp 2 \amp 1 \amp -1 \\ 4 \amp -4 \amp \phantom{-}1 \amp 5 \end{array} \end{equation*}

to upper triangular form, overwriting the zeroes that are introduced with the multipliers.

Solution
\begin{equation*} \begin{array}{r r r | r} 2 \amp -1 \amp 1 \amp 1 \\ -1 \amp 1 \amp 2 \amp 0 \\ 2 \amp -2 \amp \phantom{-}3 \amp \phantom{-}3 \end{array} \end{equation*}

###### Homework5.1.1.2.

The execution of the LU factorization algorithm with

\begin{equation*} A = \left( \begin{array}{r r r } 2 \amp -1 \amp 1 \\ -2 \amp 2 \amp 1 \\ 4 \amp -4 \amp \phantom{-}1 \\ \end{array} \right) \end{equation*}

in the video overwrites $A$ with

\begin{equation*} \left( \begin{array}{r r r } 2 \amp -1 \amp 1 \\ -1 \amp 1 \amp 2 \\ 2 \amp -2 \amp \phantom{-}3 \end{array} \right). \end{equation*}

Multiply the $L$ and $U$ stored in that matrix and compare the result with the original matrix, let's call it $\hat A \text{.}$

Solution
\begin{equation*} L = \left( \begin{array}{r r r } 1 \amp 0 \amp 0 \\ -1 \amp 1 \amp 0 \\ 2 \amp -2 \amp \phantom{-}1 \end{array} \right) \mbox{ and } U = \left( \begin{array}{r r r } 2 \amp -1 \amp 1 \\ 0 \amp 1 \amp 2 \\ 0 \amp 0 \amp \phantom{-}3 \end{array} \right). \end{equation*}
\begin{equation*} L U = \left( \begin{array}{r r r } 1 \amp 0 \amp 0 \\ -1 \amp 1 \amp 0 \\ 2 \amp -2 \amp \phantom{-}1 \end{array} \right) \left( \begin{array}{r r r } 2 \amp -1 \amp 1 \\ 0 \amp 1 \amp 2 \\ 0 \amp 0 \amp \phantom{-}3 \end{array} \right) = \left( \begin{array}{r r r } 2 \amp -1 \amp 1 \\ -2 \amp 2 \amp 1 \\ 4 \amp -4 \amp \phantom{-}1 \\ \end{array} \right) = \hat A . \end{equation*}