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  {\large \textbf{Mathematical Tripos Part IB: Easter 2001}}

  {\Large \textbf{Numerical Analysis -- Detailed Example
      1\footnote{Corrections and suggestions to these notes should
        be emailed to \texttt{a.shadrin@damtp.cam.ac.uk}. All handouts
        are available on the WWW at the URL 
        \texttt{http://www.damtp.cam.ac.uk/user/na/PartIB/Handouts.html}.}}}

  \vspace{8pt}
  \textbf{\textit{\LARGE LU factorization}}
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Let
\begin{displaymath}
  A=\left[
    \begin{array}{rrrr}
      -3 & 2 & 3 & -1\\
      6 & -2 & -6 & 0\\
      -9 & 4 & 10 & 3\\
      12 & -4 & -13 & -5
    \end{array}\right].
\end{displaymath}
Hence
\begin{Eqnarray*}
  &&\MM{\ell}_1=\left[
    \begin{array}{r}
      1\\-2\\3\\-4
    \end{array}\right],\qquad \MM{u}_1^{\CC{T}}=[
  \begin{array}{rrrr}
    -3 ,& 2, & 3, & -1
  \end{array}],\qquad A-\MM{\ell}_1\MM{u}_1^{\CC{T}}=\left[
    \begin{array}{r|rrr}
      0 & 0 & 0 & 0\\\hline
      0 & 2 & 0 & -2\\
      0 & -2 & 1 & 6\\
      0 & 4 & -1 & -9
    \end{array}\right],\\
  &&\MM{\ell}_2=\left[
    \begin{array}{r}
      0\\1\\-1\\2
    \end{array}\right],\qquad \MM{u}_2^{\CC{T}}=[
  \begin{array}{rrrr}
    0, & 2, & 0, & -2
  \end{array}],\qquad A-\MM{\ell}_1\MM{u}_1^{\CC{T}}-
  \MM{\ell}_2\MM{u}_2^{\CC{T}} =\left[
    \begin{array}{rr|rr}
      0 & 0 & 0 & 0\\
      0 & 0 & 0 & 0\\\hline
      0 & 0 & 1 & 4\\
      0 & 0 & -1 & -5
    \end{array}\right],\\
  &&\MM{\ell}_3=\left[
    \begin{array}{r}
      0 \\ 0 \\ 1 \\ -1
    \end{array}\right],\qquad \MM{u}_3^{\CC{T}}=[
  \begin{array}{rrrr}
    0, & 0, & 1, 4
  \end{array}], ,\qquad A-\MM{\ell}_1\MM{u}_1^{\CC{T}}-
  \MM{\ell}_2\MM{u}_2^{\CC{T}}-\MM{\ell}_3\MM{u}_3^{\CC{T}} =\left[ 
    \begin{array}{rrr|r}
      0 & 0 & 0 & 0\\
      0 & 0 & 0 & 0\\
      0 & 0 & 0 & 0\\\hline
      0 & 0 & 0 & -1
    \end{array}\right],\\
  &&\MM{\ell}_4=\left[
    \begin{array}{r}
      0 \\ 0 \\ 0 \\ 1
    \end{array}\right], \qquad \MM{u}_4^{\CC{T}}=[
  \begin{array}{rrrr}
    0, & 0, & 0, & -1
  \end{array}],
\end{Eqnarray*}
and we deduce that
\begin{displaymath}
   L=\left[
    \begin{array}{rrrr}
      1 & 0 & 0 & 0\\
      -2 & 1 & 0 & 0\\
      3 & -1 & 1 & 0\\
      -4 & 2 & -1 & 1
    \end{array}\right],\qquad U=\left[
    \begin{array}{rrrr}
      -3 & 2 & 3 & -1\\
      0 & 2 & 0 & -2\\
      0 & 0 & 1 & 4\\
      0 & 0 & 0 & -1
    \end{array}\right].
\end{displaymath}



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